Conc Notes S05

ARCH 3126 – Steel, Timber & Concrete – Spring 2005

Monday, March 28th

Homework #23 is due Today (Chapter 3 – Steel). Homework #24 is due Wednesday, March 30th (Chapter 12 - Steel).

The text is written for the ACI 318-95 code. The current edition, ACI 318-02, contains some major revisions and we will cover most of the difference in the two codes that pertain to homework solutions. The newest version of the code 318-05 has not been published yet.

Chapter 1 -- Introduction, Materials, and Properties

1.1 -- Reinforced Concrete Structures

The three most common building materials used today from which structures are built are wood, steel, and concrete (including prestressed). The three most common structural systems are better termed timber, structural steel, and reinforced concrete.

The primary reason that reinforced concrete is a logical union of plain concrete and steel in regards to there strength is due to the high compressive (but low tensile) strength of concrete and the high tensile strength of steel reinforcing (Draw a beam showing the location of the reinforcement in the tension zone). Also they have similar rates of thermal expansion.

1.2 -- Historical Background

Very interesting, but this is not a history class.

1.3 --Concrete

Plain concrete is made by mixing cement, fine aggregate, coarse aggregate, water, and frequently admixtures.

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1.4 -- Cement

The cement has adhesive and cohesive properties, which bond mineral fragments. Cements used in reinforced concrete construction are called hydraulic cements, which set and harden in the presence of water (Type I Portland cement). Strength is normally attained in 28 day (varies with strength). Table 1.4.1 pp. 6 shows the types of Portland cement as follows:

Type PropertiesI Ordinary constructionII Moderate sulfate resistanceIII High early strengthIV Low heat of hydrationV High sulfate resistanceK Expansive

In addition an A indicates that the concrete be air-entrained which provides durability.

1.5 -- Aggregates

Aggregate occupies about 75% of the total volume of concrete (least expensive part). Fine aggregate (sand) is material less than 3/16 in. and coarse aggregate (gravel) large than that. The nominal maximum size of coarse aggregate from ACI - 3.3.2 is governed by the clearance between the sides of a form and the adjacent bars as follows:

1. 1/5 the narrowest dimension between sides of forms2. 1/3 the depth of slabs3. 3/4 the minimum clear spacing between reinforcing bars

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Structural lightweight concrete is usually made from kiln dried aggregates of expanded shale or clay (some are natural). Typical weights are from 70 to 115 pcf (145 for normal weight). All-lightweight concrete contains both lightweight fine and coarse aggregate. Sand-lightweight concrete contains only lightweight coarse aggregate. Sand replacement is a term used to define concrete with all or part of the lightweight fine aggregate replaced with natural sand. See Figure 1.5.1 pp. 7 for some approximate unit weights of lightweight aggregate concrete.

1.6 -- Admixtures

1. The most widely used admixture in concrete is air-entraining which provides an increase in durability.

2. Accelerating admixtures will decrease the time required for curing (best for cold weather placement).

3. Water-reducing and Set-controlling admixtures may be used for higher strength (less water) and durability (also for hot weather placement).

4. Admixtures for flowing concrete are used to produce slump rates of 7½ inches or greater and increase workability (commonly termed plasticizers).

Other admixtures can produce gas, expansion, color, fungus-germ-insect protection or provide dampproofing, reduced permeability and aggregate expansion or inhibit corrosion.

1.7 -- Compressive Strength

The strength of concrete is controlled by proportioning of cement, coarse and fine aggregate, water and admixture. The most important variable in determining strength is the water to cement (w/c) ratio.

The slump test is the measure of the workability of concrete where a truncated cone-shaped 12 inch metal mold is filled with fresh concrete and the lifted off. The distance the top of the wet mass is the slump (3 to 4 inched is normally desired).

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The strength of concrete is denoted by f'c which is the standard 28 day compressive strength in psi of test cylinder 6 in. in diameter and 12 in. high. (200-mm cube test in other parts of the world). The maximum ultimate strain of concrete is 0.003 per ACI - 10.2.3. Strengths of concrete range from 3000 psi up the 18000 psi (3 to 10 normal for slabs to high strength columns).

1.8 -- Tensile Strength

The strength of concrete in tension greatly effects the extent and size of cracking in structures and is measured by the split-cylinder test. Tensile strength in flexure is known as the modulus of rupture and is governed by ACI - 11.2 as follows:

f f normalct c6 7. ' ( )

1.9 -- Modulus of Elasticity

The modulus of elasticity of concrete varies primarily with strength, but also with weight, age and size. The modulus of elasticity per ACI - 8.5.1 is as follows (also in Table 1.9.1 pp. 16 for normal weight):

1.10 -- Creep and Shrinkage

Creep is the property of concrete by which it continues to deform with time under sustained loads at unit stresses within the elastic range. Shrinkage is the property of concrete by which it continues to change in volume with time that is unrelated to load application (both rates will decrease with time).

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1.11 -- Concrete Quality Control

The specified compressive strength is deemed adequate when both of the following are occur (ACI 5.6.3.3):

1. Average of all sets of three consecutive strength test equal or exceed f'c2. No individual strength test falls below f'c by more than 500 psi when f'c

<5000 psi or by more than 0.1 f'c when f'c >5000 psi.

If these are not met in-place testing may indicate the concrete is adequate. If not re-analysis may indicate acceptance. If not load testing may be used for acceptance. If not then it must be strengthened or removed.

1.12 -- Steel Reinforcement

Steel reinforcement may consist of bars, welded wire fabric, or wires (usually deformed bars). Sizes are given in Table 1.12.1 pp. 21 (also in back of your ACI code or ACI 340 SP-7(97) REINFORCEMENT 1, (make a big copy of this) and types and strengths of steel in Table 1.12.4 pp. 23. Billet steel (ASTM A615/A615M) is newly made and is sufficiently ductile. Grade 60 is the primary material used (75 is become more popular), but grade 40 is used for smaller bars to be bent. The modulus of elasticity of steel is 29,000,000 psi (ACI 8.5.2) and prestress steel is lower (27,000,000 psi) and more variable, thus it should be obtained from the manufacturer or by test (ACI 8.5.3).

1.13 -- SI Units

You are the torch bearers for this, Federal government now and others later. We will do very little in terms of SI units.

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Tuesday, March 29th

Chapter 2 -- Design Methods and Requirements

2.1 -- ACI Building Code

The ACI Building Code Requirements for Reinforced Concrete is based partly on empirical and mostly rational data.

2.2 -- Strength Design and Working Stress Design Methods

The working stress method focuses on conditions at service load and the strength design method focuses on loads when failure may be imminent.

2.3 -- Working Stress Method

The working stress method (now referred to as alternate design method, but not in ACI 318-02) has set limits on the stresses allowed under service loads (working loads).

Alternate Design Method ACI App A of old code, not in ACI 318-02.

Some of the obstacles to the working stress method are as follows:

1. No account for different types of loads.2. Creep and shrinkage are not easily accounted for in elastic

stresses.3. Stress is not proportional to strain at concrete crushing therefore

the inherent factor of safety is unknown.

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2.4 -- Strength Design Method

The strength design method (formerly called ultimate strength method) has the service loads increased by sufficient factors to obtain the load at which failure is considered to be imminent. The strength provided must be greater than the required strength to carry these factored loads.

2.5 -- Comments on Design Methods

Although strength design is currently the philosophy employed most widely, serviceability must be maintained. Working stress is stilled required to calculate deflections and cracking of structure in service load conditions.

2.6 -- Safety Provisions--General

The two primary factors used to provide safety in the ultimate strength method of design are U the overload factors (load factors) in ACI 9.2 (new factors) or ACI C.2 (old factors) and the understrength factors (strength reduction factors) in ACI 9.3 (new factors) or ACI C.3 (old factors).

Design Strength<Required Strength

2.7 -- Safety Provisions—Traditional ACI Code – skip details

The structure must be designed for the most severe of any load combination. The load factors for some basic combinations are as follows per ACI C.2 (the text pp. 38 is old code):

(C-1)U = 1.4D + 1.7L Dead – D & Live - L(C-2)U = 0.75(1.4D + 1.7L) + (1.6W or 1.0E) Wind – W Seismic - E

1.3W w/o direction factor 1.4E if service loads

(C-3)U = 0.9D + (1.3W or 1.0E) Overturning(C-4)U = 1.4D + 1.7L +1.7H Earth pressure - H

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(C-5)U = 0.75(1.4D + 1.7T + 1.7L) Temperature - T (C-6)U = 1.4D + 1.4T

No account for D+L+(2 others), but old codes use 2/3 of basic factors.

In addition, for intermediate moment frames shear must be check with seismic load doubled per ACI 21.12(b). Applies to members resisting seismic loading (frame).

The strength reduction factors accounting for adverse variations in material strength, workmanship, dimensions, control & degree of supervision, and importance of the member to the building structure are as follows per ACI C.3 (the text pp. 39 is old code):

= 0.90 Tension controlled sections (t > 0.005) = 0.75 Compression members, spirally reinforced

Compression controlled sections (t < 0.002 i.e. balance) = 0.70 Compression members, others

Compression controlled sections (t < 0.002 i.e. balance) = 0.85 Shear and torsion = 0.70 Bearing on concrete = 0.65 Bending in plain concrete

2.8 -- Safety Provisions—ACI Appendix C Load and Strength Reduction Factors

The structure must be designed for the most severe of any load combination. These are the factors in the ACI 318-02 code. The load factors for some basic combinations are as follows per ACI 9.2 (the text pp. 40 is old appendix (typo 1.5 vs. 15)):

(9-1) U = 1.4(D+F)Dead – D & Fluid - F

(9-2) U = 1.2(D+F+T) + 1.6(L+H) + 0.5(Lr or S or R)Live - LTemperature – T

8


Earth pressure – HRoof Live - Lr

Snow – SRain - R

(9-3) U = 1.2D + 1.6(Lr or S or R) + (1.0L or 0.8W)Wind – WCan use 0.5L (not garages or L>100psf)

(9-4) U = 1.2D + 1.6W + 1.0L + 0.5(Lr or S or R)1.3W w/o directionality

(9-5) U = 1.2D + 1.0E + (1.0L or 0.2S)Seismic – E1.4E if serviceCan use 0.5L (not garages or L>100psf)

(9-6) U = 0.9D + 1.6W + 1.6H1.3W w/o directionality

(9-7) U = 0.9D + 1.0E + 1.6H1.4E if service

The strength reduction factors accounting for adverse variations in material strength, workmanship, dimensions, control & degree of supervision, and importance of the member to the building structure are as follows per the text pp. 41 pr C.3 (ASCE 7-88):

= 0.90 Tension controlled sections = 0.70 Compression members, spirally reinforced

Compression controlled sections = 0.65 Compression members, others

Compression controlled sections = 0.75 Shear and torsion = 0.65 Bearing on concrete = 0.55 Bending in plain concrete

2.9 -- Handbooks and Computer Software

Some of the useful books include the following:

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1. Strength Design Handbook - ACI 340R-97 SP-17(97)2. CRSI Handbook3. PCI Design Handbook4. ACI Detailing Manual5. PCA Design Handbook

2.10 -- Dimensions and Tolerances

These are covered in ACI 7.5 called placing reinforcing and the strength reduction factors are intended to account for this.

2.11 -- Accuracy of Computations

Every book has this section (Carry 4 digits in calculations and round answer to 3 digits).

Chapter 3 -- Strength of Rectangular Sections in Bending

3.1 -- General Introduction

Working stress methods will be covered in Chapter 4 and for now we will concentrate on the strength design method.

3.2 -- Basis of Nominal Flexural Strength

The modern analytical approach to reinforced concrete beam design should include a linear strain distribution and a nonlinear stress distribution as shown in Figure 3.2.1 pp. 48 (shown below). The following equations can be derived statically assuming a ductile failure of steel at yield stress:

k1 - fraction of stress area to rectangular blockk2 - fraction of x to centroidk3 - fraction of f’c at maximum stress

10


C k k f xbc 1 3 '

T A fs y

F T Cx 0

xA f

k k f bs y

c

1 3 '

M M T d k xc n ( )2

M A f dk

k k

A f

f bn s y

s y

c

2

1 3 '

It should be noted that only the ratio of k2/(k1k3) need be know and not the individual constants. Tests indicate that the variation is between 0.55 and 0.63 as shown in Figure 3.2.3 pp. 50.

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3.3 -- Whitney Rectangular Stress Distribution

The computation of flexural strength based on the approximately parabolic stress may be done, but the designer needs a simpler method based on static equilibrium. The Whitney rectangular stress block as shown in Figure 3.3.1 pp. 51 (shown below) is the basis for standard design. The average stress of 0.85fc' is used over a rectangular depth a=1x (ACI 10.2.7.1). The value for 1 should be taken as 0.85 for fc' < 4000 psi, and 0.05 less for each 1000 psi of fc' in excess of 4000 psi (10.2.7.3). The value of 1 need not be less than 0.65 (See ACI SP-17(97) FLEXURE 1 for this and more). This can be written as follows:

1 0 85 0 054000

10000 65

. .

'.

f c

The static derivation is as follows:

C f bac0 85. '

T A fs y

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F T Cx 0

aA f

f bs y

c

0 85. '

M M A f da

c n s y ( )2

M A f dA f

f bn s y

s y

c

0 59.

'

It should be noted that the ratio of k2/(k1k3) turn out to be 0.59 which is right in the middle between 0.55 and 0.63.

3.4 – Nominal Moment Strength Mn - Rectangular Sections Having Tension Reinforcement Only

The above definitions are for sections that are singly reinforced. Minimum concrete cover around the reinforcing per ACI 7.7 provides a minimum, but protection from fire may require more. Typical values for cast-in-place concrete include the following:

3” Cast against and permanently exposed to earth (All)2” Exposed to weather or earth (#6 & Larger)1-1/2” Beam & columns not exposed to weather or earth (All)3/4” Slab walls & joists not exposed to weather or earth

(#11 & smaller)

It should be obvious that all the concrete in tension was ignored and the effective depth d (distance from tensile steel to compressive face of concrete) was used in computation.

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The ACI Strength Design Method includes the provision for basic load factors and a strength reduction factor of 0.9 (tension controls) down to 0.65 (tied members). The basic equations for the ultimate moment caused by the load Mu and the nominal moment strength of the beam Mn is as follows:

M Mn u

The following assumptions apply (ACI 10.2)

1. The strength of members shall satisfy equilibrium and compatibility.2. Strain in the steel and concrete shall be directly proportional to the

distance from the neutral axis.3. The maximum usable strain cu at the extreme compression fiber shall

be 0.003.4. The tensile strength of the concrete is neglected5. The modulus of elasticity of steel is 29000 ksi.6. Whitney stress block applies as previously defined.

The following example is from Fall 1999 and is actually a homework problem for Fall 1997 and 2001 and Spring 2004 (work either the next or this example). This example is revised for Unified Design.

Problem 3.3 case 2 - Compute the nominal flexural strength Mn and the service moment capacity (MD + ML) assuming the total service moment is 60% live load. Use d=19.5 in, bw=12 in, 3-#10 bars, f’c=3500psi and fy=60,000psi.

Since f’c=3500psi < 4000 psi

Else OR FLEXURE 1

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OR

OR

OR

Find what value to use from t

Since a = 6.403 = 1x , x = a/1x = 6.403/0.85 = 7.533

Find t from the strain diagram by geometry as follows:

Since t > 0.004, meets code, but t < 0.005, transition zone

t = 0.005 = 0.9 tension controlst = 0.004766 = 0.8805 Our caset = 0.002 = 0.65 compression controls (tied)

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Since ML=0.6Ms, MD=0.4Ms and Mu=1.2MD+1.6ML then:

The following example is from Fall 1997 and 2001 and is actually a homework problem for Fall 1999 and 2003 and Spring 2005 (work either the previous or the this example). This example is revised for Unified Design.

Problem 3.3 case 5 - Compute the nominal flexural strength Mn and the service moment capacity (MD + ML) assuming the total service moment is 60% live load. Use d=36.25 in, bw=18 in, 8-#11 bars, f’c=4000psi and fy=60,000psi.

1 0 85 0 054000

10000 85 0 05

4000 4000

10000 85 0 65

. .

'. . . .

f c FLEXURE 1

C f ba a ac 0 85 0 85 4000 18 61200. ' .

T A ff y 8 1 56 60000 748800.

F T C C T ax 0 61200 648800

aA f

f bs y

c

0 85

8 1 56 60000

0 85 4000 1812 24

. '

.

.. OR a

748800

6120012 24.

M M A f da

ib in k ftc n s y

( ) . .

.

28 1 56 60000 36 25

12 24

222560000 1880

OR

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M M T da

C da

ib in k ftc n

( ) ( ) .

.

2 2648800 36 25

12 24

222560000 1880

OR

M A f dA f

f bin lb k ftn s y

s y

c

0 59 8 1 56 60000 36 25 0 59

8 1 56 60000

4000 1822560000 1880.

'. . .

.

M k ftn 1880

Find what value to use from t

Since a = 12.24 = 1x , x = a/1x = 12.24/0.85 = 14.40

Find t from the strain diagram by geometry as follows:

Since t > 0.004, meets code, but t < 0.005, transition zone

t = 0.005 = 0.9 tension controlst = 0.004552 = 0.8618 Our caset = 0.002 = 0.65 compression controls (tied)

Since ML=0.6Ms, MD=0.4Ms and Mu=1.2MD+1.6ML then:

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Wednesday, March 30th

Homework #24 is Today (Chapter 12 - Steel). Homework #25 is due Wednesday, April 6th (Chapters 1 & 2).

3.5 -- Balanced Strain Condition

At the balanced strain condition the maximum strain cu at the extreme concrete compression fiber just reaches 0.003 simultaneously with the tension steel reaching y = fy/Es (0.002069 for fy = 60 ksi) as shown below. The amount of steel corresponding to the balanced condition is Asb.

If the actual steel provided was more than Asb then a sudden failure would occur at ultimate loading since the concrete would fail first in a brittle manner (BAD).

If the actual steel provide was less than Asb then a gradual failure would occur at ultimate loading since the steel would fail first in a ductile manner (GOOD).

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Derive Equations 3.5.1 through 3.5.4 on pp. 55 for the balanced strain condition for a rectangular beam with tension reinforcement only (SKIP AND SHOW RESULTS) (FLEXURE 1).

A

bds (3.5.1)

x db

cu cu y

x

db cu

cu y

x

d fb

y

0 003

0 003 29000000

.

. /

x

d fb

y

87000

87000 (3.5.2)

C f b xb c b0 85 1. '

T A f bdfb sb y b y

F T Cx 0

0 85 1. 'f b x bdfc b b y

bc

y

bf

f

x

d

0851

. '(3.5.3)

bc

y y

f

f f

0 85 87000

870001

. '(3.5.4)

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3.6 -- Maximum Reinforcement Ratio

In order to assure a ductile failure the ACI limits (ACI 10.3.3) the amount of tension steel to not more than 75 % of the amount in the balanced strain condition (max = 0.75b). Table 3.6.1 pp. 57 gives maximum reinforcement ratios, also FLEXURE 1 SP-17(97). This is all from the 1995 or 1999 ACI code. The 2002 code states in 10.3.5 that if t>0.004 to ensure ductility (about 72.4% for grade 60).

A more direct way of controlling ductility is prescribe a maximum value for the neutral axis distance x at the failure imminent condition (max x = 0.75 xb).

For the 2002 code for flexural tension reinforcement at t = 0.004 the strength reduction factor turns out to be = 0.81667 using t = 0.002 for grade 60 steel. Other important strain-phi relationships are as follows:

t = 0.005 = 0.9 tension controlst = 0.002 = 0.65 compression controls (tied)t = 0.002 = 0.70 compression controls (spiral)t = 0.004 = 0.81667 maximum for singly reinforced (tied)t = 0.003667 = 0.78889 balanced condition (tied)

3.7 -- Minimum Reinforcement Ratio

This is all a very nice derivation of the amount of reinforcement required to provide the same moment capacity as the cracking moment of the concrete. This led to ACI 10.5.1 that states:

Af

fb ds

c

yw,min

'

3

but not less than

20


OR

For statically determinate T-section with the flange in tension ACI 10.5.2 that states:

Af

fb ds

c

yw,min

'

6 (old) (new)

but not less than

Af

fb ds

c

yE,min

'

3 (old) none (new)

Looks like the same requirements to me. Is it?

The amount of reinforcing may be less than the minimum per ACI 10.5.3 provided the area provided is one-third more than is required. For Slab or footings the minimum area is per ACI 7.12 with spacing not to exceed 3t or 18”.

Grade 40 or 50 As = 0.0020*Ag

Grade 60 As = 0.0018*Ag

Grade 40 or 50 As = 0.0018*Ag(60/fy)

3.8 -- Design of Rectangular Sections in Bending Having Tension Reinforcement Only Under ACI-10.3 and 10.5

The problem is to determine b, d, and As from the required value of Mn = Mu/ and the material properties fc' and fy. Since there are only two applicable equations of equilibrium, but three unknown, many possible solutions exist. Assume = 0.9, if required, and check at the end.

If the reinforcement ratio is preset then the following equation can be derived from the previously solved equilibrium conditions:

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(3.8.4b)

where

mf

fy

c

0 85. ' (3.8.4a)

In some situations the values of b and d are preset are the equations would be:

11 1

2

m

mR

fn

y(3.8.5)

The procedure to be used in strength design of rectangular sections with tension reinforcement only is as follows (Flexure 2.1 - 2.4 from ACI SP-17(97) typ., cover this NOW):

Design Table for Singly Reinforced Rectangular Beams

ACI 340R-97 SP-17(97)

FLEXURE 2.1 - 2.4 is set up to find As given bd as follows:

Enter FLEXURE 2.1 - 2.4 with Kn and interpolate to find .

A bds

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It should be noted that if Kn is above the line then is must be increased to 1.33 (<min). If 1.33 is less than minimum then use it

(<min). If 1.33 is more than minimum then use minimum (min). Also if Kn is off the bottom of the chart compression steel is required

(>max).

FLEXURE 2.1 - 2.4 is also set up to find bd given as follows:

Enter FLEXURE 2.1 - 2.4 with and interpolate to find Kn.

It should be noted that if Kn is above the line then is must be decreased to 0.75Kn (<min). Also if Kn is off the bottom of the chart

compression steel is required (>max).

bd F2 12000

FLEXURE 2.1 - 2.4 is also set up to find Mn given As and bd as follows:

A

bds

Enter FLEXURE 2.1 - 2.4 with and interpolate to find Kn.It should be noted that if Kn is above the line then is must be decreased

to 0.75Kn (<min). Also if Kn is off the bottom of the chart compression steel is required (>max).

Fbd

2

12000

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The following example is from Fall 1999 and is actually a homework problem for Fall 1997 and 2001 and Spring 2004, it has been revised for the newer version of SP-17 without in the tables (work either the next or this example). This example is revised for Unified Design.

Problem #3.8 case 2 - Design a rectangular beam with tension reinforcing only such that excessive deflections would not be expected under normal circumstances. Use the following data and include the actual beam weight (d/b=1-5-2.0[1.75], wD=0.8k/ft, wL=1.8k/ft, L=30ft, f’c=3000psi and fy=60,000psi.)

You could assume thebeam weight in lb/ft = 20 lb/ft * span in feet20 lb/ft * 30 = 600 lb/ft

@ FLEXURE 1 or 2.1

Enter Table 2.1 with to find Kn and check = 0.9

and and Fbd

2

12000

For total height add 2.5’’ and round up to h=33 (d=30.5) and b=18

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The following revised values include the beam weight

Enter Table 2.1 with Kn and interpolate to find and check = 0.9

0.00751

Choose 3-#11 from REINFORCEMENT 14 (bmax = 23.5 int.)

The following a check

Enter Table 2.1 with to find Kn and check = 0.9

Kn=456.0

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There are many answers to this question, but I will use the following

18x33 w/3-#11

The following example is from Fall 1997 and 2000 and is actually a homework problem for Fall 1999 and 2003 and spring 2005 (work either the previous or this example). This example is revised for Unified Design.

Problem #3.8 case 4 - Design a rectangular beam with tension reinforcing only such that excessive deflections would not be expected under normal circumstances. Use the following data and include the actual beam weight (d/b=1-5-2.0[1.75], wD=1.2k/ft, wL=2.5k/ft, L=30ft, f’c=4000psi and fy=60,000psi.)

@ FLEXURE 1 or 2.2

Enter Table 2.2 with to find Kn check = 0.9

and and Fbd

2

12000

26


Increase 5% add 2.5’’ and round up to h=33 (d=30.5) and b=18 inches

The following revised values include the beam weight

Enter Table 2.2 with Kn and interpolate to find and check

0.01014

Choose 4-#11 from REINFORCEMENT 14

The following a check

Enter Table 2.2 with to find Kn andcheck = 0.9

Kn=613.7

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There are many answers to this question, but I will use the following

18x33 w/4-#11

1. Assume that min < < max (I like =@ T=0.01)

2. Determine the required bd2 (from Rn)3. Choose values for b and d (reasonable proportions d/b=1.5 to 2.0)4. Determine revised (for new b and d)5. Compute As (from )6. Select reinforcement and calculate strength (Mn > Mu)7. Remember to check = 0.9 an any step

28


Thursday, March 31st

Exam # 5 – Chapter 3, 5, 6, 11 & 12 (Steel)

29


Friday, April 1st

Seminar #3 – Steel

Homework #25 is due Wednesday April 6th (Chapters 1 & 2). Homework #26 is due Friday, April 8th (Chapter 3).

30


Monday, April 4th

Homework #25 is due Wednesday April 6th (Chapters 1 & 2). Homework #26 is due Friday, April 8th (Chapter 3).

3.9 -- Practical Selection for Beam Sizes, Bar Sizes, and Bar Placement

The choice of the steel reinforcement ratio is dependent on the limitation on the deflection of the beam. Normally one-half of the maximum permissible value is adequate ( = 0.5 max = 0.375b).

Numerous table in the text and handbooks have been developed to aid in the selection of proper reinforcement. Some of the following in the text (I like to us Reinforcement 14 - ACI - SP17):Table 3.9.1 pp. 66 Total Area for Various Numbers of Reinforcing Bars

Also in REINFORCEMENT 2Table 3.9.2 pp. 67 Minimum Beam Width According to ACI Code

Also in REINFORCEMENT 9 & 10Table 3.9.3 pp. 68 Minimum Beam Width for 2 db clear spacingTable 3.9.4 pp. 68 Minimum Beam Width for 3 db clear spacingTable 3.9.5 pp. 69 Standard Types and Sizes of Bar SupportsTable 3.9.6 pp. 76 Area per Foot of Width for Various Bar Spacings

Also in REINFORCEMENT 15

The following are some guidelines for selection of beam size:

1. Use whole inches for overall dimensions (slabs 1/2 inches)2. Beam stem widths in multiple of 2 or 3 inches3. Minimum clear cover is to ties and stirrups (d not whole inches)4. Economical beam depth-to-width ratio of 1.5 to 2.05. T-shape beam flanges about 20 % of overall depth

The following are some guidelines for selection of reinforcement:

6. Maintain bar symmetry about plane of loading

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7. Use at least 2 bars for flexure (1 in joist sometimes)8. Use #11 for maximum size beam bar9. No more than 2 bar sizes and no more than two sizes apart (ok but)10. Try to use one layer of reinforcement11. Follow ACI 7.6.1 (clear space horizontally > db or 1’’) & 7.6.2

(clear space vertically > 1’’ and directly above bars below) for distances between bars

12. For layers place largest bar nearest face of beam

3.10 – Nominal Moment Strength Mn of Rectangular Sections Having Both Tension and Compression Reinforcement (SKIP THIS)

Rectangular section with both tension and compression reinforcement are called doubly reinforced sections. Compression reinforcement is used to reduce the size of the cross section and reduce deflections. This may lead to other problems such as large amounts of shear reinforcement.

The proportion of tension reinforcement above the maximum allowed that is equalized by compression reinforcement need not be reduced by the 0.75 factor.

Let the class work through Examples 3.10.1 & 3.10.2 pp. 78 - 81.

3.11 – Design of Beams Having Both Tension and Compression Reinforcement Under ACI-10.3.3 (SKIP THIS)

The compression steel will yield under the following conditions from Figure 3.11.1 pp. 83:

s y' (3.11.2)

T A f bdfs y y

C f b xc c0 85 1. '

32


C f f bds y c 0 85. ' '

F T C Cx c s 0

0 85 0851. ' . ' 'f b x f f bd bdfc y c y

xf d

f

f

fy

c

c

y

0 85

10 85

1. ''

. '

(3.11.3)

scu

xx d' ' (3.11.4)

Substituting (3.11.4) into (3.11.2) gives

cuyx

x d '

87000

xx d f y '

xd

fy

87000

87000

'(3.11.5)

Substituting (3.11.3) into (3.11.5) gives

(3.11.1)

An alternate solution is as follows (assume at yield, find x):

x d

cu cu y

'

Find the area of tension steel for this condition (Asy). If Asy > As then As' yields and s' > y . The solution is direct (F find x and M find Mn). If Asy < As then As' does not yield and s' < y . The solution is a quadratic as follows (draw strain diagram):

33


x d

cu cu s

'

' or

scu

xx d' '

T A fs y

C f b xc c0 85 1. '

C A f f A E fs s s c s s c ' . ' ' . '085 085

C A Ex

x d fs scu

c

' ' . '

0 85

F T C Cx c s 0

T C Cc s

This turn out to be a quadratic in x (solve it). Then write the moment equation as follows:

M M C da

C d dn c s

2

'

34


Tuesday, April 5th

3.12 -- Design of Sections in Bending—ACI Appendix B Unified Procedure (THIS IS WHAT WE ARE DOING)

This will be a major part of the ACI 318-02 code. Let the class work through Examples 3.12.1 & 3.12.2 pp. 90 - 91.

3.13 -- Non-Rectangular Beams (SKIP THIS)

Whitney stress block applies to these shapes per ACI 10.2.7.

The following procedures need revisions for exclusion of from tables

Design Table for Doubly Reinforced Rectangular Beams

Table A (2.1 - 2.4) is set up to find As and As' given bd as follows:

It is assumed that Kn is off the bottom (compression steel is required).

Select Kn at max from the bottom of Table A(2.1 - 2.4).

Fbd

2

12000

M FKn n1

Mn1 should be less than Mu.

A bds1 max

Calculate the additional tension steel required beyond max.

M M Mn u n2 1

With d'/d enter Table B (3.1 - 3.2) to find an'.

35


AM

a dsn

n2

2

'

A A As s s 1 2

Select c/d at max from the bottom of Table A (2.1 - 2.4) and together with d'/d enter Table D (4.1 - 4.4) to find an''. If an'>an'' then

compression steel has not yielded (As'>As2) use an'', otherwise use an'.

AM

a dn

n

'' '

2

To enter Table F (10.1.1 - 10.1.2) with '/ and d'/d to check max.

Design Table for Design of Singly Reinforced T-Beams

First assume the neutral axis is in the flange (a<t).

Fb dE

2

12000

KM

Fnu

Table A (2.1 - 2.4) with Kn find and a/d, if a/d<hf/d=t/dthen N.A. is in flange (a<t)

A b ds E

Case with neutral axis in the web (a/d>hf/d=t/d) from above (a>t)Table C (3.3) with b/bw find Knf

Table B (3.1 - 3.2) with hf/d=t/d find jf & anf

Ak j b h

asfnf f w f

nf

36

ARCH 3126 – Steel, Timber & Concrete – Spring 2005M A a dn sf nf2

M M Mnw u n 2

KM

Fnnw

Fb dw

2

12000

Table A (2.1 - 2.4) with Kn find

A b dsw w

A A As sw sf

A

b ds

E

Table E (10.2.1 - 10.2.2) with b/bw and hf/d=t/d find max, check that <max

37


Wednesday, April 6th

Homework #25 is due Today (Chapters 1 & 2). Homework #26 is due Friday, April 8th (Chapter 3). Homework #27 is due Monday, April 11 th

(Chapter 3 & 4).

Chapter 4 -- Rectangular Sections in Bending Under Service Load Condi-tions

4.1 -- General Introduction

The 2002 ACI code does not cover the alternate design or working stress method. We will skip the working stress method, which was called the alternate design method in Appendix A of the ACI code. In this chapter all references will be to the old code. Strength is associated with imminent failure at a multiple of the service load. Serviceability means satisfactory performance under service load (working stress) conditions. This performance is based on deflections and crack control.

The concept of working stress is to allow only a maximum value of concrete compressive stress (0.45fc') and steel tension stress to exist (20 or 24 ksi).

4.2 -- Fundamental Assumptions

There are four basic assumptions used in working stress for flexure:

1. Plane sections before bending remain plane after bending, that is, strain is linear across the section.

2. Stress is proportional to strain.3. Concrete does not take tension.4. No slip occurs between the steel and concrete.

38


4.3 -- Modulus of Elasticity Ratio n

The modular ratio (n) is equal to the modulus of elasticity of the steel divided by the modulus of elasticity of the concrete. ACI A.5.4 (old code) say that n may be rounded to the nearest whole number and should never be less than 6. The text recommends the use of Table 4.3.1 pp. 107 or DEFLECTION 9.

4.4 -- Equilibrium Conditions

Example 4.4.1 pp. 108 - 109 will be used to show the basic concepts of strain, stress, and equilibrium. The basic formulas are as follows from the given drawing (we will skip this method):

C f bxc1

2

T f As s

F T C 0

39

ARCH 3126 – Steel, Timber & Concrete – Spring 2005f

f

bx

As

c s

2 1

s

c

d x

x

f

f

E

En n

d x

xs

c

s s

c c

s

c

2

bx

And x

xs2

1 & 2

xA n

bx

A nd

bs s2 2 2

0

Solutions of the above equations yield the location of the neutral axis (x).

M M C dx

T dx

w

3 3

The above equation can be used to find T or C given the moment.

fC

bxc 2

fT

Ass

The above two equations should be checked against the allowable stress.

4.5 -- Method of Transformed Section

In the transformed section method the area of steel is transformed into an equivalent area of concrete by multiplying by the modular ratio (n). The static equilibrium and strain compatibility equations still apply from a sketch as follows (we will do this method):

40


A f A fs s t t

f

E

f

Es

s

t

c

These concepts and equations will be shown by example in the following sections and the moment of inertia of the transformed cracked section as follows:

bxnA d xs

2

20

Solve above location for the position of the neutral axis. The transformed cracked moment of inertia is then as follows:

Ibx

nA d xcr s 3

2

3

The equivalent concrete area and imaginary stress are as follows:

A nAt s

ff

nts

41


The actual stress are found form the bending stress theory (My/I) as follows:

fMx

Iccr

f n

M d x

Iscr

The following example is from Fall 1999 and is actually a homework problem for Fall 1997 and 2001 and Spring 2004 (work either the next or this example).

Problem 4.3 case 4 - Calculate the allowable bending moment for the beam according to the working stress (Alternate Method). Is the service moment capacity controlled by concrete or steel? Use d=19.5 in, bw=14 in, 3-#8 bars, f’c=4000psi and fy=60,000psi.

According to ACI section A.3 (Appendix A of old code) the allowable stresses are as follows:

and

The modular ratio can be calculated or found in DEFLECTION 9

The location of the neutral axis can be found by either method.One might notice that the equations are actually identical

(Equilibrium)

42


bxnA d xs

2

20 (Transformed)

YOU CAN SKIP THIS METHOD

Assuming that the moment is controlled by concrete (wrong)

From equilibrium C=T therefore

Assuming that the moment is controlled by steel (right)


43


STEEL CONTROLS

WE WILL USE THIS METHOD

Alternatively, according to the working stress method as follows:

Solve above location for the position of the neutral axis.

The transformed cracked moment of inertia is then as follows:

Assuming the concrete controls and from the bending stress equation

(My/I) as follows (wrong):

The stress in the steel is as follows:

Assuming the steel controls and from the bending stress equation (My/I) as follows:


44


STEEL CONTROLS

The following example is from Fall 1997 and 2000 and is actually a homework problem for Fall 1999 and 2003 and Spring 2005 (work either the previous or this example).

Problem 4.3 case 2 - Calculate the allowable bending moment for the beam according to the working stress (Alternate Method). Is the service moment capacity controlled by concrete or steel? Use d=19.5 in, bw=12 in, 3-#10 bars, f’c=3000psi and fy=40,000psi.

According to ACI section A.3 (Appendix A) the allowable stresses are as follows:

f f psi psic c 0 0 3000 1350.45 ' .45

and

f psi Grades 20 000 40, ( )

The modular ration can be calculated or found in DEFLECTION 9

nE

E fs

c c

29 000 000

57000

29 000 000

57000 3000

29 000 000

3122 0009 289 9

, ,

'

, , , ,

, ,.

The location of the neutral axis can be found by either method.One might notice that the equations are actually identical

bxnA d xs

2

20

45


12

29 3 1 27 19 5 0

2xx . .

6 34 29 668 655 02x x . .

x x2 5 715 111 0 . .4425

x 8 079 13 794. , .

YOU CAN SKIP THIS METHOD

Assuming that the moment is controlled by concrete

C f bx psi in in lb kc 1

2

1

21350 12 8 079 65 440 65. , .44


fC

A

k

inksi ksi OKs

s

65

3 1 271718 20

2

.44

.. ( )

M C dx

k in k in k ftw

365 19 5

8 079

31 099 8 91 65.44 .

., . .

M k ftw 91 65.

CONCRETE CONTROLS

WE WILL USE THIS METHOD

Alternatively, according to the working stress method as follows:


x 8 079.


46


Ibx

nA d x incr s 3

23

2 4

3

12 8 079

39 3 1 27 19 5 8 079 6582

.. . .

Assuming the concrete controls and from the bending stress equation

(My/I) as follows:

f

Mx

IM

f I

x

psi in

inlb in k ftc

cr

c cr 1350 6582

8 0791 099 851 91 65

4

., , .


f nM d x

I

k in in

inksi ksi OKs

cr

91099 8 19 5 8 079

65821718 20

4

. . ..

4.6 -- Investigation of Rectangular Sections in Bending with Tension Reinforcement Only

The concept of working stress is to allow only a maximum value of concrete compressive stress (0.45fc') and steel tension stress to exist (20 or 24 ksi).

The condition where both the concrete and steel reach their respective allowable stresses simultaneously is called the ideal condition (neutral axis at ideal location or balanced condition). If the steel reaches the allowable stress first (steel controls) the beam is said to be under-reinforced (neutral axis closer to compression face than the ideal location). If the concrete reaches the allowable stress first (concrete controls) the beam is said to be over-reinforced (neutral axis farther to compression face than the ideal location). The condition is shown in the following sketch.

47


4.7 -- Design of Rectangular Sections in Bending with Tension Reinforcement Only (SKIP)

This is all very nice, but nobody designs using working stress.

48


Thursday, April 7th

4.8 -- Serviceability - Deflections

Solve of the moment of inertia of the transformed cracked section as follows in the sketch:

bxn A x d nA d xs s

2

21 0 ' '

Solve above location for the position of the neutral axis. The transformed cracked moment of inertia is then as follows:

Ibx

n A x d nA d xcr s s 3

2 2

31 ' '

The following example is from Fall 1999 and is actually a homework problem for Fall 2000, 2001, 2002 and 2003 and Spring 2005 (work next example).

Problem 4.12 – For the beam shown in figure for Prob. 4.12, compute the transformed cracked section moment of inertia Icr that would be needed for a deflection calculation. Use d=36.1 in, d’=2.5 in, bw=18 in, h=40 in, 6-#7 bars (top), 10-#11 bars (bot), f’c=4000psi and fy=60,000psi.

49



bxn A x d nA d xs s

2

21 0 ' '


Ibx


2 2

31 ' '

The following example is new for Fall 2000, 2001, 2002 and 2003 and is actually a homework problem for Fall 1999 and Spring 2004 (work this example).

Problem 4.11 – For the beam of Problem 4.10, compute the transformed cracked section moment of inertia Icr that would be needed for a deflection calculation. Use d=27.3 in, d’=2.5 in, bw=16 in, h=30 in, 3-#11 bars (top), 4-#11 bars (bot), f’c=4000psi and fy=60,000psi.

50



bxn A x d nA d xs s

2

21 0 ' '


Ibx


2 2

31 ' '

4.9 -- Serviceability - Flexural Crack Control for Beams and One-Way Slabs

Go over Figure 4.9.1 pp. 122 and the use of Equation 4.9.2 pp. 122 as follows in the picture:

51


Where A=Ae/number of bars (m=As/Ab for largest bar), Ae=area of concrete surrounding bars and having the same centroid of the bars, and dc=thickness of concrete cover measured form the extreme tension fiber to the center of the bar located closest to that fiber (largest bar used).

REINFORCEMENT EXAMPLE 1 (ACI 340R-97 SP-17 (97)) – Check the following beam section for Exterior Crack Control provisions. d=18 in, bw=11 in, cc=2 in (main reinforcing), 3-#9 (1L), f’c=4000 psi, & fy=60,000 psi. WORK ONLY THIS EXAMPLE

For single layer of equal size bars

52


If bw=15in, then z=145.3 (No good just barely, but rounding to 3 digits z=145 (Ok) see REIFORCEMENT 14, which is based on 2 inches of

clear cover to main reinforcing. Other tables could have be used (REINFORMENT 8.1, 8.2 & 11)

SKIP THIS EXAMPLE

REINFORCEMENT EXAMPLE 3 (ACI 340R-97 SP-17 (97)) – Check the following beam section for Exterior Crack Control provisions. d=18 in, bw=11 in, cc=2 in (main reinforcing), As=2.0in2, f’c=4000 psi, & fy=60,000 psi.

Just using REINFORCEMENT 14 the following judgements are made:

As Bars Arrange Min bw Max bw Comment2.00 2 #9 1L 8.0 10.0 No good2.37 3 #8 1L 9.5 15.5 OK BEST2.12 2#6 & 4#5 2L 9.0 21.5 Example2.20 5 #6 2L 9.0 21.0 Example

Check of 2 #9 bars

For single layer of equal size bars

53


Check maximum bw for 5 #6 (2L) bars

For multi-layers of equal size bars find centroid of bars

First find location of top layer of bar

54


Note: REINFORCEMENT 14 is incorrect (Example is correct)?

Check maximum bw for 2 #6 & 4 #5 (2L) bars

For multi-layers of unequal size bars find centroid of bars

First find location of top layer of bar

where Ab is largest bar

55


Note: REINFORCEMENT 14 is correct (Example is correct)?

4.10 -- Serviceability - Side Face Crack Control for Large Beams

For deep beams the maximum crack width may occur on the side faces. Equation 4.10.1 pp. 126 (ACI 10.6.7) is to be used for beams greater than 3 feet deep as follows:

A dsk 0 012 30. in2/ft of height

Spaced at the MIN(d/6,12) as shown in the sketch as follows:

56


Friday, April 9th

Homework #26 is due today (Chapter 3). Homework #27 is due Monday, April 11th (Chapters 3 & 4). Homework #28 is due Wednesday, April 13th (Chapter 5).

Chapter 5 -- Shear Strength and Shear Reinforcement

5.1 -- Introduction

Failures in beams commonly referred to as "shear failures" are actually tension failures at the inclined cracks as shown below.

5.2 -- The Shear Stress Formula Based on Linear Stress Distribution (SKIP)

This is a very nice derivation of shear stress that you have seen in Strength of Materials and Elementary Steel (lets not do it again).

5.3 -- The Combined Stress Formula (SKIP)

This is just another form of the combined stress equation that was derived in Strength of Materials. Shown below is the cracking orientation along a simple beam.

57


5.4 -- Behavior of Beams Without Shear Reinforcement

An inclined crack in a beam that was previously uncracked due to flexure is known as a web-shear crack. An inclined crack forming at the top of and becoming an extension of an existing flexural crack is known as a flexure-shear crack. The sketch below illustrate cracking types near a support.

The transfer of shear in a reinforced concrete member occurs by a combination of the following mechanisms:

1. Shear resistance of the uncracked concrete, Vcz.2. Aggregate interlock force tangent to the shear crack (similar to friction)

due to irregular interlock of the aggregate, Va.3. Dowel action by the longitudinal reinforcement to transverse force, Vd.4. Arch action on relatively deep beams

58


5. Shear reinforcement resistance from stirrups, Vs.

There are four general categories of failure may be established and the corresponding a/d (shear span) ratios as follows (SKIP):

1. a/d<=1, Deep beams that behave like an tied-arch after cracking (many possible failures in Figure 5.4.5 pp. 143)

2. 1<a/d<=2.5, Short beams with shear strength in excess of the inclined cracking strength (flexure-shear crack failure)

3. 2.5<a/d<=6, Usual beams of intermediate length that form beam segments known as teeth due to flexural cracks

4. a/d>6, Long beams normally failing in yielding of the tension steel.

5.5 -- Shear Strength of Beams Without Shear Reinforcement-ACI Code

The (detailed method ACI 11.3.2.1) shear strength of concrete beams without shear reinforcing is as follows (don't worry about the derivation):

V fV d

Mb d f b dc c

w u

uw c w

19 2500 35. ' . '

The quantity Vud/Mu<1.0 in the above equation.

The (simplified method ACI 11.3.1.1) shear strength of concrete beams without shear reinforcing is as follows:

V f b dc c w2 '

The cracking in a continuous beam is somewhat similar to a simple beam as shown below

59


For lightweight concrete two methods may be used as follows (substitute the values shown for the square root of f'c):

ACI 11.2.1.1 ff

fcct

c'.

' 6 7 Values of Vc and Tc

ACI 11.2.1.2 f fc c' . '0 75 Values of Vc,Tc, and Mcr all LW

ACI 11.2.1.2 f fc c' . '0 85 Values of Vc,Tc, and Mcr Sand LW

Interpolation is allowed for partial sand replacement.

5.6 -- Function of Shear Reinforcement

Shear reinforcement can come in many types as follows:

1. Vertical stirrups (perpendicular to longitudinal reinforcement) are the most common

2. Welded wire fabric (used in double tees)3. Inclined stirrups (45o or more from the longitudinal reinforcement) is

very effective but labor intensive4. Longitudinal reinforcement bent toward the compression zone

Figure 5.6.3 pp. 151 shows the effectiveness of the various types of shear resistance devices. Concrete and stirrups are primary resistance in uncracked and cracked states respectively. Dowel action and aggregate interlock are secondary.

60


5.7 -- Truss Model for Reinforced Concrete Beam

The truss analogies shown in Figure 5.6.2 pp. 150 and Figure 5.7.1 pp. 152 show how the requirement that required shear reinforcement be placed at no more than one-half the effective depth s<=d/2.

5.8 -- Shear Strength of Beams with Shear Reinforcement

The nominal shear strength per ACI 11.1.1 is as follows:

V V V Vu n c s

The following is the strength of reinforcement ACI 11.5.6.2 & 11.5.6.3:

VA f d

ssv y

V

A f d

ss

v ysin cos

The above is with measured from the longitudinal axis.

5.9 -- Lower and Upper Limits for Amounts of Shear Reinforcement

The minimum amount of shear reinforcement per ACI 11.5.5.3 is as follows (only when required and spaced at not more than d/2 or 24 inches ACI 11.5.4.1) (NEW in 2002 code):

This corresponds the following strength of reinforcement (only for the absolute maximum) for the limiting case is as follows:

VA f d

sb ds

v yw 50

61


Minimum shear reinforcement must be provided whenever the factored shear exceeds one-half the concrete strength per ACI 11.5.5.1 in all but the following conditions:

1. Slabs and footings2. Concrete joist construction3. Beams with the largest of the following:

a. h<10inchesb. h<2.5*flange thicknessc. h<0.5*web width

To ensure that the reinforcement amount not be to high ACI 11.5.6.8 states that the following limit:

V f b ds c w8 '

In addition the minimum ACI 11.5.4.3 requires a minimum spacing of d/4 or 12 inches whenever:

V f b ds c w4 '

62


Monday, April 11th

Homework #27 is due today (Chapters 3 & 4). Homework #28 is due Wednesday, April 13th (Chapter 5). Exam #6 is Thursday, April 14 th (Chapters 1-5 - Concrete)

5.10 -- Critical Section for Nominal Shear Strength Calculations

The critical location for shear may be taken at d away from the face of the support provided the following conditions are met per ACI 11.1.3:

1. It is a compressive reaction (column not girder)2. No concentrated load within d from face of support

The following sketch shows when you can and cannot go d from the support face for the critical shear.

63


5.11 -- ACI Code Provisions for Shear Strength of Beams

The nominal concrete strength equations are given here again (simplified and detailed).

The following categories are listed in the text for reinforcement (the text table 5.11.1 is more detailed on pp. 160 & 161):

1. VV

uc

2 none required

2. V

V Vcu c2

minimum required (not slab, footing, joist, etc.)

3. V V V Vc u c s min minimum required all members

4. V V V V f b dc s u c c w min '4 calculated required

5. V f b d V V f b dc c w u c c w 4 8' ' reduce min. spacing

6. V f b d Vc c w u 8 ' cannot do

The maximum value for the square root of f'c is 100 unless the minimum reinforcement is used. The old code is as follows (11.1.2.1):

Af b s

f

b s

fvc w

y

w

y

'

500050 150 Ignore this

Also the shear strength of the concrete may be increased 10% for concrete joist construction.

The code sections are summarized in table 5.11.1 pp. 160 & 161.

The following are actually homework problems from Spring 2004 (work these example).

64


Problem 5.5 case 2 - Design and detail on the beam the vertical U stirrup (specified their size, dimension their locations, and show them on a side view of the beam)Case span(ft) bw(in.) d(in.) As(in2) wD wL f’c fy2 26 18 29.6 6#9 1.6 2.7 4,000 60,000

Calculate the slab/beam weight and factored loads, assume h=32in

Calculate the maximum and minimum factored shears in the span

Determine the equation for the shear at any point along the span

Find the critical shear at d from the support face (9in + 29.6in = 3.2167 ft)

Determine nominal concrete shear strength

65


Since Vmin<Vc/2 or (14.04<25.27), location to terminated minimum stirrups must be found and no stirrups required at mid-span.

Determine maximum stirrup spacing, minimum shear reinforcing and the nominal shear capacity with minimum reinforcing.

Use 14 in

Since Vnmin<Vd or (71.48<71.57), find required shear (spacing) a critical location

)12(94.13 inuseins

Capacity at 12 in spacing need only be found to check if reduced spacing is required.

Reduced spacing not required

66


Determine the location where 14 in. spacing and no stirrups are adequate and then determine the final stirrup spacing

From 3 in from face to 38.8 in from support center use 12 in spacingFrom 38.8 in to 133.1 in from support center use 14 in spacing

1@3,3@12,[email protected] center (o.c.) each end (EE)(10 in space left each side of mid-span)

Problem 5.12 - The beam of the accompanying figure carries a live load of 2.7 kips/ft in addition to the weight of the slab and beam. Design and detail on the beam the vertical U stirrup spacing for #3 stirrups. Use f’c=3000 psi and fy=40,000 psi. Use the ACI Code simplified procedure with constant Vc. This example is per ACI 318-99. It has been revised for ACI 318-02 at the end (skip to revisions).

67


Calculate the slab/beam weight and factored loads

w w w k ft k ft k ftu d l 1 1 7 1 0 5859375 1 7 2 7 5.4 . .4 . / . . / .4103125 /


V

w L k ft ftku

max

.410 /.

2

5 23

262 22

V

w L k ft ftkl

min

. . /.

8

1 7 2 70 23

813 20

Determine the equation for the shear at any point along the span68


V x VV V

Lx x x

max

max min

/.

. .

.. .

262 22

62 22 13 20

11 562 22 4 263

Find the critical shear at d from the support face (6 in + 22 in = 2.333 ft)

V k2 333 62 22 4 263 2 333 52 27. . . . .


V f b d lb kc c w 2 0 85 2 3000 12 22 24 582 24 58' . , .

V kkc

2

24 58

212 29

..

Since Vmin>Vc/2, minimum stirrups will be required to mid-span


Ab s

fypsi

in in

psiUokv

wmin . . # 50 50

12 11

400000 165 0 22 3

V VA f d

s

in psi in

inlb ks s

v ymin , .

.. 11

2

0 850 22 40000 22

1114960 14 96

V V V k k kn s cmin min . . . 14 96 24 58 39 54

Since Vnmin<Vd, find required shear a critical location

V k V V V k V kd s c s s 52 27 24 58 27 69. . .

V kA f d

s

in ksi in

s ssv y 27 69 0 85

0 22 40 22 164 562

. .. .

69


s in use in5 94 5. ( )

Use one other spacing, say 8 in, and calculate the nominal shear strength

V V VA f d

s

in ksi in

inkn c s

v y, , . . .

..8 8

8

2

24 58 24 58 0 850 22 40 22

84515

Determine the location where 8 in. and 11 in. spacing are adequate and then determine the final stirrup spacing

V k V x x x ft inn8 54515 62 22 4 263 4 004 48 04 . . . . .

V k V x x x ft inn11 839 54 62 22 4 263 5 320 63 84 . . . . .


From 63.84 in from support center to mid-span use 11 in spacing

1@3,8@5,2@8,Balance(7)@11in.on center (o.c.) each end (EE)(8 in space at mid-span)

The following is the revised solution for ACI 318-02


w k ft

in in in in

in ftk ftbm

0 15

4 5 69 21 12

1440 5859375

2 2. /

.

/. /


70



Find the critical shear at d from the support face (6 in + 22 in = 2.333 ft)


Since Vmin>Vc/2, minimum stirrups will be required to mid-span


Ab s

fypsi

in in

psiUokv

wmin . . # 50 50

12 11

400000 165 0 22 3

Since Vnmin<Vd, find required shear a critical location

71


Capacity at 5 in spacing and maximum reinforcing check as follows:


Use one other spacing, say 8 in, and calculate the nominal shear strength

Determine the location where 8 in. and 11 in. spacing are adequate and then determine the final stirrup spacing


From 69.63 in from support center to mid-span use 11 in spacing

1@3,10@5,2@8,Balance(6)@11in.on center (o.c.) each end (EE)(8 in space each side of mid-span)

These are new examples from Spring 2004. and use them for Homework this Spring 2005 (skip these examples)

72


Problem 5.5 case 1 - Design and detail on the beam the vertical U stirrup (specified their size, dimension their locations, and show them on a side view of the beam)Case span(ft) bw(in.) d(in.) As(in2) wD wL f’c fy1 20 18 36.3 5#10&5#116 10 4,000 40,000

Use the ACI Code simplified procedure with constant Vc.

Calculate the slab/beam weight and factored loads (assume h=40in)



Find the critical shear at d from the support face (9 in + 36.3 in = 3.775 ft)


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Since Vmin>Vc/2, (40.00>39.99)minimum stirrups will be required to mid-span


Use 18 in.

Use #4 16in.

Since Vnmin<Vd, (89.215<165.1) find required shear a critical location

Capacity at 4 in spacing and maximum reinforcing check as follows:


Use two other spacing, say 8 & 12in and calculate the nominal shear strength74


Determine the location where 8, 12, & 16 in. spacing are adequate and then determine the final stirrup spacing

1@3,16@4,2@8,1@12 & [email protected] center (o.c.) each end (EE)(last stirrup at mid-span)

Problem 5.11 - Design and detail on the beam the vertical U stirrup Use no spacing less than 3in.

span(ft) bw(in.) d(in.) wD(k/ft) wL(k/ft) f’c fy

16 14 21.5 1.6 3.0 3,000 60,000

Use the ACI Code simplified procedure with constant Vc.



75



Find the critical shear at d from the support face (6 in + 21.5 in = 2.292 ft)


Since Vmin<Vc/2, (9.6<12.36) minimum stirrups will not be required at mid-span


Use 10 in.

Use #3 10in.

Since Vnmin>Vd, (46.015>41.11) will work for rest of beam. 76


Determine the location where 10 in. spacing are

Use 1@3, 9@10 in spacing

5.12 -- Working Stress Method - ACI Code, Appendix A

The (detailed method of the old ACI Appendix A A.7.4.4) shear strength of concrete beams without shear reinforcing is as follows (SKIP THIS):

v fVd

Mfc c w c ' . '1300 1 9

The quantity Vd/M<1.0 in the above equation.

The (simplified method of the old ACI Appendix A A7.4.1) shear strength of concrete beams without shear reinforcing is as follows:

v fc c1 1. '

These are compare to the actual shear stress as follows:

The area of shear reinforcement is as follows per (A7.5.6.2):

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With the maximum spacing the same as strength design (d/2,24) per ACI A7.5.4.1 as is the minimum shear reinforcing per ACI A7.5.3.3 and the max spacing shall be cut in half when:

v v fc c 2 '

In addition the maximum shear stress is such that (ACI A7.5.6.8):

v v fc c 4 4. '

5.13 -- Shear Strength of Beams - Design Examples

Cover these in class with special note on the factored shear at mid-span of a simple beam and shear at d from the face of the support or anywhere. Also note the drawing of capacity and shear diagrams.

Problem 5.14 – For a rectangular beam of 14in. width and effective depth 22.5 in. with f’c=4000 psi and fy=60,000 psi, determine the maximum factored shear Vu for this beam for the following conditions. This example is per ACI 318-99. It has been revised for ACI 318-02 at the end (skip to revisions).

(a) When no stirrups are used.(b) When minimum percentage of shear reinforcement (#3 U stirrups) is

used according to ACI-11.5.5.3; specify the spacing to be used.(c) When maximum percentage of stirrups is used (#4 U stirrups); specify

the spacing to be used.


No stirrups are required if Vu<Vc/2

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(a) Vu=16.9 kips with no stirrups

Determine maximum stirrup spacing and minimum shear reinforcing per ACI 11.5.5.3.

Check 11 in. #3 U stirrup spacing

Calculate nominal shear strength for #3 U stirrups at 11 in. o.c.

V V V k k kn s cmin min . . . 14 96 24 58 39 54

(b) Vu=56.8 kips with #3 U stirrups at 11 in. o.c.

Determine maximum shear reinforcing per ACI 11.5.6.8.

Determine #4 U stirrup spacing


79


(c) Vu=149 kips with #4 U stirrups at 4 in. o.c.

The following is the revised solution for ACI 318-02.


No stirrups are required if Vu<Vc/2

(a) Vu=14.9 kips with no stirrups

Determine maximum stirrup spacing and minimum shear reinforcing per ACI 11.5.4.1 & 11.5.5.3.

Check 11 in. #3 U stirrup spacing (since f’c<4444psi use limit)


80


(b) Vu=50.1 kips with #3 U stirrups at 11 in. o.c.

Determine maximum shear reinforcing per ACI 11.5.6.9.

Determine #4 U stirrup spacing


(c) Vu=131 kips with #4 U stirrups at 4 in. o.c.

5.14 -- Shear Strength of Members Under Combined Bending and Axial Load

Axial compression increases shear strength, while axial tension reduces it (why is that?). The following graph shows the strength in shear verses the axial compression.

81


The (detailed method ACI 11.3.2.2) shear strength of concrete beams without shear reinforcing and subjected to axial compression may be as follows:

V fV d

Mb d f b d

N

Ac cw u

uw c w

u

g

19 2500 35 1

500. ' . '

This is the same equation except that for Mu we use Mm (if Mm<0 then just use the limiting value)as follows:

M M Nh d

m u u

4

8

The quantity Vud/Mu<1.0 is not valid for this equation.

The (simplified method ACI 11.3.1.2 shear strength of concrete beams without shear reinforcing and subjected to axial compression is as follows:

VN

Af b dc

u

gc w

2 1

2000'

82


The (detailed method ACI 11.3.2.3) shear strength of concrete beams without shear reinforcing and subjected to axial tension is as follows:

VN

Af b dc

u

gc w

2 1

500'

In this same Nu is negative 11.3.1.3 (why is this?).

The (simplified method ACI 11.3.1.3 shear strength of concrete beams without shear reinforcing and subjected to axial tension is as follows:

ZERO

Figure 5.14.2 pp. 177 shows the effect on the shear strength of concrete subjected to axial compression and tension.

The code sections are summarized in table 5.14.1 pp. 178.

83


Tuesday, April 12th

Chapter 6 -- Development of Reinforcement

6.1 -- General

The development length is roughly proportional to the bar diameter. In smaller bars failure is primarily due to pullout (smaller lugs), while in larger bars failure is primarily due to splitting of the surrounding concrete (larger lugs). Splitting depends on the ability of the concrete to resist tension with favorable conditions of increased concrete cover, larger spacing between bars, and confinement by ties or stirrups.

6.2 -- Development Length

The derivation of development length must include failures due to pullout (friction on the surface area) and splitting (bearing per unit length).

6.3 -- Flexural Bond

As bending moment varies along the span the tensile force in the steel also varies; this induces a longitudinal interaction between the steel and the surrounding concrete known as flexural bond (normally not significant).

6.4 -- Failure Modes

Deformed bars were created to change the behavior pattern so that there would be less reliance on friction or adhesion and more reliance on the bearing of the lugs against the concrete. The figure below illustrates the bearing and splitting forces on the concrete (show this).

84


The following figure shows the various types of splitting failures to the closest face of the concrete or other bars. Close spacing and side cover cause horizontal splitting, while close bottom or top cover would cause V-notch splitting.

This final figure shows the splitting influence area surrounding typical bars.

6.5 -- Reasons for Not Using Localized Surface Stress in Strength Design

Many situations occur where the localized resistance (flexural bond) does not provided adequate safety against splitting (Ignored anyway).

85


6.6 -- Moment Capacity Diagram - Bar Bends and Cutoffs

The moment capacity diagram (like shear) must fall outside the actual factored moment diagram). The bars develop their capacity over some length Ld, in addition other criteria need to be considered. Capacity must be provided a distance d or 12db beyond cutoff locations per ACI 12.10.3 (except at ends of simple spans and cantilevers). Cutoffs in a tension zone per ACI 12.10.5 require other considerations (covered later). The following figure shows a moment capacity bound with d or 12db and Mn=Asfy(d-a/2).

86


6.7 -- Development Length for Tension Reinforcement - ACI Code

6.8 – Modifications , , and to the Bar Development Length Equations of ACI-12.2.2

The general outline of the procedure for determining development length is as follows (major changes to the 1995 Code):

Compute development length Ld per ACI 12.2.2 OR 12.2.3 but not less than 12” as follows and in REINFORCEMENT 17.1.1 & 17.1.2 (NEW use ATR not size).

Draw a section showing sc and cc.

ACI 12.2.2 No. 6 and smaller bars and deformed wires

No. 7 and larger bars

Clear spacing of bars being developed or spliced not less than db, clear cover not less than db, and stirrups or ties throughout ld not less than the code minimum

orClear spacing of bars being developed or spliced not less than 2db and clear cover not less than db

RememberACI-11.5.5

smax=min(d/2,24”)Avmin=50bws/fy

Also for columnsACI-7.10.5

smax=min(h,16dbl,48dbt)

Other cases

ACI-12.2.4 gives the following factors:

87


= reinforcement location factor1.3 for top bars (12’’ of concrete below a horizontal bar)1.0 for all others

= coating factor1.5 for epoxy-coated bars (clear cover < 3db and clear spacing < 6db)1.2 all other epoxy-coated bars1.0 for all others

x need not exceed 1.7

= reinforcement size factor0.8 for No. 6 and smaller bars or deformed wires1.0 for all others

= lightweight aggregate factor1.3 for lightweight aggregate used and fct not specified1.0 for normal weightor as follows if fct is specified, but not less than 1.0:

6 7.'f

fc

ct

ACI 12.2.3 is the following equation:

in which per ACI-12.2.4

c K

dtr

b

2 5.

with

88


c = the smaller or 1/2 the center-to-center spacing or center to nearest concrete surface of the bars being developed (draw this).

and

KA f

sntrtr yt

1500

in which

Atr = total cross-sectional area of all transverse reinforcement which is within the spacing s and which crosses the potential plane of splitting through the reinforcement being developed

fyt = specific yield strength of transverse reinforcement

s = maximum spacing of transverse reinforcement within ld

n = number of bars or wires being developed along the plane of splitting

The length may be reduced for excess reinforcement per ACI 12.2.5 (multiply by Asred’d/Asprov’d).

Useful tables include REINFORCEMENT 17.1 & 17.2 or tables 6.7.1 & 6.7.2 pp. 233 & 234.

In addition ACI 12.10.3 specifies that reinforcement extend 12db or d beyond where it is no longer required (except at the ends of cantilevers and simple spans).

89



Homework #29 is due Monday, April 18th (Chapter 6).

Problem # O’H 6 – Draw the moment capacity bound for the exterior span of the continuous frame shown using the requirements of (a) ACI-12.2.2 and (b) ACI-12.2.3. This is in accordance with ACI 318-99. The revised solution for ACI 318-02 will follow this example.

Check requirements of ACI-12.2.2 CASE 1 (ignoring stirrups)

Clear cover for all bars is 2in

cc>db cc= 2.000 > db= 1.000 (#8)1.128 (#9)1.410 (#11)

Clear spacing for each bar group as follows (4-#8):

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sc>2db sc= 2.000 > 2db= 2.000 (4-#8)8.000 > 2.000 (2-#8)

7.744 > 2.256 (2-#9)1.829 > 2.256 (4-#9)

> 1.128 =db Stirrups?7.180 > 2.820 (2-#11)1.453 > 2.820 (4-#11)

> 1.410 =db Stirrups?

Check requirements of ACI-12.2.2 CASE 1 (including stirrups)

Minimum stirrup requirements per ACI-11.5.5 & 11.5.5.3

smax=min(d/2,24)=min(27.5/2,24)=min(13.75,24)=13.75 in. (12 in. used)

Avmin=50bws/fy=50(14)12/40000=0.21>0.40 (#4U)

4-#11 have #4U at 12in and sc=1.453>db=1.410 CASE 1

4-#9 may have #4U at 24in if ld>7’-0” (Take a chance use) CASE 1

Development lengths per ACI12.2.2 CASE 1 (=1.0)

For (#8) ld=47.43 (#9) ld=53.51 (#11)ld=66.88

Bot bars (#9) ld=53.51Top bars (#8) ld=61.66 =1.3 (#11)ld=86.95

Since ld for the #9’s is less than 7’-0’’, they are CASE 1

Only the #11 top bars will not develop over 7’-0’’ (84)

91


Have not done 90o hook yet but for #8 ldh=19.0 with a 16db tail (16in).

This is 19.0/47.43 (40%) of the #8 capacity (tail has 60%)

Calculate the nominal moment capacities of the various bar groups

Bars 2-#8 4-#8 2-#9 4-#9 2-#11 4-#11Mn 188 363 236 450 358 662

Draw moment capacity bound shown below.

For top 4-#11

For top 2#11 first left side

For top 2#11 last left side

92


93


Thursday, April 14th

Exam # 6 – Chapters 1 – 5 - Concrete

94


Friday, April 15th

Homework #29 is due Monday, April 18th (Chapter 6). Homework #30 is due Wednesday, April 20th (Chapter 7).

6.9 -- Development Length for Compression Reinforcement

The basic development for reinforcement in compression is as follows (12.3.2) or REINFORCEMENT 17.4:

L df

fd fdb b

y

cb y 0 02 0 0003.

'.

The basic development length in compression can be reduced for enclosure by ties and excess reinforcement by the multipliers as follows:

1. Spiral ties at least 1/4 inch in diameter with not more than a 4 inches pitch or # 4 ties at not more than 4 inches multiply by 0.75 per ACI 12.3.3.2 .

2. Excess reinforcement multiply by the following per ACI 12.3.3.1:

a.requiredA

providedAs

s

1 0.

The modified development length cannot exceed 8 inches per ACI 12.3.1.

6.10 -- Development Length for Bundled Bars

Bundled bar require that the development length be increase 20 % for three-bar bundles and 33 % for four-bar bundles. For determination of all the needed modification factors the bundles should be treated as a single bar a diameter equal derived from the equivalent area (12.4.2).

95


6.11 -- Development Length for Tension Bars Terminated in a Standard Hook

Standard hooks are defined in ACI 7.1 and 7.2. The hook reduces the longitudinal length to develop the bar force as shown below.

The basic development length for 60000 psi steel per ACI 12.5.2 is as follows (Table 6.11.2 pp. 243 for 60000 psi) or REINFORCEMENT 18.1 (coming soon):

= 1.2 for coating factor

= 1.3 for lightweight aggregate factor

The following modifications can or must be done to the development length of a hook:

96


1. 180o hook on #11 bars or smaller with side cover>=2.5db and

additional for 90o hook 2 inches of cover beyond hook extension multiply by 0.7 per ACI 12.5.3.2.

2. # 11 bars or smaller with enclosed in ties or stirrups spaced along the length at not more than 3db of the bar multiply by 0.8 per ACI 12.5.3.3. Either the bar or the tail of a 90O hook as shown below.

3. # 11 bars or smaller with enclosed in ties or stirrups spaced along the length at not more than 3db of the bar multiply by 0.8 per ACI 12.5.3.3. Along the bar of a 180O hook.

4. Excess reinforcement multiply by the following per ACI 12.5.3.4:

a.requiredA

providedAs

s

1 0.

The final development length of the hook shall not be less than 8db or 6 inches per ACI 12.5.1

6.12 -- Bar Cutoffs in Negative Moment Region of Continuous Beams

An additional requirement apply whenever reinforcement is terminated in negative moment regions of a continuous beam as follows:

1. At least one-third of the total area of reinforcement required at the support must be developed beyond the inflection point by a minimum length as follows per ACI 12.12.3.

97


a. db. 12dbc. 1/16 of the clear span length

An additional requirement apply whenever reinforcement is terminated in the tension zone of a member per ACI 12.10.5 with one of the following satisfied (this applies to any type of member).

1. Factored shear must not exceed two-thirds of the nominal shear strength.

V Vu n2

3

2. Excess stirrup area is provided to give the following strength in excess of what is required:

V psi b ds w 60

sd

b

8

bscut

sleft

A

A

3. For #11 bars or smaller the following two conditions could be satisfied:V Vu n

3

4

2M Mu n

If these cannot be satisfied the reinforcement should be terminated in or bent up into the compression zone.

6.13 -- Bar Cutoffs in Positive Moment Region of Continuous Beams

An additional requirement apply whenever reinforcement is terminated in positive moment regions of a continuous beam as follows:

98


1. At least one-fourth (one-third for simple spans) of the total area of reinforcement required must embedded at least 6 inches into the support per ACI 12.11.1 (If part of a lateral system it must be developed per ACI 12.11.2).

The same additional requirement applies to reinforcement terminated in the tension zone of a member per ACI 12.10.5 as previously described.

If these cannot be satisfied the reinforcement should be terminated in or bent up into the compression zone.

6.14 -- Bar Cutoffs in Uniformly Loaded Cantilever Beams (SKIP)

This is a very nice example and outline of what has been previously presented.

6.15 -- Development of Reinforcement at Simple Supports and at Points of Inflection

The additional requirement of ACI 12.11.3 is as follows for simple supports (shear value at support and La is beyond centerline of support):

This also applies to point of inflection (shear value at inflection and La is beyond inflection).

6.16 -- Development of Shear Reinforcement

All shear reinforcement should be terminated as close as possible to the compression and tension faces of the member by a standard hook per ACI 12.13 or lapped 1.3Ld (More to it, but this is the important stuff).

99


6.17 -- Tension Lap Splices

The tension lap splice length as defined in ACI 12.15.1 is either a Class A or B with factors of 1.0 or 1.3 to Ld or 12 inches minimum as follows:

1. Class B when more than 50 % is needed and more than 50 % is spliced

6.18 -- Welded Tension Splices and Mechanical Connections

Bars larger than #11 may not be lapped spliced (14 or 18 can be spliced to 11), but must be welded or mechanically spliced per ACI 12.14.2.1. In addition, tension tie members must not be lapped spliced. A tension tie member is has large tension forces that cause the entire cross section to be in tension and have high tension stresses in the reinforcement.

6.19 -- Compression Lap Splices

The basic compression lap splice length per ACI 12.16.2 is as follows, but not less than 12 inches (Table 6.19.1 pp. 264)

1. Lap=0.0005fydb for fy<= 60000psi

2. Lap=(0.0009fy-24)db for fy> 60000psi

In addition, if the compressive strength of the concrete is less than 3000 psi the length must be increased by one-third.

Decreases in the lap length are permitted for confinement per ACI 12.17.2.4 or 12.17.2.5 (closed ties).

6.20 -- Compression End Bearing Connections, Welded Splices, and Mechanical Connections

Compression end bearing is allowed per ACI 12.16.4.

100


6.21 -- Splices for Members under Compression and Bending

This will be presented later in columns but the basic criteria is if it looks like compression and it talks like compression and it walks like compression, its compression or if it looks like tension and it talks like tension and it walks like tension, its tension (it could be both).

6.22 -- Design Examples

Study these mainly for procedure to follow.

BEAM ANALYSIS SUMMARY - FLEXURE AND SHEAR (OLD SKIP)

I. Factored Loadsa. wu = 1.4xwd + 1.7xwlb. wul = 1.7xwl

II. Factored Shear, VuA. Maximum, Vmax = wuxL/2B. Minimum, Vmin = wulxL/8C. Equation, VX = Vmax - [(Vmax - Vmin)/(L/2)]xXD. Critical, Vd, distance d from support faceE. Plot curve with all values and locations

III. Factored Moment, MuA. Maximum, Mmax = wuxL2/8B. Equation, MX = (wuxX/2)x(L - X/2)C. Plot curve with all values and locations

IV. Nominal Shear Strength, Vn, = 0.85A. Concrete, Vc

1. Minimum, Vc/2B. Reinforcing, Vs

1. Maximum spacing, smax = min(d/2 , 24'')2. Minimum area Av = 50 bws/fy3. Capacities, Vsmin = Avfyd/s

C. Combined, Vn = Vc + Vs

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D. Plot curve with all values and locationsE. Check values and locations vs. Factored Shear

1. Use equation for exact locationsF. Determine adequacy

V. Nominal Flexural Strength, Mn, = 0.9A. Development length, Ld

1. Ld per 12.2.2 or 12.2.3B. Capacity, Mn

1. Flexure formula or tablesa. Check minimumb. Check maximumc. Check preferred

2. Note is bars are fully developeda. Find percent developed if needed

C. Plot curve with all values and locationsD. Check values and locations vs. Factored Moment

1. Use equation for exact locationsE. Determine adequacy

VI. AnswersA. Plots should clearly shown results

1. Studio quality2. Values and locations

B. Note other problems in solution

BEAM ANALYSIS SUMMARY - FLEXURE AND SHEAR (NEW)

I. Factored Loadsa. wu = 1.2xwd + 1.6xwlb. wul = 1.6xwl

II. Factored Shear, VuA. Maximum, Vmax = wuxL/2B. Minimum, Vmin = wulxL/8C. Equation, VX = Vmax - [(Vmax - Vmin)/(L/2)]xX

102


D. Critical, Vd, distance d from support faceE. Plot curve with all values and locations

III. Factored Moment, MuA. Maximum, Mmax = wuxL2/8B. Equation, MX = (wuxX/2)x(L - X/2)C. Plot curve with all values and locations

IV. Nominal Shear Strength, Vn, = 0.75A. Concrete, Vc

1. Minimum, Vc/2B. Reinforcing, Vs

1. Maximum spacing, smax = min(d/2 , 24'')2. Minimum area Av = 50 bws/fy3. Capacities, Vsmin = Avfyd/s

C. Combined, Vn = Vc + VsD. Plot curve with all values and locationsE. Check values and locations vs. Factored Shear

1. Use equation for exact locationsF. Determine adequacy

V. Nominal Flexural Strength, Mn, = 0.9 (TC) to 0.65 (CC)A. Development length, Ld

1. Ld per 12.2.2 or 12.2.3B. Capacity, Mn

1. Flexure formula or tablesa. Check minimumb. Check maximumc. Check preferred

2. Note is bars are fully developeda. Find percent developed if needed

C. Plot curve with all values and locationsD. Check values and locations vs. Factored Moment

1. Use equation for exact locationsE. Determine adequacy

VI. Answers103


A. Plots should clearly shown results1. Studio quality2. Values and locations

B. Note other problems in solution

104


Monday, April 18th

Homework # 29 is due Today (Chapter 6). Homework # 30 is due Wednesday, April 20th (Chapter 7). Homework # 31 is due Friday, April 22nd (Chapter 8).

Chapter 7 -- Continuity in Building Frames of Reinforced Concrete

7.1 -- Common Building Frames

Reinforced concrete building construction commonly has floor slabs, beams, girder, and columns continuously placed to form a monolithic system. Figure 7.1.1 pp. 292 show these as follows (show floor system and section):

1. Nine slab spans continuous over ten supporting beams (section A-A).2. Three beams spans continuous over four supporting girders or

monolithic with four supporting columns (difference in end restraint).3. Three girder spans monolithic with four supporting columns (not

uniformly loaded).

7.2 -- Positions of Live for Moment Envelope

The influence of load on the maximum and minimum (positive and negative) bending moments must be included in any analysis of a continuous system (Figure 7.2.1(b) & (c)) as follows:

1. For maximum positive moment within a span, load that span and all other alternates spans (primary load case since moments are same sign as dead load for positive moment).

2. For maximum negative moment within a span, load the two spans adjacent to that span and all other alternates spans (complement to case 1 and is the secondary load case for negative moment).

3. For maximum negative moment at a support, load the two spans adjacent to that support and all other alternates spans (primary load case since moments are same sign as dead load for negative moment).

105


4. For maximum positive moment at a support, load the two spans beyond each of the two spans adjacent to that support and all other alternates spans (complement to case 3 and is the secondary load case for positive moment).

ACI 8.9 requires only the primary cases to be considered. Primary load cases must always be considered and secondary load cases need to be considered in the following cases:

1. High live to dead load ratios in which the live load moments exceed the dead load moments causing stress reversals.

2. Whenever lateral loading is considered and may cause stress reversals.3. Long spans where partial span loading should be considered to produce

the true moment envelope.

7.3 -- Method of Analysis

Even though factored loads per ACI 9.2 are used in analysis for ultimate failure, elastic methods of analysis are permitted by the code in ACI 8.3 (stated in next section). Many types of elastic analysis exist including the following:

1. Matrix methods based on member stiffness of the entire structure (Analysis II).

2. Moment distribution based on a slope deflection equations for a portion or the entire structure (Analysis I).

3. Approximate method, such as the ACI moment coefficients, used in normal construction.

In any analysis method the principles and laws of Statics must be met, that is, the sum of the moments on any particular span must be the same as the statical sum for a simple span (the height of the moment diagrams are the same):

106


M M M MS O L R 1

2

This can be seen in Figure 7.3.3 pp. 301, which is the result of moment distribution on three of equal six continuous spans.

7.4 -- ACI Moment Coefficients

ACI 8.3 specifies the following methods of analysis and consideration:

1. The theory of elastic analysis is to be used in analyzing frames or continuous construction (ACI 8.3.1).

2. Except for prestressed concrete, approximate methods of frame analysis are permitted for buildings of usual types of construction, spans, and story heights (ACI 8.3.2).

3. In lieu of frame analysis, the following approximate moments and shears are permitted for design of continuous beams and one-way slabs provided:a. There are two or more spans,b. Spans are approximately equal, with the larger of two adjacent

spans not greater than the shorter by more than 20 percent,c. Loads are uniformly distributed,d. Unit live load does not exceed three times unit dead load, ande. Members are prismatic.

Positive moment

End spans

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Discontinuous end unrestrained wuln2/11

Discontinuous end integral with support wuln2/14

Interior spans wuln2/16

Negative moment at exterior face of first interior support

Two spans wuln2/9

More than two spans wuln2/10

Negative moment at other faces of interior supports wuln2/11

Negative moments at face of all supports for:

Slabs with spans not exceeding 10 ft; Beams where ratio of sum of column stiffness to beam stiffness exceed eight at each end of the span

wuln2/12

Negative moment at interior face of exterior support for members built integrally with supports

Where support is a spandrel beam wuln2/24

Where support is a column wuln2/16

Shear in end members at face of first interior support 1.15wuln/2

Shear at face of all other supports wuln/2

The clear span, ln, is defined as the clear span of that span for positive moment in the span or shear, and the average clear span of the two adjacent spans for negative moment at a support.

108


7.5 -- ACI Moment Diagrams

Figures 7.5.1 through 7.5.4 pp. 305 - 308 show the moment diagrams and corresponding shear diagrams for the ACI moment coefficient derived from static equilibrium, derive part of 7.5.3 (a&b) & 7.5.4 (a&b) shown below.

We will derive the maximum positive moment case for and exterior span. Assume that the moment at the exterior support and the maximum positive moment occurs under full load (live and dead), because they do.

109


Remember from statics that the change in moment is equal to the shear and the slope of the shear is equal to the load, then the following relationship can be written to locate the location of maximum positive moment (zero shear).

Similarly the location of zero moment (inflection point) can be found from the maximum positive moment location.

The reaction at the left (VL) can be found since the change in shear is equal to the area under the load.

Now we can use the two equations of statics to find the other shear (VR) and moment (MR).

The following diagram show the moment diagrams for an exterior and interior span for the maximum in the positive and negative zone.

110


If we look at just the exterior span for both cases as follows.

111


Now combine the two diagrams to produce maximum effects as follows.

112


Finally ignore the values that don’t control as follows.

This can be done for any span and for any shear diagram as well.

113


7.6 -- Shear Envelope for Design

When the ACI moment coefficients are used, it is generally assumed that the shear diagrams just discussed are used. One could just used the ACI shear coefficients along with the load on the span (will not always check statically).

Problem # O’H 7 – Draw the ACI moment coefficient diagrams for the exterior and interior spans of the continuous frame shown using the wd=1.0 k/ft & wl=1.2 k/ft.

Statical Moment (not really necessary)

Moments in exterior span (-Left, +Middle & -Right)

Moments in interior span (-Left, +Middle & -Right)

Location of maximum and minimum values is shown below, with the locations calculations in the drawing.

114


Note: that for a complete analysis to be used in design, the shear envelopes should be drawn as well.

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Tuesday, April 19th

Chapter 8 -- Design of One-Way Slabs

8.1 -- Definition

A slab that is twice as long as it is wide is normally considered a one-way slab (two-way slabs are discussed in Chapter 16). A one-way slab will span in the shorter direction and transmit load to supports at each end of the shorter span.

8.2 -- Design Methods

Normally one-way slab design will fall under the conditions for use of the ACI moment Coefficients (ACI 8.8.3).

8.3 -- Thickness of Slab

A one-way slab is normally designed using a imaginary strip 12 in. wide. The thickness of the slab (also beams) depends on the deflection, bending, and shear requirements. ACI Table 9.5(a) governs the minimum thickness for slab (and beams) supporting construction not likely to be damaged by larger deflection as follows:

Solid one-way slabs Beams or ribbed one-way slabsSimply supported l/20 l/16One end continuous l/24 l/18.5Both ends continuous l/28 l/21Cantilever l/10 l/8

The span length, l, as defined in ACI 8.7 is as follows:

1. Span lengths of members not built integrally with support shall be considered the clear span plus the depth of the member but need not exceed the distance between center of supports (ACI 8.7.1).

116


2. In analysis of frames or continuous construction for determination of moments, span length shall be taken as the distance center-to-center of supports (ACI 8.7.2).

3. For beams built integrally with supports, design on the basis of moments at faces of support is permitted (ACI 8.7.3).

4. Solid or ribbed slabs built integrally with supports, with clear spans not more than 10 ft. are permitted to be analyzed as continuous slabs on knife edge supports with spans equal to the clear span of the slab and width of beams otherwise neglected (ACI 8.7.4).

If the slab (or beam) supports construction likely to be damaged by larger deflection, then the deflection requirements of Table 9.5(b) must be met as follows:

Roofs and damage unlikely live l/180Floors and damage unlikely live l/360Roof or floors and damage likely live + long time l/480Roof or floors and damage not likely live + long time l/240

It could be noted that the requirements for supporting construction likely to be damaged by larger deflection are about twice as stringent as those that do not. The use of Table 9.5(a) could then be used with an equivalent height equal to one the provides twice the stiffness.

h h 2 33

The flexural strength can be used as a design criteria by using a desired reinforcement ratio to determine a thickness. Normally shear will not control the design and need only be checked.

Example 8.3.1 pp. 315 - 317 shows the three basic criteria (deflection, flexure, and shear) in design of one-way slabs.

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8.4 -- Choice of Reinforcement

Reinforcement in one-way slabs may be placed in configuration similar to Figure 8.4.1 pp. 317 (straight and truss bars), but normally only straight bars are used and proper development lengths and embedment checked.

Some other limitations that affect the choice or reinforcing are as follows:

1. Minimum amount shall not be less than shrinkage and temperature reinforcement (ACI 7.12).

2. The principal reinforcement shall not be spaced at more than 3 times the slab thickness nor 18 in. (ACI 7.6.5).

Example 8.4.1 pp. 318 - 321 shows the total design process including development lengths, but straight bars probably should be used.

8.5 -- Continuity Analysis

When the conditions to use the ACI moment coefficient are not met, an elastic analysis is required. Example 8.5.1 show a continuity analysis on the previous example and compares the results.

8.6 -- Shrinkage and Temperature

ACI 7.12.1 requires that temperature and shrinkage reinforcement be used normal to flexure reinforcement (flexure must also meet this), and ACI 7.12.2.1 give the following areas or required reinforcement based on the gross concrete area (bh not bd), but not less than 0.0014:

1. Grade 40 or 50 0.00202. Grade 60 0.00183. Greater than grade 60 0.0018*60000/fy

ACI 7.12.2.2 states that this reinforcement shall be spaced at not more than 5 times the slab thickness nor 18 in.

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8.7 -- Bar Details

Figure 8.7.1 show typical one-way slab detail that are consistent with the ACI moment coefficients complements of the ACI Detailing Manual.

Problem 8.1 – Design for a warehouse a continuous one-way slab supported on beams 12 ft on centers as shown in the figure for Prob. 8.1. Assume that the beam stems are 12 in. wide. The dead load is 25 psf in addition to the slab weight, and the live load is 200 psf. Assume the reinforcement limits of ACI-10.3.3 apply. Use f’c=3000 psi, fy=40,000 and the strength method. Use ACI coefficients if permissible and use only straight reinforcing bars.

Determine the slab thickness required per ACI Table 9.5(a).

h=L/24=(12ft*12in/ft)/24=6.00in.

Since fy<60,000 psi multiply h by (0.04+fy/100,000)

H=6.00in*(0.04+40,000/100,000)=6.00in(0.8)=4.80in

Rounding up to the nearest ½ in, h=5in.

Since the minimum clear cover for slabs is ¾ in., use d=h-1.25in.-3.75in. (this will worth all the way up to a #8 bar).

119


Next the factored load (including slab weight) and clear span can be found.

Calculate the ACI coefficients for the clear span and factored load.

Moments in exterior span (-Left, +Middle & -Right)

Moments in interior span (-Left, +Middle & -Right)

Determine the required area of steel by equation of FLEXURE 2.1 then select the reinforcement (be sure to check minimum flexural, as well as, temperature and shrinkage steel). The following spread sheet shows the results of the flexural calculations. Remember to check = 0.9 (it is).

The dimensions consistent with Figure 8.7.1 on pp. 325 are given in the spread sheet and shown on the final design drawing below.

120


121


122



Homework #30 is due today (Chapter 7). Homework # 31 is due Friday, April 22nd (Chapter 8). Homework # 32 is due Monday, April 25th (Chapter 9).

Chapter 9 -- T-Sections in Bending

9.1 -- General

Almost all cast-in-place concrete structures involve the use of beams built monolithically with the slab. These beam are treated a T-beams primarily for positive moment capacity with the slab in the compression zone and as rectangular beams for negative moment with the beam web in the compression zone.

9.2 -- Comparison of Rectangular and T-Sections

If the depth of the effective stress block (Whitney stress block) of a T-section falls within the flange (or slab), the section is treated as a rectangular beam with a width equal to the effective width of the slab.

If the depth of the effective stress block of a T-section falls below the flange, the section is treated as a T-section and the computation of the nominal moment capacity is a little different than a rectangular beam (shown in later section).

9.3 -- Effective Flange Width

The compressive stress in the flange of a T-beam varies as a distance from the web, as shown in Figure 9.3.1 pp. 330. This width, known as the effective width bE, has been simplified by the ACI codes as the smallest of the following per ACI 8.10.2:

1. bE = L/4 (L = beam span length per ACI 8.7)2. bE = bw + 16t (bw = beam web width & t = slab thickness)

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3. bE = bw + 1/2 clear distance to next beam on each side

For L-sections the effective width is the smaller of the following per ACI 8.10.3:

1. bE = bw + L/122. bE = bw + 6t 3. bE = bw + 1/2 clear distance to next beam

For isolated T-sections the effective width is as follows per ACI 8.10.4:

1. bE <= 4bw2. t >= 1/2bw

Transverse reinforcement shall be placed in the top of the slab and be capable of carrying the factored loads on the effective width as if it were a cantilever (spaced at 5t nor 18''). Minimum as previously specified or based on stem width.

9.4 -- Investigation of T-Section in Bending - Strength Method

As stated previously, if the depth of the effective stress block of a T-section falls within the flange the section is treated as a rectangular beam with a width equal to the effective width of the slab. This is illustrated in Case 1: a <= t [Figure 9.4.1(a)] pp. 333. The computation for moment is as follows (very similar to a normal rectangular beam):

M C da

T da

n ( ) ( )2 2

C f b ac E0 85. '

T A fs y

aA f

f bs y

c E

0 85. '

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If the depth of the effective stress block of a T-section falls below the flange , the section is treated as a T-section and the compression area of the concrete is divided into two areas. This is illustrated in Case 2: a > t [Figure 9.4.1(b)] pp. 333. First the portion that has a depth of a and width bw, and second the portion that has a depth t and a width bE - bw. The computation for moment is as follows (somewhat similar to a normal rectangular beam):

M C da

C dt

n 1 22 2( ) ( )

C f b ac w1 0 85 . '

C f b b tc E w2 0 85 . '

T A fs y

aT C

f bc w

2

0 85. '

a

A f f b b t

f bs y c E w

c E

0 85

0 85

. '

. '

Study example 9.4.1 pp. 334 - 335 for effective width and Case 1 and example 9.4.2 pp. 335 - 336 for Case 2.

9.5 -- Maximum Tension Reinforcement Permitted in T-Sections

The literal wording of ACI 10.3.3 (max <= 0.75b) will not be identical to a maximum x = 0.75 xb for non-rectangular sections. Example 9.5.1 pp. 336 - 337 show the following four items:

1. (a) Location of xb by strain compatibility2. (b) Calculation of Asb from two couple approach (a>t)

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3. (c) Calculation of max = 0.75b =Asb/bEd4. Ductility comparison of x/xb fin real x with 0.75Asb

9.6 – T-sections According to ACI-Appendix B Unified Procedure

We will skip this sections.

9.7 -- Design of T-Sections in Bending - Strength Method

The actual design, like investigation, of T-section can fall into two cases. In example 9.7.1 pp. 340 - 341 the required moment is greater than that of just the concrete flange area in compression and the steel in tension. This leads to the solution of a quadratic equation to determine the depth of the stress block, a (Case 2). In example 9.7.2 pp. 341 - 342 the required moment is less than that of just the concrete flange area in compression and the steel in tension. This leads to a solution just like a rectangular beam to determine the depth of the stress block, a (Case 1). Assume that the flange is just in full compression as follows (USE TABLES):

M C dt

n ( )2

C f b tc E0 85. '

If Mu<Mn solve as rectangular with b=bEIf Mu>Mn solve as 2 couple as follows (quadratic)

A b aw1

A b b tE w2

M f A da

f A dt

n c c 0 852

0 8521 2. ' ( ) . ' ( )

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Once the quadratic is solved for a, all the forces and area of steel can be determined. The following is a design procedure using the tables in the library (ACI SP 17):

First assume the neutral axis is in the flange (a<t).

FLEXURE 2.1 - 2.4 with Kn find and a/dif a/d<hf/d=t/d then N.A. is in flange (a<t)

A b ds E

Case with neutral axis in the web (a/d>hf/d=t/d) from above (a>t)FLEXURE 3.1 - 3.2 with hf/d=t/d find jf & anf

FLEXURE 3.3 with b/bw find Knf

Fb dw

2

12000

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FLEXURE 2.1 - 2.4 with Kn find

A b dsw w

A A As sw sf

A

b ds

E

FLEXURE 10.2.1 - 10.2.2 with b/bw and hf/d=t/d find maxcheck that <max

9.8 -- Investigation of T-Sections in Bending - Working Stress Method

T-sections must be investigated by the working stress method for crack control and deflections. Crack control can be done very similar to a rectangular beam (working stress) or just use a working stress in the steel of 60 % yield. Deflection computations only require the same knowledge of strength of material as rectangular beam using the transformed cracked section moment of inertia. This is shown in example 9.8.1 pp. 343 - 344. It is interesting to note that the exclusion of the concrete area in the web below the flange has little effect on the solution.

Problem # O’H 9 – Completely design the continuous beam of problem # O’H 7 for flexure and shear. Assume the beam is interior (beams on either side) and the following data (similar to old homework #7):

Materials Beam Size Other

Fy = 60 ksi (flexure) t = L/80 (up to nearest ½“) cc = 2” (flexure)Fy = 40 ksi (shear) h = L/16 (up to nearest in.) #3 U-stirrupsf’c = 4 ksi bw = hcol + 2” ACI 12.2.2 (no stir.)Interior exposure d = h -2.5” s = L/4

ACI shear and moment diagrams shown below from problem #O’H 7

128


Span data calculations are as follows:

h = L/16 = (42ft*12in/ft)/16 = 31.5 in. Round to h = 32 in.d = h – 2.5 = (32 – 2.5)in = 29.5 in.bw = hcol + 2 = (18 + 2)in = 20 in.t = L/80 = (42ft*12in/ft)/80 = 6.3 in. Round to t = 6.5 in.s = L/4 = 42ft/4 = 10.5 ft = 126in

For negative moments beam is rectangular and using FLEXURE 2.2 for As and REINFORCEMENT 14 for bar selection. Check = 0.9 (it is).

F = bwd2/12000 = 20*(29.5)2/12000 = 1.450 min=0.00333 (all values do exceed this)

Exterior 1st Interior 2nd InteriorMu 319.8 511.8 465.2

Mn = Mu/ 355.3 568.7 516.9Kn = Mn/F 245.0 392.2 356.4

(FLEX 2.2) 0.00424 0.00697 0.00629As = bwd 2.50 4.11 3.71

Use (REINF 14) 3#9 3#11 3#10

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As (REINF 14) 3.00 4.68 3.81 = As/bwd 0.005085 0.00793 0.00646

Kn (FLEX 2.2) 291.4 442.5 365.5Mn = KnF 380.3 577.4 477.0

For positive moments beam will be assumed rectangular with b=bE (check a/d<t/d=6.5/29.5=0.2203) and using FLEXURE 2.2 for As and REINFORCEMENT 14 for bar selection.

bE = Minimum of the following (124 in.):

1. bE = L/4 = 42ft/4 = 10.5 ft. = 126 in.2. bE = bw + 16t = (20 + 16*6.5)in = 124 in. 3. bE = s = 126 in. (bw + 1/2 clear distance to next beam on each side)

F = bEd2/12000 = 124*(29.5)2/12000 = 8.993min=0.00333 <w=As/(bwd) (all values exceed this & check =0.9 it is).

Exterior InteriorMu 365.5 319.8

Mn = Mu/ 406.1 355.3Kn = Mn/F 45.2 39.5

(FLEX 2.2) 0.00076 0.00066a/d 0.0134<0.2203 0.0117<0.2203

As = bEd 2.78 2.41<w=As/(bwd) 0.00471 0.00408

Use (REINF 14) 3#9 3#9As (REINF 14) 3.00 3.00

= As/bEd 0.00082 0.00082Kn (FLEX 2.2) 48.8 48.8Mn = KnF 395.0 395.0

Development length calculations using ACI-12.2.2 worst case (3#11)

sc = (20-2(2)-3(1.41))/2 = 5.882 in > 2db = 2.82 in (CASE 1 for all)

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From REINFORCEMENT 17.1 (Ld)

#9 bottom 53.51 (use 54”) top 1.3*53.51 = 69.56 (use 70”)#10 top 1.3*60.24 = 78.31 (use 79”)#11 top 1.3*66.88 = 86.94 (use 87”)

Run top bars beyond point of zero moment by max(12db,d)Max (12*1.41,29.5) in = 29.5 in

From moment diagrams the point of zero moments are as follows:

Mextleft 79.8 in cut at 109.3 9’-2”Mextright 116.2 in cut at 145.7 12’2”Mint 116.1 in cut at 145.6 12’-2”

Moment bounds below with flexural reinforcing shown.

131


132


Thursday, April 21st

Shear calculations


Exterior leftExterior right

Interior

Find the critical shear at d from the support face (29.5 in = 2.458 ft)


Interior



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Since Vnmin>Vd, only need minimum at critical location

Location where 14 in. spacing is no longer needed (no stirrups)


Interior

From 3 in from face to these distances use 14 in spacing

1@3,11@14in. left end and 1@3,11@14in. right end Exterior span 1@3,[email protected] each Interior span

The final design is shown below with flexural and shear reinforcing, also shown is the shear capacity vs. required strength diagram.

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T-Beam Transformed Cracked Moment of Inertia

Assume that the location of the neutral axis is in the flange (x<t). The moment of the areas of the transformed cracked section about the neutral axis is as follows:

Solve above location for the position of the neutral axis. Then check that the assumption is correct (x<t). Do one on the following based on assumption check.

If (x<t), the transformed cracked moment of inertia is then as follows:

If (x>t), then the assumption was wrong and the location of the neutral is as follows (ignore web):


Or more exactly if (x>t), then the assumption was wrong and the location of the neutral is as follows (includes web):


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Problem # O’HNO 9 – Calculate the Transformed Cracked Moment of Inertia for the positive moment region of the beam in problem # O’H 9.

Required data from previous problem as follows:

d = 29.5 in bw = 20 in t = 6.5 in bE = 124infy = 60 ksi f’c = 4 ksi (n=8) As = 3.00 in2 (3-#9’s)

Assume rectangular (x<t)

Assumption is correct (x<t)

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Friday, April 22nd

Homework #31 is due today (Chapter 8). Homework # 32 is due Monday, April 25th (Chapter 9). Homework # 33 is due Wednesday, April 27th (Chapter 10).

Chapter 10 -- Continuous Slab-Beam-Girder and Concrete Joist Floor Systems

10.1 -- Introduction

In most case the design of slabs and beams can be performed with only the analysis for the ACI moment coefficients since they are uniformly loaded and usually meet the other criteria of loading, span, etc. The deign of girders usually cannot be done this way because they normally have concentrated loads for the beam. For preliminary design of girder the statical moment due to the load could be factored in a similar manner to the ACI moment coefficients.

10.2 -- Size of Beam Web

The size of the beam web for continuous T-shaped sections is usually controlled by flexure and shear at the exterior face of the first interior support for equal spans. The depth should meet the requirements of ACI Tables 9.5(a) and 9.5(b).

Mw l

orw l

uu n u n

2 2

10 9 (2 span)

Mw l

orw l

uu n u n

2 2

14 11 (unrestrained)

Vw l

uu n1 152

. (0.6wl)

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The amount of negative flexural reinforcement should be limited to less than max, but this would give a value of less than 0.9. Therefore we will used the value of at the =0.9 limit or t=0.005. The amount of positive flexural reinforcement should be limited to less than 0.5max. For simplicity we will use the value corresponding to t=0.01, this is approximately the same. These new values have been added to the FLEXURE tables. The amount of shear reinforcement should be limited to 3 times the nominal concrete strength.

From the previous criteria a selection can be made on the depth, width, and approximate reinforcement in the beam as shown in example 10.2.1 pp. 349 - 351.

10.3 -- Continuous Frame Analysis for Beams

This will be skipped due to lack of academic background in analysis.

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Monday, April 25th

Homework #32 is due today (Chapter 9). Homework # 33 is due Wednesday, April 27th (Chapter 10).

10.4-- Choice of Longitudinal Reinforcement in Beams

The choice or reinforcement requires both the areas of steel and the development lengths. Example 10.4.1 pp. 357 - 364 is a rather elaborate flexure problem.Homework #35 is due today (Chapter 9). Homework #36 is due Friday, December 5th (Chapter 10).

10.5 -- Shear Reinforcement in Beams

The complete design or shear reinforcement is given in example 10.5.1 pp. 364 - 366.

10.6 -- Details of Bars in Beams

For typical conditions of equal spans and uniform loads where the ACI moment coefficients apply, standard details such as those provided by the ACI Detailing Manual may be used (similar to slabs on pp. 325). The library has a copy of this reference. Example 10.6.1 pp. 367 - 372 does an extensive illustration of the design.

10.7 -- Size of Girder Web

As discussed earlier, the preliminary design of girders can be performed with the statical moment due to the load factored in a similar manner to the ACI moment coefficients. The depth from ACI Tables 9.5(a) and 9.5(b) should be increased 25 - 50 %. This can be summarized as follows:

MM

orM

us s8

10

8

9 (2 span)

MM

orM

us s8

14

8

11 (unrestrained)

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V Vu s1 15. (1.2V)

Just as with beams, the amount of negative flexural reinforcement should be limited to less than max, but this would give a value of less than 0.9. Therefore we will used the value of at the =0.9 limit or t=0.005. The amount of positive flexural reinforcement should be limited to less than 0.5max. For simplicity we will use the value corresponding to t=0.01, this is approximately the same. These new values have been added to the FLEXURE tables. The amount of shear reinforcement should be limited to 3 times the nominal concrete strength

Design Summary for One-Way Reinforced Concrete Systems

f'c = 4 ksi, fy = 60 ksi (revise Rn for other strengths)I. Choose column layout (rectangular, if possible, L/W = 1.25 to 1.5)II. Choose span direction for beams

A. Long direction for overall uniform structural depthB. Short direction for minimum depth under maximum area

III. Choose beam spacing (L/4 for optimum T-beam with uniform spacing)IV. Determine slab thickness (Table 9.5(a) L/24 or L/28 interior)

A. Use center-to-center spacing of beams for spanB. Revise, if needed, per Table 9.5(b) for construction likely to be

damaged by large deflections as follows:h h 2 33

V. Determine beam size for slab and superimposed loadingA. Determine beam thickness (Table 9.5(a) L/18.5 or L/21 int.)

1. Use center-to-center- spacing of girders for span2. Revise, if needed, per Table 9.5(b)

B. Positive reinforcement at approx. t = 0.011. +Mu = wuL2/14 (wuL2/16 int)

2. Rn = Kn = 601.5 = Mu/bEd2 (FLEXURE 2.1-2.4 )

C. Negative reinforcement at approx. t = 0.005

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1. -Mu = wuL2/10 (wuL2/11 int)

2. Rn = Kn = 911.0 = Mu/bwd2 (FLEXURE 2.1-2.4 )

D. Shear strength at approximately 3Vc = Vn1. Vu = 1.15wuL/2 (wuL/2 int)2. Vu = Vn = 3Vc = 6 Square root (f'c)bwd

E. Choose beam width based on C & D above and check flange width based on B.

VI. Determine girder size for loading from static moment (Ms) due to beam reactions

A. Determine girder thickness (Table 9.5(a) L/18.5 or L/21 int.)1. Use center-to-center spacing of columns for span2. Revise, if needed, per Table 9.5(b)3. For girders spanning short direction

a. Increase 25% to 75% for L/W = 1 to 24. For girders spanning long direction

a. Increase 25% to 0% for L/W = 1 to 2B. Positive reinforcement at approx. t = 0.01

1. +Mu = 8Ms/14 (8 Ms /16 int)

2. Rn = Kn = 601.5 = Mu/bEd2 (FLEXURE 2.1-2.4 )

C. Negative reinforcement at approx. t = 0.0051. -Mu = 8Ms/10 (8 Ms /11 int)

2. Rn = Kn = 911.0 = Mu/bwd2 (FLEXURE 2.1-2.4 )

D. Shear strength at approximately 3Vc = Vn1. Vu = 1.15Vs (Vs int)2. Vu = Vn = 3Vc = 6 Square root (f'c)bwd

E. Choose beam width based on C & D above and check flange width based on B.

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Problem # O’H 10 – Perform two preliminary designs for a reinforced concrete one-way slab/beam/girder system with 32 ft by 48 ft bays. One design has the beams spanning the long direction and the other has the beams spanning the short direction. The system is not supporting construction likely to be damage by large deflection. All sizes should be rounded up to the nearest inch with the exception of the slab thickness (round up to nearest ½ inch). Number of spaces should be rounded up to the nearest whole number. Use the following data for design:

Materials Superimposed loading OtherFy = 60 ksi (flexure) wl = 100 psf h = d + 2.5” (beam/girder)f’c = 4 ksi wd = 25 psf h = d + 1.25” (slab)

I. Choose column layout (rectangular, if possible, L/W = 1.25 to 1.5)

Given: L = 48 ft & W = 32 ft (L/W=48/32=1.50)II. Choose span direction for beams

A. Long direction for overall uniform structural depthWill show this in all calculation

B. Short direction for minimum depth under maximum areaWill not show calculations, just solutions

III. Choose beam spacing (L/4 for optimum T-beam with uniform spacing)L/4=48 ft/4=12ft N=32 ft/12 ft=2.67 spaces Use 3 S=32 ft/3 spaces=10.67ft 3 spaces @ 10’-8’’ (6@8’-0’’)IV. Determine slab thickness (Table 9.5(a) L/24)A. Use center-to-center spacing of beams for span

L=10’-8’’ h=10.67 ft (12 in/ft)/24=5.33 in tslab=5.5ind=h-1.25 in=5.5 in–1.25 in=4.25 in (tslab=4.0 &d=2.75 in)

B. Revise, if needed, per Table 9.5(b) for construction likely to be damaged by large deflections as follows:

h h 2 33

Not needed: supporting construction not likely to be damagedV. Determine beam size for slab and superimposed loading

wu=1.2(25 psf +5.5 in(150 pcf)/(12 in/ft)+1.6(100 psf)=272.5 psfwu=272.5 lb/ft2(10.67 ft)=2906.7 plf (250 psf & 2000 plf)

142


A. Determine beam thickness (Table 9.5(a) L/18.5)1. Use center-to-center- spacing of girders for span

L=48 ft h=48 ft(12 in/ft)/18.5=31.14 in hbeam=32ind=h-2.5 in=32 in-2.5 in=29.5 in (hbm=21 & d=18.5 in)

2. Revise, if needed, per Table 9.5(b)Not needed: same reason as IV.B

B. Positive reinforcement at approx. t = 0.011. +Mu = wuL2/14

+Mu=(2.9067klf)(48 ft)2/14=478.4 k-ft

2. Rn = Kn = 601.5 = Mu/bEd2

601.5=478.4 k-ft/[(0.9)(bE)(29.5 in)2/12000]bE=12.18 in (9.47 in)

C. Negative reinforcement at approx. t = 0.0051. -Mu = wuL2/10

+Mu=(2.9067 klf)(48 ft)2/10=669.7 k-ft

2. Rn = Kn = 911.0 = Mu/bwd2

911.0=669.7 k-ft/[(0.9)(bw)(29.5 in)2/12000]bw=11.26 in (8.76 in)

D. Shear strength at approximately 3Vc = Vn1. Vu = 1.15wuL/2

Vu=1.15(2.9067 klf)(48 ft)/2=80.22 k2. Vu = Vn = 3Vc = 6 Square root (f'c)bwd

80.22 k(1000 lb/k)=0.75(6)[(4000)1/2bw(29.5 in)bw=9.55 in (6.99 in)

E. Choose beam width based on C & D above and check flange width based on B.bW=max(11.26,9.55)=11.26 bw=12in (bw=9)bE=min(L/4,bw+16t,S)=100 in (73 in) Okay

VI. Determine girder size for loading from static moment (Ms) due to beam reactionswu=2.9067 plf+1.2(150 pcf)(32 in-5.5 in)(12 in)/(144 in2/ft2)=3304.2 plfPu=wuL=3304.2 plf(48ft)=158602 lb=158.6 k (70.12 k)

143


For two concentric point loads M=Pa & V=PMs=(158.6 k)(10.67 ft)=1692 k-ft (2524 k-ft)Vs=(158.6 k)=158.6 k (210.36 k)

A. Determine girder thickness (Table 9.5(a) L/18.5 or L/21 int.)1. Use center-to-center spacing of columns for span

L=32 ft h=32 ft(12 in/ft)/18.5=20.76 in (31.14)2. Revise, if needed, per Table 9.5(b)

Not needed: same reason as IV.B3. For girders spanning short directiona. Increase 25% to 75% for L/W = 1 to 2

L/W=1.5 increase 50%h=1.5(20.76 in)=31.14 in hgird=32ind=h-2.5 in=32 in-2.5 in=29.5 in

4. For girders spanning long directiona. Increase 25% to 0% for L/W = 1 to 2

(L/W=1.5 increase 12.5%) (hgird=36 & d=33.5)B. Positive reinforcement at approx. t = 0.011. +Mu = 8Ms/14

+Mu=8(1692 k-ft)/14=966.9 k-ft

2. Rn = Kn = 601.5 = Mu/bEd2

601.5=966.9 k-ft/[(0.9)(bE)(29.5)2/12000]bE=24.62 in (28.49 in)

C. Negative reinforcement at approx. t = 0.0051. -Mu = 8Ms/10

-Mu = 8(1692 k-ft)/10=1354 k-ft

2. Rn = Kn = 911.0 = Mu/bwd2

911.0=1354 k-ft/[(0.9)(bw)(29.5)2/12000]bw=22.76 in (26.34 in)

D. Shear strength at approximately 3Vc = Vn1. Vu = 1.15Vs

Vu=1.15(158.6 k)=182.4 k2. Vu = Vn = 3Vc = 6 Square root (f'c)bwd

182.4 k(1000 lb/k)=0.75(6)[(4000)1/2bw(29.5 in)

144


bw=21.72 in (21.14 in)E. Choose beam width based on C & D above and check flange

width based on B.bW=max(22.76,21.72)=22.76 bw=23in (bw=27)bE=min(L/4,bw+16t,S)=96 in (91 in) Okay

Summary of size

Slab thickness 5.5 in. (4 in.)48’ long beams 12X32 with 3 spaces @ 10’-8’’ (32’-9X21w/6@8’-0’’)32’ long girders 23X32 (48’-27X36)

10.8 -- Continuous Frame Analysis for Girders

This will be skipped due to lack of academic background in analysis.

145


Tuesday, April 26th

10.9 -- Choice of Longitudinal Reinforcement in Girders

The typical conditions of equal spans and uniform loads where the ACI moment coefficients apply usually do not apply the girders. The reinforcement must be calculated and detailed by mechanics and ACI code. Example 10.9.1 pp. 379 - 386 does an complete illustration of the design.

10.10 -- One-Way Joist Floor Construction

One-way concrete joist floor construction, sometimes called "ribbed slab construction" consist of regularly spaced ribs monolithically built with a top floor slab and arranged to span in one direction (could be two-way or waffle slab).

Normally removable and reusable form fillers (pans) are used in the spaces between the joist. The standard pan widths are 20 & 30 in. and depths of 6, 8, 10, 12, 14, 16, & 20 in. The sides of the pan are taper at a slope of 1/12 (Figure 10.10.1 pp. 387). Common slab thicknesses are 3 & 4.5 in. Also tapered end pans can be used to increase shear resistance by increasing the end width 4 & 5 in for 20 & 30 in pans tapered to the normal width 3 ft away from the end (Figure 10.11.1 pp. 390).

The code requirements for joist construction are as follows per ACI 8.11:

1. Rib width not less than 4 in. and depth not more than 3.5 times the minimum rib width per ACI 8.11.2.

2. Clear spacing between rib not to exceed 30 in per ACI 8.11.33. For removable forms, slab thickness not less than 2 in. nor 1/12 the

clear distance between ribs per ACI 8.11.6.14. Reinforce slab for flexure but not less than 7.12 (shrinkage &

temperature) per ACI 8.11.6.2146


5. 10 percent increase in shear strength (code Chapter 11) per ACI 8.11.8.

10.11 -- Design of Concrete Joist Floors

The design of concrete joist floor system involves the slab, joist, and girders. All three components usually can be analysis using the ACI moment coefficients. Example 10.11.1 pp. 387 - 392 show some important steps in design as follows:

1. Slab normally designed as a fixed end member (M=wl2/12 and plain concrete strength (modulus of rupture 0.65*5*square root (f'c)).

2. Welded wire fabric (Table 10.11.1 pp. 389) is normally used in the slab with more steel perpendicular to joist (closer spacing) and less steel parallel to the joist (farther spacing).

3. Depth per ACI Table 9.5(a) with any modifications necessary.4. Only 3/4 in clear cover required as minimum per ACI 7.7.1.5. 10 percent increase in shear strength per ACI 8.11.8 and tapered end

pans (Figure 10.11.1 pp. 390) normally allows for no shear reinforcement (when use normally ladder stirrups).

6. Average width used for weight calculation, but minimum width for shear and flexure (-M and bE for +M).

7. Truss bars still in common use in joist construction.

The following example needs to be revised for ACI 318-02.

Problem # O’HNO 10 – Completely design the a interior span of a continuous joist using 1/5 the values (1/5 of the loads, shears and moments) of old homework #6 – group #1 for flexure and shear (Do old homework #8 – problem #2 – group #1). The supporting girder width is equal to the width of the column, the joist are spaced at 3’-0” on center and the slab is 4.5” thick. The joist height is equal to L/26 rounded up, such that h = the nearest available pan depth + the 4.5” slab (Verify all sizes). Use only two bars (each) for the positive and negative flexural reinforcement and tapered end pans if required. Use the additional following data (all other data is the same as the previously mentioned problem):

147


Materials Joist Size OtherFy = 60 ksi (flexure) h = d +1.5” cc = 1” (flexure)f’c = 4 ksi Use no stirrups

The following shows 1/5 the values of ACI shear and moment diagrams from homework #6 – group #1:

Span data calculations are as follows:

h = L/26 = (22*12in/ft)/26 = 10.15 in.hpan = 10.15in - 4.5in = 5.65 Round to hpan = 6 in. h = 10.5ind = h – 1.5 = (10.5 – 1.5)in = 9 in.bw = 6 in. (given at joist bottom)tslab = 4.5 in. (given)s = 36in. (given)wd = 0.20 k/ft wl = 0.40 k/ft wu = 0.96 k/ft (given)

148


Beam web widths 6 in at bottom (–M and V conservative)bw+2*(1/12)hpan= 7 in at top

6.5 in average (V and weight calculations)bE 36 in effective (+M)

Will work this problem like a true design (nothing assumed)

Slab design ln = 36in – 7in = 29in

wu = 0.96 k/ft given on joist (use 1/3 or 1 ft/3 ft for slab)

wu = 0.32 k/ft (for slab)

Mu = wuln2/12 = 0.32 k/ft (29 in/12 in/ft)2/12 = 0.1557 k-ft/ft

Mn = fr(bh2/6) = 0.65[5(f’c)½](bh2/6) > Mu

h = {[6Mu]/[(0.65)(5)(b)(f’c)½]}½ = [6(0.1557k-ft/ft)(12000lb-in/k-ft)]/[(0.65)(5)(12in)(4000psi)½]}½

h = 2.132 in < 4.5 in (OKAY)

Atemp = 0.0018Ag = 0.0018 (4.5in)(12)in/ft = 0.0972 in2/ft

Use 4 X 12 – W3.5 X W2 WWF (Table 10.11.1 pp. 389)OR #3 @ 13.5 in o.c. (REINFORCEMENT 15)

Joist design ln = 21.0833 ft wu = 0.96 k/ft d = 9 in

For negative moment joist is rectangular (bw = 6 in) and using FLEXURE 2.2 for As and REINFORCEMENT 14 (conservative) for bar selection.

F = bwd2/12000 = 6*(9)2/12000 = 0.04050 min=0.00333

149


InteriorMu 38.79

Mn = Mu/ 43.10Kn = Mn/F 1064.2

(FLEX 2.2) 0.02203As = bwd 1.19

Use (REINF 14) 2#7As (REINF 14) 1.2

= As/bwd 0.02222Kn (FLEX 2.2) 1071.0Mn = KnF 39.04

For positive moments beam will be assumed rectangular with b=bE (check a/d<t/d=4.5/9=0.5000) and using FLEXURE 2.2 for As and REINFORCEMENT 14 for bar selection.

bE = Minimum of the following (36 in.):

3. bE = L/4 = 22ft/4 = 5.5 ft. = 66 in.4. bE = bw + 16t = (7 + 16*4.5)in = 79 in.

3. bE = s = 36 in. (bw + 1/2 clear distance to next beam on each side)

F = bEd2/12000 = 36*(9)2/12000 = 0.2430min=0.00333 <w=As/(bwd)

InteriorMu 26.67

Mn = Mu/ 29.63Kn = Mn/F 121.9

(FLEX 2.2) 0.00207a/d 0.0366<0.50003

As = bEd 0.68<w=As/(bwd) 0.01259

Use (REINF 14) 2#6As (REINF 14) 0.88

150


= As/bEd 0.00272Kn (FLEX 2.2) 158.8Mn = KnF 34.74

Development length calculations using ACI-12.2.2 (bottom 2#6)

sc = (6-2(1)-2(0.75)) = 2.5 in > 2db = 1.5 in (CASE 1 for all)

From REINFORCEMENT 17.1 (Ld)

#6 bottom 28.46#7 top 1.3*41.50 = 53.95

Run top bars beyond point of zero moment by max(12db,d)=(12(0.875),9))=10.5 in

From moment diagrams the point of zero moments are as follows:

Mint 60.44 + 10.5 = 70.94 in < ld

Moment bounds could be now drawn (skip it).

Shear calculations


Interior


Since Vmax > Vc used tapered end pans.

151


Find location where normal width will work

The final design shown be shown below with flexural reinforcing and tapered end pans, also show the shear capacity vs. required strength diagram (skip it).

10.12-- Redistribution of Moments - Introduction to Limit Analysis

ACI 8.4 allows for redistribution on negative moment in continuous nonprestressed flexural members analyzed by elastic theory. This due to the fact that as the ultimate load state is reached the ends of the member must first yield, then the center of the span must yield for failure to occur. The percentage of the negative moment that can be redistributed to the positive moment is as follows:

20 1

'

b

percent

In addition redistribution can only be made when the section at which moment is reduced is designed so that the following occurs:

152



Chapter 13-- Members in Compression and Bending

Homework #33 is due Today (Chapter 10).

13.1 -- Introduction

Concrete compression members are rarely subjected to pure axial load. Bending moments are induced by the following:

1. End restraint caused by monolithic construction.2. Accidental eccentricity from imperfections in construction.3. Unbalanced floor loading.4. Eccentric loads caused by cranes or other concentrations5. Lateral loading due to wind and seismic

ACI 8.8 provides the following simplifying assumptions:

1. The far ends of a column may be considered fixed in continuity analysis for gravity loads.

2. Maximum bending moment in a column is due to factored loads on a single adjacent span (also include axial loads on roof and all floors above).

3. The loading causing maximum ratio of bending to axial load shall also be considered.

13.2 -- Types of Columns

A column is defined as a member used to primarily support axial compression with a height to least lateral dimension ratio or 3 or greater (shorter members are called pedestals) per ACI 2.1. Reinforced concrete columns are classified according to the manner in which the longitudinal reinforcement is laterally supports as tied (separated ties) or spirally reinforced (continuously wrapped ties). Sketch plans and elevations of each. Composite columns can be made up of structural shapes encased in concrete or visa versa with or without reinforcement.

153


13.3 -- Behavior of Axially Loaded Columns

The nominal strength of an axially loaded column takes the general form of concrete, steel, and spiral strength as follows:

P k f A f A k f An c c c y st s sy sp '

The terms will be derived later but are essentially the same as for flexure (except the spirals).

13.4 -- Safety Provisions

The overload factors that primarily apply are 1.4D + 1.7L, 0.75(1.4D+ 1.7L + 1.7W), and 0.9D + 1.3W. The understrength factors are 0.70 for tied columns and 0.75 for spirally reinforced columns. For combined compression and bending the understrength factor varies from the given values at some value of compression (given later) to 0.9 at pure bending.

13.5 -- Concentrically Loaded Short Columns

The maximum nominal axial strength for a concentrically loaded short column is as follows as you like (statics)

P f A A f Ac g st y st0 085 . '

P A f fg c g y g0 0 85 1 . '

P A f f fg c g y c0 0 85 0 85 . ' . '

13.6 -- Strength Interaction Diagram

When combined axial compression and bending moment act on a member (low slenderness) the strength is governed by the material (pure statics and strength). Depending on the ratio of Moment to Axial load the strain diagram can fall into two distinct categories:

154


1. Compression control region where the concrete reaches a strain or 0.003 prior to tension steel yield strain.

2. Tension controls region where the steel strain is greater than yield when the concrete strain reaches 0.003.

The point at which the concrete strain reaches 0.003 at the same time the steel strain reaches yield is known (of coarse) as the balanced condition. Figure 13.6.1 pp. 459 shows what is known as a typical strength interaction diagram for axial compression and bending moment about one axis (the pure mechanics).

It should be notes that the moment can be represented as the axial force at an eccentricity from the plastic centroid (resultant static system). This eccentricity varies with the ratio bending moment to axial compression from 0 at pure compression to infinity at pure bending.

The reading assignment for Friday, November 2nd is Section 13.13 through 13.17 per the course outline.

13.7 -- Length Effects

The effects of slenderness must be considered in reinforced concrete column design. The slenderness ratio kLu/r must be calculated (k was discussed in steel design) and the moment must be amplified (covered later) if required. Slenderness may be neglect in the following cases per ACI 10.12.2 and 10.13.2 for braced and unbraced systems respectively:

kL

r

M

Mu b

b

34 12 1

2

kL

ru 22

13.8 -- Lateral Ties

ACI 7.10.5 prescribes the following conditions for size and placement of lateral ties (Figure 13.8.1 pp. 463):

155


1. Minimum ties size of #3 for #10 or smaller longitudinal bars and #4 for #11, #14, #18, and bundled longitudinal bars.

2. Maximum spacing not to exceed 16 longitudinal bar diameters, 48 tie bar diameters, or the least column dimension.

3. Ties arranged so that every corner and alternate longitudinal bar shall have lateral support by the corner of a tie (included angle not more than 135O and no bar shall be farther than 6 in. clear on either side from a tied bar.

4. Where the bars are located around the periphery of a circle, a complete circular tie may be used.

The following figure shows the arrangement of ties:

156


157


13.9 -- Spiral Reinforcement and Longitudinal Bar Placement

Spiral reinforcement (unlike ties) provide the column with the ability to absorb considerable deformation prior to failure (spring). The strength of the spiral reinforcement is as follows:

P f A f A f A An sy sp sy s c c g c 2 0 2 0 0 75. . . '

Where s is the ratio of the volume of spiral reinforcement, in one loop, to the volume of the core, over length s. The minimum code provision in ACI 10.9.3 is as follows:

s

g

c

c

sy

A

A

f

f

0 45 1.

'

ACI 7.10.4 prescribes the following conditions for size and placement of spiral reinforcement:

1. Minimum size of 3/8 in (could be smooth).2. Clear spacing at least 1 in., but not more than 3 in.3. Anchorage by 1½ extra turn of the spiral at each end.4. Minimum splices of 48 spiral bar diameters, 12 inches, or welded.5. Minimum clear cover per ACI 7.7.1 of 1½ inch.

The minimum spacing between longitudinal bars in a spirally reinforced column per ACI 7.6.3 (1½ bar diameters or 1½ inches).

13.10 -- Limits on Percentage of Reinforcement

The percentage of total longitudinal reinforcement must be between 1 and 8% per ACI 10.9.1. A lower percentage (down to 0.5%) may be used per ACI 10.8.4 in a reduced section (not less than ½ the total area) can be shown to carry the required loading (1% steel for reduced section).

158


13.11 -- Maximum Strength in Axial Compression - ACI Code

The maximum axial load nominal strength per ACI 10.3.5 is as follows for tied and spirally reinforced concrete columns (really minimum eccentricity requirement):

P P f A A f An c g st y st(max) ( . ) . ( . ) . ' 0 80 0 70 0 80 0 850

P P f A A f An c g st y st(max) ( . ) . ( . ) . ' 0 85 0 75 0 85 0 850

The effect of this maximum is to truncate the strength interaction diagram, which will be shown later when is included. When the slenderness ratio is high enough to require consideration of length effects a minimum eccentricity of (0.6 + 0.03h) is used to evaluate the magnified factored moment (M2b) per ACI 10.12.3.2 (magnification covered later).

13.12 -- Balanced Strain Condition - Rectangular Sections

The balanced strain conditions represents the dividing point between the compression controls and tension controls regions of the strength interaction diagram. The solution for the nominal axial load and bending moment comes directly from static once the neutral axis is located from strain compatibility (similar to just bending). The following is a summary of the process (similar for known strain relationships):

Strain compatibility relationship (find x for other strains)

x db

cu cu y

x x db

cu

b

s

'

Static summation of forces (note signs for other cases and f's if steel compression is not yielded)

159


P C C T f x b A f f A fb c s c b s y c s y 085 0851. ' ' . '

Static summation of moments (note signs for other cases)

M Pe C dx

d C d d d Td f x b dx

d A f f d d d A f db b b cb

s c bb

s y c s y

11

1

20 85

20 85' ' ' ' ' ' ' . ' ' ' ' . ' ' ' ' ' '

It should be noted that the summation of moments is about the plastic centroid. For symmetric bar arrangements (good idea) this would be the same as the elastic centroid (d'' = 0.5(d-d')).

13.13 -- Investigation of Strength in Compression Controls Region - Rec-tangular Sections

In this region the tension steel has not reached yield. The actual strain in the tension steel must be determined from the strain diagram and then the stress/force in the steel can be found (similar to the compression steel in other cases).

13.14 -- Investigation of Strength in Tension Controls Region - Rectangu-lar Sections

In this region the tension steel has reached yield. The actual strain in the tension steel is therefore known and then the stress/force in the steel is simply that at yield (still must find compression steel strain/stress/force as in other cases).

13.15 -- Design for Strength - Region I, Minimum Eccentricity

This region is one in which the design selection can be made primarily on axial compression. Size the column based on 3 to 4 % reinforcement ratio with only axial compression. Check using actual axial compression and bending moment.

160


13.16 -- Design for Strength - Region II, Compression Controls (emin < e < eb)

This region is one in which the design selection should be made based on both axial compression and bending moment. Size the column based on 2 to 3 % reinforcement ratio with only axial compression. Check using actual axial compression and bending moment.

13.17 -- Design for Strength - Region III, Tension Controls (e > eb)

This region is one in which the design selection can be made primarily on bending moment. Size the column based on 1 to 2 % reinforcement ratio with only axial compression (could use beam design equations). Check using actual axial compression and bending moment.

Cover the preliminary design of beam-columns, just for fun.

161


Thursday, April 28th

Chapter 20 – Footings – Chapter 8 - BOWLES

CHAPTER 8 - SPREAD FOOTING DESIGN – (BOWLES)

8-1 Footings - Classification and Purpose

A footing use to carry a single column is called a spread, single, isolated, or spot footing. The function is to spread the load laterally to the soil so that the stress intensity is reduced to a value that the soil can safely carry. Wall or continuous footings serve a similar purpose.

Spread footings with tension reinforcing may be called two-way or one-way, depending on whether the steel used for bending runs both ways or just one direction. They may be either uniform thickness, stepped, or sloped. Pedestals may be provided when a steel or wood column interfaces with a footing.

Footings are designed to resist the full dead load and either all or part of the live load under normal conditions. The normal condition should have the resultant located at the centroid of the footing. In addition, the footing may be required to resist wind or earthquake loads in combination with the normal loads. The resultant would not necessarily be located at the centroid.

8-2 Allowable Soil Pressures in Spread Footing Design

The allowable soil pressure is obtained as the worst case of bearing capacity and settlement. The value will have a suitable safety factor included (2 to 5 for cohesionless soils and 3 to 6 for cohesive soils). Eccentric load or moment is usually not included in the allowable soil pressure. Increases in soil pressure for transient loads should be verified by the geotechnical engineer.

162


8-3 Assumptions Used in Footing Design

Although the stress distribution is not uniform due to rigidity of the footing and soil, the assumption that it is uniform is usually used. Codes that cover footing design included ACI, ASSHTO, and AREA (will use ACI).

8-4 Reinforced Concrete Design - USD

The latest revision of the ACI Standard Building Code Requirements For Reinforced Concrete (ACI 318-95), hereinafter termed the Code, places almost total emphasis on strength-design (USD) methods. Some components of USD that must be included are as follows:

1. Load factors

2. Strength reduction factors

3. Statics USD equations

4. Maximum and minimum steel ratios

5. Reinforcing clearances

6. Tension development lengths

7. Wide beam and two-way shear

8. Connection to column or wall

8-5 Structural Design of Spread Footings

163


Steps in square or rectangular spread footing design with a concentrically loaded column and no moments are:

1. Compute the footing-plan dimensions B x L or B using the allowable soil pressure. A rectangular footing may have a number of satisfactory solutions.

BP

qa

or

2. Convert the allowable soil pressure to an ultimate value for use in USD. Obtain Pu by applying appropriate load factors to the given design loading.

qP

Buu2 or q

P

BLuu

3. Obtain the allowable two-way action shear stress and compute the required effective footing depth.

164


d vq

d vq

c BL cq

cu

cu u2 2

4 2 4

or

d vq

d vq

c B cq

cu

cu u2 2 2

4 2 4

withv fc c4 '

4. If the footing is rectangular, check wide beam shear or compute the required effective depth for wide beam shear (use larger).

d

L c

v

qc

u

2 2

1

withv fc c2 '

5. Compute the required steel for bending and use the same amount each way for square footings. Use the effective d to the intersection of the two bar layers for square footings if d > 12 in. For d < 12 in. and for rectangular footings use the actual d for the two directions. The bending moment is computed at the critical section shown (face). Check the steel ratio verses minimum and maximum steel ratios.

Mq B L c

uu

2 2 2

2

or EL

B

2

1distribution

Mq L B c

uu

2 2 2

2

165


6. Compute the column bearing and use dowels for bearing if the allowable bearing stress is exceeded. Minimum dowels must always be provided. If dowels are required, the length must be adequate for compression bond.

7. Detail the design.

Design of Square Spread Footings

Design a square spread footing using f’c = 3 ksi and fy = 60 ksi on soil with and allowable bearing pressure of 4.5 ksf. An 18-inch square column is concentrically place on the footing with 295 kips of dead load and 295 kips of live load.

Determine footing size from service load and allowable soil pressure

Determine factored soil pressure for reinforced footing design

Determine effective depth of footing for two-way shear

166


Determine effective depth of footing for one-way shear

Determine required flexural reinforcing

since <min increase 1/3

min use or

Select appropriate reinforcing for exterior crack control and 2-in clear cover using Reinforcement 15 and Reinforcement 11 (cracking.xls)

#6@5½ in o.c. or 24#6’s ea. way 11’-6” x 11’-6” x 30 in deep footing

Alternatively #7@8 in o.c. would violate crack control

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Friday, April 29th

Seminar #4 - Concrete

168


Friday, May 6th

Exam #7 from 10:00 to 11:50 a.m. (Chapters 6, 7, 8, 9, &10).

Make-up Exam from 4:00 to 5:50 p.m.

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Documents

Conc Notes S05