Chapter 11, Problem 11.3.23P

The truss ABC shown in the figure supports a vertical load W at joint B. Each member is a slender circular steel pipe (E = 30,000 ksi) with an outside diameter of 4 in. and wall thickness 0.25 in. The distance between supports is 23 ft. Joint B is restrained against displacement perpendicular to the plane of the truss.

Determine the critical value Wcr of the load.

  

To determine

The critical of the load.

Answer to Problem 11.3.23P

The critical load is $14.41\text{k}$ .

Explanation of Solution

Given information:

The modulus of elasticity of the steel pipe is $30000\text{ksi}$ , the outside diameter of the steel pipe is $4\text{in}$ , the thickness of the pipe is $0.25\text{in,}$ and the distance between the two supports is $23\text{ft}$ .

Write the expression for the inside diameter of steel pipe.

  $d_i=d_o-2t$   .........(I)

Here, the outside diameter is $d_o$ and the thickness is $t$ .

Write the expression for the moment of inertia of hollow of steel pipe.

  $I=\frac{\pi }{64}\left({\left(d_o\right)}^4-{\left(d_i\right)}^4\right)$   .........(II)

Here, the outside diameter is $d_o$ and the inside diameter is $d_i$ .

The following figure shows the free body diagram.

  

  Figure-(1)

Write the expression for the length of the steel pipe $\text{AB}$ .

  $L_{AB}=L\left(\frac{\sin {\theta }_2}{\sin \left(180°-{\theta }_1-{\theta }_2\right)}\right)$   .........(III)

Here, the distance between two supports is $L$ , the angle at support $\text{C}$ is ${\theta }_2,$ and the angle at support $\text{A}$ is ${\theta }_1$ .

Write the expression for the length of the steel pipe $\text{BC}$ .

  $L_{BC}=L\left(\frac{\sin {\theta }_1}{\sin \left(180°-{\theta }_1-{\theta }_2\right)}\right)$   .........(IV)

Here, the distance between two supports is $L$ , the angle at support $\text{C}$ is ${\theta }_2,$ and the angle at support $\text{A}$ is ${\theta }_1$ .

Write the expression for critical load of pipe of both end pinned.

  $P_{cr-AB}=\frac{{\pi }^{2}EI}{L_{AB}^2}$   .........(V)

Here, modulus of elasticity is $E$ and moment of inertia is $I$ .

Write the expression for critical load of pipe of both end pinned joined.

  $P_{cr-BC}=\frac{{\pi }^{2}EI}{L_{BC}^2}$   .........(VI)

Here, modulus of elasticity is $E$ and moment of inertia is $I$ .

The following figure shows the free body diagram taking joint at $\text{B}$ .

  

  Figure-(2)

Write the expression for force acting along y-direction.

  $\begin{array}{l}\uparrow {\displaystyle \sum F_y}=0\\ F_{AB}\sin \left({\theta }_1\right)-F_{BC}\sin \left({\theta }_2\right)=W\\ F_{AB}\sin \left({\theta }_1\right)-F_{BC}\sin \left({\theta }_2\right)-W=0\end{array}$   .........(VII)

Write the expression for force acting along x-direction.

  $\begin{array}{l}\uparrow {\displaystyle \sum F_x}=0\\ F_{AB}\cos \left({\theta }_1\right)-F_{BC}\cos \left({\theta }_2\right)=\\ F_{AB}\cos \left({\theta }_1\right)-F_{BC}\cos \left({\theta }_2\right)=0\end{array}$   .........(VIII)

Write the expression for critical load for member AB.

  $W_{cr}=\min \left(W_{cr-AB},W_{cr-BC}\right)$   .........(IX)

Calculation:

Substitute $4\text{in}$ for $d_o$ and $0.25\text{in}$ for $t$ in Equation (I).

  $\begin{array}{c}{d}_i=4\text{in}-\left(2\times 0.25\text{in}\right)\\ =4\text{in}-0.5\text{in}\\ =\text{3}\text{.5}\text{in}\end{array}$

Substitute $4\text{in}$ for $d_o$ and $3.5\text{in}$ for $d_i$ in Equation (II).

  $\begin{array}{c}I=\frac{\pi }{64}\left({\left(4\text{in}\right)}^4-{\left(3.5\text{in}\right)}^4\right)\\ =0.049\left(256{\text{in}}^4-150.0625{\text{in}}^{\text{4}}\right)\\ =5.200{\text{in}}^{\text{4}}\end{array}$

Substitute $40°$ for ${\theta }_1$ and $55°$ for ${\theta }_2$ in Equation (III).

  $\begin{array}{c}{L}_{AB}=\left(23\text{ft}\right)\left(\frac{\sin 55°}{\sin \left(180°-40°-55°\right)}\right)\\ =\left(23\text{ft}\right)\left(\frac{\sin 55°}{\sin 85°}\right)\\ =\left(23\text{ft}\right)\left(0.822\right)\\ =18.912\text{ft}\end{array}$

Substitute $40°$ for ${\theta }_1$ and $55°$ for ${\theta }_2$ in Equation (IV).

  $\begin{array}{c}{L}_{BC}=\left(23\text{ft}\right)\left(\frac{\sin 40°}{\sin \left(180°-40°-55°\right)}\right)\\ =\left(23\text{ft}\right)\left(\frac{\sin 40°}{\sin 85°}\right)\\ =\left(23\text{ft}\right)\left(0.64\right)\\ =14.841\text{ft}\end{array}$

Substitute $30000\text{}\text{ksi}$ for $E$ , $5.200{\text{in}}^4$ for $I$ and $18.912\text{ft}$ for $L_{AB}$ in Equation (V).

  $\begin{array}{c}{P}_{cr-AB}=\frac{{\pi }^2\left(30000\text{ksi}\right)\left(5.200{\text{in}}^{\text{4}}\right)}{{\left(18.912\text{ft}\right)}^2}\\ =\frac{{\pi }^2\left(30000\text{ksi}\left(\frac{1\text{k}/{\text{in}}^{\text{2}}}{\text{1}\text{ksi}}\right)\right)\left(5.200{\text{in}}^{\text{4}}\right)}{{\left(18.912\text{ft}\left(\frac{12\text{in}}{1\text{ft}}\right)\right)}^2}\\ =\frac{{\pi }^2\left(30000\text{k}/{\text{in}}^{\text{2}}\right)\left(5.200{\text{in}}^{\text{4}}\right)}{{\left(226.944\text{in}\right)}^2}\\ =29.59\text{}\text{k}\end{array}$

Substitute $30000\text{}\text{ksi}$ for $E$ , $5.200{\text{in}}^4$ for $I$ and $14.841\text{ft}$ for $L_{BC}$ in Equation (VI).

  $\begin{array}{c}{P}_{cr-BC}=\frac{{\pi }^2\left(30000\text{ksi}\right)\left(5.200{\text{in}}^{\text{4}}\right)}{{\left(14.841\text{ft}\right)}^2}\\ =\frac{{\pi }^2\left(30000\text{ksi}\left(\frac{1\text{k}/{\text{in}}^{\text{2}}}{1\text{ksi}}\right)\right)\left(5.200{\text{in}}^{\text{4}}\right)}{{\left(14.841\text{ft}\left(\frac{12\text{in}}{1\text{ft}}\right)\right)}^2}\\ =\frac{{\pi }^2\left(30000\text{k}/{\text{in}}^{\text{2}}\right)\left(5.200{\text{in}}^{\text{4}}\right)}{{\left(178.092\text{}\text{in}\right)}^2}\\ =48.55\text{k}\end{array}$

Substitute $40°$ for ${\theta }_1$ and $55°$ for ${\theta }_2$ in Equation (VII).

  $\begin{array}{c}{F}_{AB}\sin \left(40°\right)-F_{BC}\sin \left(55°\right)-W=0\\ 0.642F_{AB}-0.819F_{BC}-W=0\end{array}$   .........(X)

Substitute $40°$ for ${\theta }_1$ and $55°$ for ${\theta }_2$ in Equation (VIII).

  $\begin{array}{l}{F}_{AB}\cos \left(40°\right)-F_{BC}\cos \left(55°\right)=0\\ 0.766F_{AB}-0.5735F_{BC}=0\\ 0.766F_{AB}=0.5735F_{BC}\\ F_{BC}=1.33F_{AB}\end{array}$

  $F_{AB}=0.75F_{BC}$

Substitute $1.33F_{AB}$ for $F_{BC}$ in Equation (X).

  $\begin{array}{l}0.642F_{AB}-0.819\left(1.33F_{AB}\right)-W=0\\ W=0.642F_{AB}-\left(1.12917F_{AB}\right)\\ W=-0.48717F_{AB}\end{array}$

Neglecting the negative sign as it shows the compression.

  $W=0.48717F_{AB}$   .........(XI)

Substitute $P_{cr-AB}$ for $F_{AB}$ for member $\text{AB}$ in Equation (XI).

  $W_{cr-AB}=0.48717P_{cr-AB}$   .........(XII)

Substitute $0.75F_{BC}$ for $F_{AB}$ in Equation (X).

  $\begin{array}{l}0.642\left(0.75F_{BC}\right)-0.819F_{BC}-W=0\\ W=0.48F_{BC}-\left(0.819F_{BC}\right)\\ W=-0.3375F_{BC}\end{array}$

Neglecting the negative sign as it shows the compression.

  $W=0.3375F_{BC}$   .........(XIII)

Substitute $P_{cr-BC}$ for $F_{BC}$ in for member BC in Equation (XIII).

  $W_{cr-BC}=0.3375P_{cr-BC}$   .........(XIV)

Substitute $29.59\text{}\text{k}$ for $P_{cr-AB}$ in Equation (XII).

  $\begin{array}{c}{W}_{cr-AB}=0.48717\left(29.59\text{k}\right)\\ =14.41\text{k}\end{array}$

Substitute $48.55\text{}\text{k}$ for $P_{cr-BC}$ in Equation (XIV).

  $\begin{array}{c}{W}_{cr-BC}=0.3375\left(48.55\text{k}\right)\\ =16.38\text{k}\end{array}$

Substitute $14.41\text{k}$ for $W_{cr-AB}$ and $16.38\text{k}$ for $W_{cr-BC}$ in Equation (IX).

  $\begin{array}{c}{W}_{cr}=\min \left(14.41\text{k},16.38\text{k}\right)\\ =14.41\text{k}\end{array}$

Conclusion:

The critical load is $14.41\text{k}$ .