   # Determine the reactions and the force in each member of the truss shown in Fig. P13.55 due to a temperature drop of 25 °C in members AB, BC, and CD and a temperature increase of 60°C in member EF. Use the method of consistent deformations. FIG. P13.55, P13.56

#### Solutions

Chapter
Section
Chapter 13, Problem 55P
Textbook Problem
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## Determine the reactions and the force in each member of the truss shown in Fig. P13.55 due to a temperature drop of 25 °C in members AB, BC, and CD and a temperature increase of 60°C in member EF. Use the method of consistent deformations. FIG. P13.55, P13.56

To determine

Calculate the support reactions and the member forces of the truss using method of consistent deformation.

### Explanation of Solution

Given information:

The temperature changes (ΔT)AB in member AB is 25°C.

The temperature changes (ΔT)BC in member BC is 25°C.

The temperature changes (ΔT)CD in member BC is 25°C.

The temperature changes (ΔT)EF in member EF is 60°C.

The modulus of elasticity of members is 200GPa.

The area (A) of the members is 3,000mm2.

The coefficient of thermal expansion (α) is 1.2(105)/°C.

Apply the sign conventions for calculating reactions, forces and moments using the three equations of equilibrium as shown below.

• For summation of forces along x-direction is equal to zero (Fx=0), consider the forces acting towards right side as positive (+) and the forces acting towards left side as negative ().
• For summation of forces along y-direction is equal to zero (Fy=0), consider the upward force as positive (+) and the downward force as negative ().
• For summation of moment about a point is equal to zero (Matapoint=0), consider the clockwise moment as negative and the counter clockwise moment as positive.

Method of joints:

The negative value of force in any member indicates compression (C) and the positive value of force in any member indicates tension (T).

Condition for zero force members:

1. 1. If only two non-collinear members are connected to a joint that has no external loads or reactions applied to it, then the force in both the members is zero.
2. 2. If three members, two of which are collinear are connected to a joint that has no external loads or reactions applied to it, then the force in non-collinear member is zero.

Calculation:

Find the degree of indeterminacy of the structure using the equation:

i=(m+r)2j

Here, m is the number of members, r is the number of equilibrium equations, and j is the number of joints.

Substitute 10 for m, 3 for r, and 6 for j.

i=(10+3)2(6)=1312=1

Hence, the truss is internally indeterminate to the first degree

Select the member BF as redundant. The primary truss obtained by removing the member BF.

Apply the 1 kN unit tensile member force for member BF.

Consider the notation of member forces as uBF.

Sketch the primary truss subjected to unit tensile force in member BF as shown in Figure 1.

Refer Figure 1.

The reactions of the truss are zero. Hence, member forces AB, AE, CD, and DF are zero.

The arrow towards joint refers compression members and the arrow away from the joint refers tension member.

Find the length of member CE and BF.

LBF=LCE=LBC2+LBE2=82+62=10m

Find the member forces when 1 kN applied for member force of BF using proportion of length.

The force in member BC and EF is same due to symmetry of truss.

Find the force in member BC and EF due to unit force.

uBCLBC=uBFLBFuBC8=110uBC=0.8kN

The force in member BE and CF is same due to symmetry of truss.

Find the force in member BE due to unit force.

uBELBE=uBFLBFuBE6=110uBE=0.6kN

Find the force in member CE due to unit force.

uCELBE=uBFLBFuCE10=110uCE=1kN

The deflection of member BF (ΔBFO) due to member AB, AE, BE, CF, CD, and DF are zero

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