04 Truss- Method of Joints and Sections

Truss: Method of Joints and Sections

Theory of Structure - I

Department of Civil EngineeringUniversity of Engineering and Technology, Taxila, Pakistan

Lecture Outlines

Method of Joints

Zero Force Members

Method of Sections

Method of Joints

If a truss is in equilibrium, then each of its joints must also be in equilibrium

The method of joints consists of satisfying the equilibrium conditions for the forces exerted “on the pin” at each joint of the truss

Truss members are all straight two-force members lying in the same plane The force system acting at each pin is coplanar

and concurrent (intersecting) Rotational or moment equilibrium is automatically

satisfied at the joint, only need to satisfy ∑ Fx = 0, ∑ Fy = 0

Procedure for Analysis

Draw the free-body diagram of a joint having at least one known force and at most two unknown forces (may need to first determine external reactions at the truss supports)

Establish the sense of the unknown forces Always assume the unknown member forces

acting on the joint’s free-body diagram to be in tension (pulling on the “pin”)

Assume what is believed to be the correct sense of an unknown member force

In both cases a negative value indicates that the sense chosen must be reversed

Orient the x and y axes such that the forces can be easily resolved into their x and y components

Apply ∑ Fx = 0 and ∑ Fy = 0 and solve for the unknown member forces and verify their correct sense

Continue to analyze each of the other joints, choosing ones having at most two unknowns and at least one known force

Members in compression “push” on the joint and members in tension “pull” on the joint

Mechanics of Materials and building codes are used to size the members once the forces are known

The Method of Joints

Cy = 500 N

500 NB

Joint B

yFx = 0:+

500 - FBCsin45o = 0FBC = 707.11 N (C)

Fy = 0:+

- FBA + FBCcos45o = 0FBA = 500 N (T)

Ax = 500 N

Ay = 500 N

Cy = 500 N

Ax = 500 N

Ay = 500 N

Joint A

Fx = 0:+

500 - FAC = 0FAC = 500 N (T)

Zero Force Members

Truss analysis using the method of joints is greatly simplified if one is able to determine those members which support no loading (zero-force members)

These zero-force members are used to increase stability of the truss during construction and to provide support if the applied loading is changed

If only two members form a truss joint and no external load or support reaction is applied to the joint, the members must be zero-force members.

If three members form a truss for which two of the members are collinear, the third member is a zero-force member provided no external force or support reaction is applied.

Zero-force members Frequently the analysis can be simplified by identifying

members that carry no load two typical cases are found

When only two members form a non-collinear joint and there is no external force or reaction at that joint, then both members must be zero-force

If either TCB or TCD ≠ 0, then C cannot be in equilibrium, since there is no restoring force towards the right.

Hence both BC and CD are zero-load members here.

When three members form a truss joint for which two members are collinear and the third is at an angle to these, then this third member must be zero-force in the absence of an external force or reaction from a

support

TBCTAB

Here, joint B has only one force in the vertical direction.

Hence, this force must be zero or B would move (provided there are no external loads/reactions)

Also TAB = TBC

While zero-force members can be removed in this configuration, care should be taken any change in the loading can lead to the member carrying a

load the stability of the truss can be degraded by removing the

zero-force memberP

You may think that we can remove AD and BD to make a triangle …

This satisfies the statics requirements

However, this leaves a long CE member to carry a compressive load. This long member is highly susceptible to failure by buckling.

FCB Fx = 0: FCB = 0+

Fy = 0: FCD = 0+

Fy = 0: FABsin = 0, FAB = 0 +

Fx = 0: FAE + 0 = 0, FAE = 0+

Example 3-4

Using the method of joints, indicate all the members of the truss shown in the figure below that have zero force.

SOLUTION

Joint D

Fy = 0:+ FDCsin = 0, FDC = 0

Fx = 0:+ FDE + 0 = 0, FDE = 0

EFEF = 0Fx = 0:+

Joint E

FHAFHB

Joint H

Fy = 0:+ FHB = 0

GGx FGF

FGAJoint G

Fy = 0:+ FGA = 0

Method of Sections

Based on the principle that if a body is in equilibrium, then any part of the body is also in equilibrium

Procedure for analysis Section or “cut” the truss through the members

where the forces are to be determined Before isolating the appropriate section, it may be

necessary to determine the truss’s external reactions (then 3 equs. of equilibrium can be used to solve for unknown member forces in the section).

Draw the free-body diagram of that part of the sectioned truss that has the least number of forces acting on it

Establish the sense of the unknown member forces

Apply 3 equations of equilibrium trying to avoid equations that need to be solved simultaneously Moments should be summed about a point that

lies at the intersection of the lines of action of two unknown forces

If two unknown forces are parallel – sum forces perpendicular to the direction of these unknowns

The Method of Sections

2 m 2 m 2 m

100(2) - FBC(2) = 0FBC = 100 N (T)

Fy = 0:+

-100 + FGCsin45o = 0FGC = 141.42 N (T)

100(4) - FGF(2) = 0FGF = 200 N (C)

Example 3-6

Determine the force in members GF and GD of the truss shown in the figure below. State whether the members are in tension or compression. The reactions at the supports have been calculated.

6 kN 8 kN 2 kN

Ax = 0

Ay = 9 kN Ey = 7 kN

B C DE

3 m 3 m 3 m 3 m

3 m4.5 m

2 kN Ey = 7 kN

Section a-a

6 kN 8 kN 2 kN

Ax = 0

Ay = 9 kN Ey = 7 kN

B C DE

3 m 3 m 3 m 3 m

3 m4.5 m

SOLUTION a

FFGsin26.6o(3.6) + 7(3) = 0, FFG = -17.83 kN (C)

- 7(3) + 2(6) + FDGsin56.3o(6) = 0, FDG = 1.80 kN (C)

Example 3-7

Determine the force in members BC and MC of the K-truss shown in the figure below. State whether the members are in tension or compression. The reactions at the supports have been calculated.

B C D E FG

M N O P

5.34 kN 6.67 kN 8 kN 7.11 kN12.9 kN

4.6 m 4.6 m 4.6 m 4.6 m 4.6 m 4.6 m

B C D E FG

M N O P

5.34 kN 6.67 kN 8 kN 7.11 kN12.9 kN

4.6 m 4.6 m 4.6 m 4.6 m 4.6 m 4.6 m

SOLUTIONa

Section a-a

5.34 kN12.9 kN

FBC(6) - 12.9(4.6) = 0, FBC = 9.89 kN (T)

B C D E FG

M N O P

5.34 kN 6.67 kN 8 kN 7.11 kN12.9 kN

4.6 m 4.6 m 4.6 m 4.6 m 4.6 m 4.6 m

C D E FG

6.67 kN 8 kN 7.11 kN

4.6 m 4.6 m 4.6 m 4.6 m

9.89 kN

+ MK -FCMcos33.1o(6) - 9.89(6) - 8(4.6) + 7.11(18.4) = 0 FCM = 6.90 kN (T)

Practice Problems

Book Structural Analysis by R.C. Hibbeler Chapter no. 03

Examples and exercise questions related to Joint Method Zero Force Members Section Method

04 Truss- Method of Joints and Sections

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