Chapter 2, Problem 2.7.7P

-7 The truss A BC Shawn in the figure supports a horizontal load P₁= 300 lb and a vertical load P₂= 9001b. Both bars have a cross-sectional area A = 2.4 in² and are made of steel with E = 30 X 10⁶ psi.

(a) Determine the strain energy U₁of the truss when the load P₁acts alone (P₂= 0).

(b) Determine the strain energy U₂when the load P₂acts alone (P₁= 0).

(c) Determine the strain energy U₃when both loads act simultaneously.

  

To determine

The strain energy of the truss when load $P_1$act alone.

Answer to Problem 2.7.7P

The strain energy of the truss when load $P_1$act alone.

Explanation of Solution

Given Information:

The length of the link BC is $60\text{in}$, area of the cross-section of the link is $2.4{\text{in}}^2$, and the young’s modulus of elasticity is $30\times {10}^6\text{psi}$.

The following figure shows the forces acting at joint $B$:

  

       Figure-(1)

Write the expression for the length of the link AB.

  $L_{AB}=\frac{L_{BC}}{\cos 30°}$....... (I)

Here, the length of the link BC is $L_{BC}$.

Write the equilibrium equation for the horizontal forces.

  $-F_{AB}\cos 30°-F_{BC}+P_1=0$....... (II)

Here, the force applied on the link $AB$is $F_{AB}$, the force applied on the link $BC$is $F_{BC}$, and horizontal force acting at $B$is $P_1$.

Write the equilibrium equation for the vertical forces.

  $F_{AB}\sin 30°-P_2=0$....... (III)

Here, vertical force acting at $B$is $P_2$.

Write the expression for the strain energy of the truss.

  $U=\frac{F_{AB}^2{L}_{AB}}{2AE}+\frac{F_{BC}^2{L}_{BC}}{2AE}$....... (IV)

Here, area of the cross-section of the link is $A$, and the young’s modulus of elasticity is $E$.

Calculation:

Substitute $60\text{in}$for $L_{BC}$in Equation (I).

  $\begin{array}{c}{L}_{AB}=\frac{60\text{in}}{\cos 30°}\\ =69.28\text{in}\end{array}$

Substitute $0\text{lb}$for $P_2$in Equation (III).

  $\begin{array}{l}{F}_{AB}\sin 30°-0\text{lb}=0\\ F_{AB}=0\text{lb}\end{array}$

Substitute $0\text{lb}$for $F_{AB}$, and $300\text{lb}$for $P_1$in Equation (II).

  $\begin{array}{l}-\left(0\text{lb}\right)\cos 30°-F_{BC}+\left(300\text{lb}\right)=0\\ F_{BC}=300\text{lb}\end{array}$

Substitute $300\text{lb}$for $F_{BC}$, $0\text{lb}$for $F_{AB}$, $69.28\text{in}$for $L_{AB}$, $30\times {10}^6\text{psi}$for $E$, and $2.4{\text{in}}^2$for $A$, and $60\text{in}$for $L_{BC}$in Equation (IV).

  $\begin{array}{c}U=\frac{\left(0\text{lb}\right)\left(69.28\text{in}\right)}{2\left(2.4{\text{in}}^2\right)\left(30\times {10}^6\text{psi}\right)}+\frac{{\left(300\text{lb}\right)}^2\left(60\text{in}\right)}{2\left(2.4{\text{in}}^2\right)\left(30\times {10}^6\text{psi}\right)}\\ =\frac{{\left(300\text{lb}\right)}^2\left(60\text{in}\right)}{2\left(2.4{\text{in}}^2\right)\left(30\times {10}^6\text{psi}\right)}\\ =0.0375\text{lb}\cdot \text{in}\end{array}$

Conclusion:

The strain energy of the truss when load $P_1$act alone is $0.0375\text{lb}\cdot \text{in}$.

To determine

The strain energy of the truss when load $P_2$act alone.

Answer to Problem 2.7.7P

The strain energy of the truss when load $P_2$act alone is= $2.57\text{lb}\cdot \text{in}$.

Explanation of Solution

Calculation:

Substitute $900\text{lb}$for $P_2$in Equation (III).

  $\begin{array}{l}{F}_{AB}\sin 30°-900\text{lb}=0\\ F_{AB}=\frac{900\text{lb}}{\sin 30°}\\ F_{AB}=1800\text{lb}\end{array}$

Substitute $1800\text{lb}$for $F_{AB}$, and $0\text{lb}$for $P_1$in Equation (II).

  $\begin{array}{l}-\left(1800\text{lb}\right)\cos 30°-F_{BC}+\left(0\text{lb}\right)=0\\ -\left(1800\text{lb}\right)\cos 30°-F_{BC}=0\\ F_{BC}=-1558.84\text{lb}\end{array}$

Substitute $-1558.84\text{lb}$for $F_{BC}$, $1800\text{lb}$for $F_{AB}$, $69.28\text{in}$for $L_{AB}$, $30\times {10}^6\text{psi}$for $E$, and $2.4{\text{in}}^2$for $A$, and $60\text{in}$for $L_{BC}$in Equation (IV).

  $\begin{array}{c}U=\frac{{\left(1800\text{lb}\right)}^2\left(69.28\text{in}\right)}{2\left(2.4{\text{in}}^2\right)\left(30\times {10}^6\text{psi}\right)}+\frac{{\left(-1558.84\text{lb}\right)}^2\left(60\text{in}\right)}{2\left(2.4{\text{in}}^2\right)\left(30\times {10}^6\text{psi}\right)}\\ =\frac{{\left(1800\text{lb}\right)}^2\left(69.28\text{in}\right)+{\left(-1558.84\text{lb}\right)}^2\left(60\text{in}\right)}{\left(4.8{\text{in}}^2\right)\left(30\times {10}^6\text{psi}\right)}\\ =2.57\text{lb}\cdot \text{in}\end{array}$

Conclusion:

The strain energy of the truss when load $P_2$act alone is = $2.57\text{lb}\cdot \text{in}$.

To determine

The strain energy of the truss when both load acts simultaneously.

Answer to Problem 2.7.7P

The strain energy of the truss when both load acts simultaneously is $2.22\text{lb}\cdot \text{in}$.

Explanation of Solution

Calculation:

Substitute $900\text{lb}$for $P_2$in Equation (III).

  $\begin{array}{l}{F}_{AB}\sin 30°-900\text{lb}=0\\ F_{AB}=\frac{900\text{lb}}{\sin 30°}\\ F_{AB}=1800\text{lb}\end{array}$

Substitute $1800\text{lb}$for $F_{AB}$, and $300\text{lb}$for $P_1$in Equation (II).

  $\begin{array}{l}-\left(1800\text{lb}\right)\cos 30°-F_{BC}+\left(300\text{lb}\right)=0\\ -\left(1800\text{lb}\right)\cos 30°-F_{BC}=-300\text{lb}\\ F_{BC}=-1558.84\text{lb}+300\text{lb}\\ F_{BC}=-1258.84\text{lb}\end{array}$

Substitute $-1258.84\text{lb}$for $F_{BC}$, $1800\text{lb}$for $F_{AB}$, $69.28\text{in}$for $L_{AB}$, $30\times {10}^6\text{psi}$for $E$, and $2.4{\text{in}}^2$for $A$, and $60\text{in}$for $L_{BC}$in Equation (IV).

  $\begin{array}{c}U=\frac{{\left(1800\text{lb}\right)}^2\left(69.28\text{in}\right)}{2\left(2.4{\text{in}}^2\right)\left(30\times {10}^6\text{psi}\right)}+\frac{{\left(-1258.84\text{lb}\right)}^2\left(60\text{in}\right)}{2\left(2.4{\text{in}}^2\right)\left(30\times {10}^6\text{psi}\right)}\\ =\frac{{\left(1800\text{lb}\right)}^2\left(69.28\text{in}\right)+{\left(-1258.84\text{lb}\right)}^2\left(60\text{in}\right)}{\left(4.8{\text{in}}^2\right)\left(30\times {10}^6\text{psi}\right)}\\ \approx 2.22\text{lb}\cdot \text{in}\end{array}$

Conclusion:

The strain energy of the truss when both load acts simultaneously is = $2.22\text{lb}\cdot \text{in}$.