Chapter 11, Problem 11.3.27P

A space truss is restrained at joints O, A_,B, and C as shown in the figure. Load F is applied at joint A and load 2F acts downward at joint C. Each member is a slender, circular pipe (E = 10,600 ksi) with an outside diameter of 3.5 in. and wall thickness of 0.25 in. Length variable L = 11 ft. Determine the critical value of load variable P (kips) at which member OB fails by Euler buckling.

  

To determine

The critical value of load $P$ at which member $OB$ fails by Euler buckling.

Answer to Problem 11.3.27P

The critical value of load $P$ at which member $OB$ fails by Euler buckling is $\underset{\_}{109415821.2\text{kips}}$ .

Explanation of Solution

Given information:

Young’s modulus of each member is $10600\text{ksi}$ , outside diameter of each member is $3.5\text{in}$ , wall thickness of each member is $0.25\text{in}$ and length variable is $11\text{ft}$ .

Write the expression for force equilibrium at joint $O$ in $x$ -direction.

  $O_x=F_{OB}$

Here, force generated in member $OB$ is $F_{OB}$ and reaction force in $x$ -direction at point $O$ is $O_x$ .

Write the expression for force equilibrium at joint $O$ in $y$ -direction.

  $O_y=F_{OC}$

Here, force generated in member $OC$ is $F_{OC}$ and reaction force in $y$ -direction at point $O$ is $O_y$ .

Write the expression for force equilibrium at joint $O$ in $z$ -direction.

  $O_Z=0$

Here, the reaction force in $z$ direction at point $O$ is $O_z$ .

Write the expression for moment equilibrium equation at point $B$ along $y$ -axis.

  $O_z\times L+\left(A_x\times 0.8L\right)+PL=0$   ......(I)

Here, the length variable of the truss is $L$ , reaction force in $x$ -direction at point $A$ is $A_x$ and load variable of the truss is $P$ .

Substitute $0$ for $O_z$ in equation (I).

  $\begin{array}{l}0\times L+\left(A_x\times 0.8L\right)+PL=0\\ \left(A_x\times 0.8L\right)+PL=0\\ A_x\times 0.8L=PL\\ A_x=\frac{PL}{0.8L}\end{array}$

  $A_x=-1.25P$

Write the expression for moment equilibrium equation at point $C$ along $z$ -axis.

  $\begin{array}{c}{B}_y\times L=0\\ B_y=0\end{array}$

Here, reaction force in $y$ -direction at point $B$ is $B_y$ .

Write the expression for moment equilibrium equation at point $O$ along $y$ -axis.

  $\left(-B_z\times L\right)+\left(0.8L\times A_x\right)=0$   ......(II)

Here, reaction force in $z$ -direction at point $B$ is $B_z$ .

Substitute $-1.25P$ for $A_x$ in equation (II).

  $\begin{array}{l}\left(-B_z\times L\right)+\left(0.8L\times \left(-1.25P\right)\right)=0\\ -B_z\times L=0.8L\times 1.25P\\ B_z=-0.8\times 1.25P\\ B_z=-P\end{array}$

The figure below shows the free body diagram of joint $A$ .

  

  Figure-(1)

Write the expression for force equilibrium at joint $A$ in $x$ -direction.

  $A_x+F_{AB}\cos \left({\theta }_{ABO}\right)=0$   ......(III)

Here, the force generated in member $AB$ is $F_{AB}$ and the angle between member $AB$ and member $OB$ is ${\theta }_{ABO}$ .

Write the expression for the angle between member $AB$ and member $OB$ .

  ${\theta }_{ABO}={\cos }^{-1}\left(\frac{L}{\sqrt{L^2+{\left(0.8L\right)}^2}}\right)$

Substitute ${\cos }^{-1}\left(\frac{L}{\sqrt{L^2+{\left(0.8L\right)}^2}}\right)$ for ${\theta }_{ABO}$ and $-1.25P$ for $A_x$ in equation (III).

  $\begin{array}{l}-1.25P+F_{AB}\cos \left({\cos }^{-1}\left(\frac{L}{\sqrt{L^2+{\left(0.8L\right)}^2}}\right)\right)=0\\ -1.25P+F_{AB}\times \frac{L}{\sqrt{L^2+{\left(0.8L\right)}^2}}=0\\ -1.25P+F_{AB}\times \frac{L}{\sqrt{L^2+0.64L^2}}=0\\ -1.25P+F_{AB}\times \frac{L}{1.28L}=0\end{array}$

  $\begin{array}{c}{F}_{AB}=\frac{1.25P\times 1.28L}{L}\\ F_{AB}=1.6P\end{array}$

Write the expression for force in member $OB$ .

  $F_{OB}=F_{AB}\cos {\theta }_{ABO}$   ......(IV)

Substitute $1.6P$ for $F_{AB}$ and ${\cos }^{-1}\left(\frac{L}{\sqrt{L^2+{\left(0.8L\right)}^2}}\right)$ for ${\theta }_{ABO}$ in equation (IV).

  $\begin{array}{c}{F}_{OB}=1.6P\times \cos \left({\cos }^{-1}\left(\frac{L}{\sqrt{L^2+{\left(0.8L\right)}^2}}\right)\right)\\ =1.6P\times \frac{L}{\sqrt{L^2+{\left(0.8L\right)}^2}}\\ =1.6P\times \frac{L}{1.28L}\\ =1.25P\end{array}$

Write the expression for moment of inertia for circular pipe.

  $I=\frac{\pi }{64}\left(d_{\circ }{}^4-d_i{}^4\right)$   ......(V)

Here, outer diameter of circular pipe is $d_{\circ }$ and inner diameter of circular pipe is $d_i$ .

Write the expression for inner diameter of circular pipe.

  $d_i=d_{\circ }-2t$   ......(VI)

Here, the wall thickness of the circular pipe is $t$ .

Write the expression for critical load in member $OB$ .

  $P_{cr}=\frac{{\pi }^{2}EI}{L^2}$   ......(VII)

Here, the young’s modulus of members is $E$ .

Substitute $F_{OB}$ for $P_{cr}$ in equation (VII).

  $F_{OB}=\frac{{\pi }^{2}EI}{L^2}$   ......(VIII)

Substitute $1.25P$ for $F_{OB}$ in equation (VIII).

  $\begin{array}{c}1.25P=\frac{{\pi }^{2}EI}{L^2}\\ P=\frac{{\pi }^{2}EI}{1.25\times L^2}\end{array}$   ......(IX)

Calculation:

Substitute $0.25\text{in}$ for $t$ and $3.5\text{in}$ for $d_{\circ }$ .

  $\begin{array}{l}{d}_i=3.5\text{in}-2\times 0.25\text{in}\\ d_i=3\text{in}\end{array}$

Substitute $3.5\text{in}$ for $d_{\circ }$ and $3\text{in}$ for $d_i$ in equation (V).

  $\begin{array}{c}I=\frac{\pi }{64}\left({\left(3.5\text{in}\right)}^4-{\left(3\text{in}\right)}^4\right)\\ =\frac{\pi }{64}\left({\left(150.0625\text{in}\right)}^4-{\left(81\text{in}\right)}^4\right)\\ =\frac{\pi }{64}\times \left(464047556.5\right)\\ =22778881.16{\text{in}}^4\end{array}$

Substitute $10600\text{ksi}$ for $E$ , $22778881.16{\text{in}}^4$ for $I$ and $11\text{ft}$ for $L$ in equation (IX).

  $\begin{array}{c}P=\frac{{\pi }^2\times 10600\text{ksi}\times 22778881.16{\text{in}}^4}{1.25\times {\left(11\text{ft}\right)}^2}\\ =\frac{{\pi }^2\times 10600\text{ksi}\left(\frac{{10}^3\text{lb}/{\text{in}}^2}{1\text{ksi}}\right)\times 22778881.16{\text{in}}^4}{1.25\times {\left(11\text{ft}\times \left(\frac{12\text{in}}{1\text{ft}}\right)\right)}^2}\\ =\frac{{\pi }^2\times 10600\times {10}^3\text{lb}/{\text{in}}^2\times 22778881.16{\text{in}}^4}{1.25\times {132}^2{\text{in}}^2}\\ =\frac{2.383076585\times {10}^{15}\text{lb}\cdot {\text{in}}^2}{21780{\text{in}}^2}\end{array}$

  $\begin{array}{l}=1.094158\times {10}^{11}\text{lb}\\ \text{=}1.094158\times {10}^{11}\text{lb}\times \left(\frac{{10}^{-3}\text{kips}}{1\text{lb}}\right)\\ =109415821.2\text{kips}\end{array}$

Conclusion:

The critical value of load $P$ at which member $OB$ fails by Euler buckling is $\underset{\_}{109415821.2\text{kips}}$ .