## Chapter 10

## Q. 10.5

Assume that all members of the truss shown in Figure 10.21 have same axial rigidity EA. Calculate the total strain energy of the truss. Assume linear elastic behaviour of all members of the truss.

## Step-by-Step

## Verified Solution

We have to determine the forces carried by each member of the truss.

Figure 10.22(a) above shows the free-body diagram of the entire truss and clearly the support reactions are P/2. Figures 10.21(b) and (c) show free-body diagrams of joints A and D, respectively.

Now, from joint A

\sum F_y=0 \Rightarrow \frac{T_{ AC }}{\sqrt{2}}=-\frac{P}{2}

Therefore, T_{ AC }=-\frac{P}{\sqrt{2}}=\frac{P}{\sqrt{2}}(\text { compressive })

and \Sigma F_x=0 \Rightarrow T_{ AD }=-T_{ AC } \cos 45^{\circ}=\frac{P}{2} \text { (tensile) }

Again, from joint D free-body diagram [refer Figure 10.22(c)],

\sum F_x=0 \Rightarrow T_{ BD }=T_{ AD }

Therefore, T_{ BD }=\frac{P}{2}

and \sum F_y=0 \Rightarrow T_{ CD }=P(\text { tensile })

From symmetry, T_{ BC }=\frac{P}{\sqrt{2}} \text { (compressive) }

Thus, the total strain energy of the truss is

U=2\left(U_{ AC }+U_{ AD }\right)+U_{ CD }

This is because members AC and BC are carrying same load and also members AD and BD have identical loading. Therefore,

U=2\left[\frac{(P / \sqrt{2})^2 \sqrt{2} L}{2 A E}+\frac{(P / 2)^2 L}{2 A E}\right]+\frac{P^2 L}{2 A E}

=\frac{P^2 L}{\sqrt{2} A E}+\frac{P^2 L}{4 A E}+\frac{P^2 L}{2 A E}=\frac{P^2 L}{A E}\left\lgroup \frac{3}{4}+\frac{1}{\sqrt{2}}\right\rgroup