BEAMS

1

BEAMS

SHEAR AND MOMENT

2

Beam Shear

¨ Shear and Moment Diagrams

¨ Vertical shear: tendency for one part of a beam to move vertically with respect to an adjacent part

3

Beam Shear

¨ Magnitude (V) = sum of vertical forces on either side of the section– can be determined at any section along the

length of the beam¨ Upward forces (reactions) = positive¨ Downward forces (loads) = negative

¨ Vertical Shear = reactions – loads(to the left of the section)

4

Beam Shear

¨ Why?¨ necessary to know the maximum value of

the shear– necessary to locate where the shear changes

from positive to negative• where the shear passes through zero

¨ Use of shear diagrams give a graphical representation of vertical shear throughout the length of a beam

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Beam Shear –

¨ Simple beam – Span = 20 feet– 2 concentrated loads

¨ Construct shear diagram

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Beam Shear – Example 1

1) Determine the reactions

Solving equation (3):


Figure 6.7a =>

)'20()'161200()'68000(03

1200800002

01

2..

1

2..

1

RM

RRF

F

lblb

lblby

x

..

..

2

....2

360,320

200,67

200,19000,48'20

lbft

ftlb

ftlbftlb

R

R

.

1

...1

840,5

360,3200,1000,8lb

lblblb

R

R

7

Beam Shear – Example 1 (pg. 64)

¨ Determine the shear at various points along the beam

.)18(

.)8(

.)1(

360,3200,1000,8480,5

160,2000,8480,5

480,50480,5

lbx

lbx

lbx

V

V

V

8


¨ Conclusions– max. vertical shear = 5,840 lb.

• max. vertical shear occurs at greater reaction and equals the greater reaction (for simple spans)

– shear changes sign under 8,000 lb. load• where max. bending occurs

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¨ Simple beam – Span = 20 feet– 1 concentrated load– 1 uniformly distr. load

¨ Construct shear diagram, designate maximum shear, locate where shear passes through zero

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¨ Determine the reactions



)'24()'166000()]'6)('12000,1[(03

000,6)'12000,1(02

01

2..

1

2..

1

RM

RRF

F

lbftlb

lbftlby

x

..

..

2

....2

000,724

000,168

000,96000,72'24

lbft

ftlb

ftlbftlb

R

R

.

1

...1

000,11

000,7000,6000,12lb

lblblb

R

R

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Shear and Moment Diagrams

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.

)24(

.)16(

.)16(

.)12(

.)2(

.)1(

000,7000,6000,112000,11

000,7000,6000,112000,11

000,1)000,112(000,11

000,1)000,112(000,11

000,9)000,12(000,11

000,10)000,11(000,11

lbx

lbx

lbx

lbx

lbx

lbx

V

V

V

V

V

V

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¨ Conclusions– max. vertical shear = 11,000 lb.

• at left reaction– shear passes through zero at some

point between the left end and the end of the distributed load• x = exact location from R1

– at this location, V = 0

feetx

xV

11

)000,1(000,110

14


¨ Simple beam with overhanging ends– Span = 32 feet– 3 concentrated loads– 1 uniformly distr. load

acting over the entire beam

¨ Construct shear diagram, designate maximum shear, locate where shear passes through zero

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)'20(42

32)'32500()'28000,2()'6000,12()'4000,4(04

)'20(8232)'32500()'24000,4()'14000,12()'8000,2(03

)'32500(000,4000,12000,202

01

1.....

2

2.....

1

2...

1.

RM

RM

RRF

F

ftlblblblb

ftlblblblb

ftlblblblby

x

..

..

2

........2

800,1820

000,376

000,128000,96000,168000,16'20

lbft

ftlb

ftlbftlbftlbftlb

R

R

...

1

........1

200,15'20

000,304

000,192000,56000,72000,16'20

lbftlb

ftlbftlbftlbftlb

R

R

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Determine the reactions



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.......

)32(

.......)28(

......)28(

......)22(

.....)22(

.....)8(

....)8(

000,4'32500000,12000,2800,18200,15

000,6'28500000,12000,2800,18200,15

800,12'28500000,12000,2200,15

800,9'22500000,12000,2200,15

200,2'22500000,2200,15

200,9'8500000,2200,15

000,6'8500000,20

lbftlblblblblbx

lbftlblblblblbx

lbftlblblblbx

lbftlblblblbx

lbftlblblbx

lbftlblblbx

lbftlblbx

V

V

V

V

V

V

V

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¨ Conclusions– max. vertical shear = 12,800

lb.• disregard +/- notations

– shear passes through zero at three points • R1, R2, and under the 12,000lb.

load

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Bending Moment¨ Bending moment: tendency of a beam to bend due

to forces acting on it

¨ Magnitude (M) = sum of moments of forces on either side of the section– can be determined at any section along the length of the

beam

¨ Bending Moment = moments of reactions – moments of loads

• (to the left of the section)

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Bending Moment

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Bending Moment – Example 1 ¨ Simple beam

– span = 20 feet– 2 concentrated loads– shear diagram from

earlier

¨ Construct moment diagram

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Bending Moment – Example 1

1) Compute moments at critical locations– under 8,000 lb. load &

1,200 lb. load

....

)'16(

...)'6(

440,13)'10000,8()'16840,5(

040,350)'6840,5(ftlblblb

x

ftlblbx

M

M

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Bending Moment – Example 2¨ Simple beam

– Span = 20 feet– 1 concentrated load– 1 uniformly distr. Load– Shear diagram

¨ Construct moment diagram

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Bending Moment – Example 2

1) Compute moments at critical locations– When x = 11 ft. and under

6,000 lb. load

...

)'16(

...)'11(

000,56421212000,1)'16000,11(

500,6021111000,1)'11000,11(

ftlbftlblbx

ftlbftlblbx

M

M

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Negative Bending Moment

¨ Previously, simple beams subjected to positive bending moments only– moment diagrams on one side of the base line

• concave upward (compression on top)

¨ Overhanging ends create negative moments• concave downward (compression on bottom)

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Negative Bending Moment

¨ deflected shape has inflection point– bending moment = 0

¨ See example

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Negative Bending Moment - Example

– Simple beam with overhanging end on right side

• Span = 20’• Overhang = 6’• Uniformly distributed load

acting over entire span

– Construct the shear and moment diagram

– Figure 6.12

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)'20(226)'26600(03

)'26600(02

01

2.

1

2.

1

RM

RRF

F

ftlb

ftlby

x

1) Determine the reactions

- Solving equation (3):

- Solving equation (4):

..

..

2

..2

140,1020

800,202

800,202'20

lbft

ftlb

ftlb

R

R

.

1

..1

460,5

140,10600,15lb

lblb

R

R

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2) Determine the shear at various points along the beam and draw the shear diagram

.)20(

.)20(

.)10(

.)1(

600,3)60020(140,10460,5

540,6)60020(460,5

540)60010(460,5

860,4)6001(460,5

lbx

lbx

lbx

lbx

V

V

V

V

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3) Determine where the shear is at a maximum and where it crosses zero– max shear occurs at the right

reaction = 6,540 lb.

feetx

xV

1.9

)600(460,50

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4) Determine the moments that the critical shear points found in step 3) and draw the moment diagram

...

)20(

...)1.9(

800,10220'20600)'20460,5(

843,2421.9'1.9600)'1.9460,5(

ftlbftlblbx

ftlbftlblbx

M

M

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4) Find the location of the inflection point (zero moment) and max. bending moment

¨ since x cannot =0, then we use x=18.2’¨ Max. bending moment =24,843 lb.-ft.

feetfeetx

x

xxxxM

xxM

xxxM ftlb

2.18;0600

460,55460

)300(2

)0)(300(4)460,5(460,5

460,5300300460,50

2600460,50

2600)460,5(0

2

22

2

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Rules of Thumb/Review

¨ shear is dependent on the loads and reactions– when a reaction occurs; the shear “jumps” up by the

amount of the reaction– when a load occurs; the shear “jumps” down by the

amount of the load¨ point loads create straight lines on shear diagrams¨ uniformly distributed loads create sloping lines of

shear diagrams

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Rules of Thumb/Review

¨ moment is dependent upon the shear diagram– the area under the shear diagram = change in the

moment (i.e. Ashear diagram = ΔM)

¨ straight lines on shear diagrams create sloping lines on moment diagrams

¨ sloping lines on shear diagrams create curves on moment diagrams

¨ positive shear = increasing slope¨ negative shear = decreasing slope

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Typical Loadings

¨ In beam design, only need to know:– reactions– max. shear– max. bending moment– max. deflection

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BEAMS