COLLEGE OF ENGINEERING DEPARTMENT OF CIVIL & ARCHITECTURAL ENGINEERING

CVEN 220 : ANALYSIS OF STRUCTURES

WAEL I. ALNAHHAL, Ph. D., P. Eng

Spring, 2015

Ch.3: TRUSSES

Trusses in Building

Trusses Types for Building

Truss Bridge

Truss Bridge Types

Classification of coplanar Truss Simple Truss

Compound Truss

Complex Truss

Determinacy For plane truss

If b+r = 2j If b+r > 2j

statically determinate statically indeterminate

Where b = number of bars r = number of external support reaction j = number of joints

Stability For plane truss If b+r < 2j unstable The truss can be statically determinate or indeterminate (b+r >=2j) but unstable in the following cases: External Stability: truss is externally unstable if all of its reactions are concurrent or parallel

Internal stability:

Internally stable Internally unstable

Internally unstable

Classify each of the truss as stable , unstable, statically determinate or indeterminate.

b = 19 r = 3 j = 11

b + r = 22

2 j = 2(11) = 22

Statically determinate & Stable

b = 15 r = 4 j = 9

b + r = 19

2 j = 2(9) = 18

Statically indeterminate & Stable

b = 9 r = 3 j = 6

b + r = 12

2 j = 2(6) = 12

Statically determinate & Stable

b + r = 15 2 j = 2(8) = 16 Unstable

b = 12 r = 3 j = 8

STEPS FOR ANALYSIS 1. If the support reactions are not given, draw a FBD of the entire

truss and determine all the support reactions using the equations of equilibrium.

2. Draw the free-body diagram of a joint with one or two unknowns. Assume that all unknown member forces act in tension (pulling the pin) unless you can determine by inspection that the forces are compression loads.

3. Apply the scalar equations of equilibrium, FX = 0 and FY = 0, to determine the unknown(s). If the answer is positive, then the assumed direction (tension) is correct, otherwise it is in the opposite direction (compression).

4. Repeat steps 2 and 3 at each joint in succession until all the required forces are determined.

The Method of Joints

The Method of Joints

Example 1 Solve the following truss

FAB 8 cos 30 = 0 FAB = 6.93KN T = 0;

4 + FAG sin 30 = 0 FAG = 8KN C = 0;

Fx

Fy

= 0 FGF = 6.50KN C 8 3sin 30 + FGF

= 0; FGB 3cos 30 = 0 FGB = 2.6KN C

= 0;

Fx

Fy

FBC + 2.6 cos 60 + 2.6 cos 60 6.93 = 0

FBC = 4.33KN T

= 0; FBF sin 60 2.6 sin 60 = 0 FBF = 2.6KN T

= 0;

Fx

Fy

Problem Determine the force in each member

Zero Force Member 1- If a joint has only two non-colinear members and there is no external load or support reaction at that joint, then those two members are zero-force members

Zero Force Member 2- If three members form a truss joint for which two of the members are collinear and there is no external load or reaction at that joint, then the third non-collinear member is a zero force member.

Example 2 Find Zero force member of the following truss

Method of Section

The Method of Section

1. Decide how you need to cut the truss. This is based on: a)where you need to determine forces, and, b) where the total number of unknowns does not exceed three (in general).

2. Decide which side of the cut truss will be easier to work with (minimize the number of reactions you have to find).

3. If required, determine the necessary support reactions by drawing the FBD of the entire truss and applying the EofE.

The Method of Section STEPS FOR ANALYSIS

4. Draw the FBD of the selected part of the cut truss. We need to indicate the unknown forces at the cut members. Initially we assume all the members are in tension, as we did when using the method of joints. Upon solving, if the answer is positive, the member is in tension as per our assumption. If the answer is negative, the member must be in compression. (Please note that you can also assume forces to be either tension or compression by inspection as was done in the previous example above.)

5. Apply the equations of equilibrium (EofE) to the selected cut section of the truss to solve for the unknown member forces. Please note that in most cases it is possible to write one equation to solve for one unknown directly.

Example 2 Solve the CF & GC members in the truss

FCF sin 30(12) + 300(6.93) = 0 FCF = 346.4Ib C = 0; M E

M A = 0;

FGC (12) 300(6.93) 346.4 sin 30(12) = 0 FGC = 346.4Ib T

Example 3 Solve the GF & GD members in the truss

= 0 FGD sin 56.3(6) 7(3) + 2(6) = 0 FGF = 1.8kN C

= 0 FGF cos 26.6(3) + 7(3) = 0 FGF = 7.83kN C M D M o

Example 4 Solve the ED & EB members in the truss

Example 5 Solve the BC & MC members in the K-truss

= 0 2900(15) + FBC (20) = 0 FBC = 2175Ib T M L

= 0 FMB = 1200Ib T Fy At Joint B

= 0 2900 1200 +1200 FML FML = 2900Ib T Fy For the total cutting part

= 1532Ib C

= 1532Ib T FMK FMC

F + F = 0 13 MC 13 MK

Fx Fy

= 0 F + F = 0 13 MC 13 MK

= 0 2900 1200

At Joint M

2 2

3 3

Problem 2 Solve All members

Problem 3 Solve All members

Problem 4

Solve members CH , CI

Example 4 Compound Truss Solve the truss

= 0 5(4) + 4(2) + FHG (4 sin 60) = 0 FHG = 3.46kN C M C

Joint A FAB & FAI Joint H FHI & FHJ Joint I FIJ & FIB Joint B FBC & FBJ Joint J FJC

Example 5 Compound Truss

Example 5 Compound Truss Solve the truss

Joint A FAE

Joint B FEB

& FAB

After Solve FAE

Joint A FAF & FAG

Complex Truss

Si = Si + xsi

Force in members

'

x = ? S = S ' + xs = 0 EC EC EC

Complex Trusses

= +

SEC + x sEC =

X x

x = SEC sEC

P

A

B

C

D F

E r + b = 2j,

3 9 2(6)

Determinate Stable

FEC P

A

B

C

D F

E

= FAD

SAD

sEC P

A

B

C

D F

E

Si = S i + x si

i

Si = Si + xsi '

x = ?

S = S ' + xs = 0 EC EC EC

Member S ' si xsi Si AB AC AF FE BE ED FC EC

Example 5 Complex Truss

P

b + r < j unstable truss b + r = 3j

> 3j

statically determinate-check stability

statically indeterminate-check stability b + r

Determinacy and Stability

Space Truss

x, y, z, Force Components

Space Truss

l = x2 + y2 + z 2

F = F ( x) l x

F = F ( y) l y

F = F ( z) l z

F = F 2 + F 2 + F 2 x y z

Support Reactions x z

y

short link

Fy

x

z y

x z

y

roller

x z

Fz

y

x z

slotted roller y constrained

in a cylinder

Fx

x z

Fz

y

x z

y

ball-and -socket

Fx

x

z

Fz

y

Zero Force Member 1- If all but one of the members connected to a joint lie on the same plane, and provided no external load act on the joint, then the member not lying in the plane of the other members must subjected to zero force.

Fz = , FD =

Zero Force Member 2- If it has been determined that all but two of several members connected at a joint support zero force, then the two remaining members must also support zero force, provided they dont lie a long the same line and no external load act on the joint.

Fz = ,

Fy = ,

FB =

FD =

Example 6 Space Truss

Example 7 Space Truss

Copyright 2009 Pearson Prentice Hall Inc.

Slide Number 1Trusses in BuildingTrusses Types for BuildingSlide Number 4Slide Number 5Slide Number 6Slide Number 7Truss BridgeTruss Bridge TypesSlide Number 10Slide Number 11Slide Number 12Slide Number 13Compound TrussComplex TrussDeterminacyStabilityInternal stability:Slide Number 19Slide Number 20Slide Number 21Slide Number 22Slide Number 23Slide Number 24Slide Number 25Slide Number 26Zero Force MemberZero Force MemberSlide Number 29Slide Number 30Slide Number 31Slide Number 32Slide Number 33Slide Number 34Slide Number 35Slide Number 36Slide Number 37Slide Number 38Slide Number 39Slide Number 40Slide Number 41Slide Number 42Slide Number 43Slide Number 44Slide Number 45Problem 4

Solve members CH , CISlide Number 47Slide Number 48Slide Number 49Example 5 Compound TrussSlide Number 51Slide Number 52Slide Number 53Slide Number 54Complex TrussSlide Number 56Slide Number 57Example 5 Complex TrussSlide Number 59Slide Number 60Slide Number 61Slide Number 62Slide Number 63Slide Number 64Zero Force MemberZero Force MemberSlide Number 67Slide Number 68Slide Number 69Slide Number 70Slide Number 71Slide Number 72Space TrussSlide Number 74Slide Number 75Slide Number 76