Chapter 2, Problem 2.5.25P

A capped cast-iron pipe is compressed by a brass rod, as shown. The mil is turned until it is just snug, then add an additional quarter turn to pre-compress the cast-iron pipe. The pitch of the threads of the bolt ap = 52 mils (a mil is one-thousandth of an inch). Use the numerical properties provided.

(a) What stresses a and a_rwill be produced in the cast-iron pipe and brass rod. respectively, by the additional quarter turn of the nut?

(b) Find the bearing stress a_hbeneath the washer and the shear stress t₍in the steel cap.

  

To determine

The stress produced in pipe and bar.

Answer to Problem 2.5.25P

The stress produced in pipe is= $-0.196\text{ksi}$.

The stress produced in bar is = $3.4204\text{ksi}$.

Explanation of Solution

Given information:

The pitch of the bolt is $2\text{m}\cdot \text{ins}$, diameter of washer is $\frac{3}{4}\text{in}$, thickness of cap is $1\text{in}$, outer diameter of cast iron pipe is $6\text{in}$, inner diameter of cast iron pipe is $5.625\text{in}$, length of cast iron is $4\text{ft}$, diameter of brass rod is $\frac{1}{2}\text{in}$, modulus of elasticity of steel is $30000\text{ksi}$, modulus of elasticity of brass is $14000\text{ksi}$, modulus of elasticity of cast iron is $12000\text{ksi}$and the number of turns are $\frac{1}{4}$.

Write the expression for area of rod.

  $A_r=\frac{\pi }{4}{d}_r^2$........ (I)

Here, area of rod is $A_r$, diameter of rod is $d_r$.

Write the expression for area of pipe.

  $A_p=\frac{\pi }{4}\left(d_o^2-d_i^2\right)$....... (II)

Here, area of pipe is $A_p$, outer diameter is $d_o$and the inner diameter is $d_i$.

Write the expression for displacement between cut ends.

  ${\delta }_{net1}=-np$....... (III)

Here, displacement between cut ends of the rod is ${\delta }_{net1}$, number of turns is $n$and the pitch is $p$.

Write the expression for relative displacement between cut ends.

  ${\left({\delta }_{net}\right)}_2=F\left(\frac{L_{ci}+2t_c}{E_b{A}_r}+\frac{L_{ci}}{E_C{A}_p}\right)$....... (IV)

Here, relative displacement between cut ends is ${\left({\delta }_{net}\right)}_2$, length of cat iron is $L_{ci}$, thickness of cast iron is $t_c$.

Write the compatibility Equation for displacement.

  ${\delta }_{net1}+{\delta }_{net2}=0$....... (V)

Write the expression for force on rod.

  $F_r=F$........ (VI)

Here, force on rod is $F_r$.

Write the expression for force on pipe.

  $F_p=-F$....... (VII)

Here, force on pipe is $F_p$.

Write the expression for stress on cast iron pipe.

  ${\sigma }_p=\frac{F_p}{A_P}$....... (VIII)

Here, stress on cast iron pipe is ${\sigma }_p$.

Write the expression for stress in rod.

  ${\sigma }_r=\frac{F_r}{A_r}$....... (IX)

Here, stress on rod is ${\sigma }_r$.

Calculation:

Substitute $\frac{1}{2}\text{in}$for $d_r$in Equation (I).

  $\begin{array}{c}{A}_r=\frac{\pi }{4}{\left(\frac{1}{2}\text{in}\right)}^2\\ =\frac{\pi }{4}\left(\frac{1}{4}{\text{in}}^2\right)\\ =0.19635{\text{in}}^{\text{2}}\end{array}$

Substitute $6\text{in}$for $d_o$and $5.625\text{in}$for $d_i$in Equation (II).

  $\begin{array}{c}{A}_p=\frac{\pi }{4}\left({\left(6\text{in}\right)}^2-{\left(5.625\text{in}\right)}^2\right)\\ =\frac{\pi }{4}\left(36{\text{in}}^2-31.645{\text{in}}^2\right)\\ =3.42{\text{in}}^{\text{2}}\end{array}$

Substitute $\frac{1}{4}$for $n$and $2\text{m}\cdot \text{ins}$for $p$in Equation (III).

  $\begin{array}{c}{\delta }_{net1}=-\frac{1}{4}\left(2\text{m}\cdot \text{ins}\right)\\ =-\frac{1}{4}\left(2\text{m}\cdot \text{ins}\right)\frac{\left(0.026\text{in}\right)}{1\text{m}}\\ =-13\times {10}^{-3}\text{in}\end{array}$

Substitute $4\text{ft}$for $L_{ci}$, $1\text{in}$for $t_c$, $14000\text{ksi}$for $E_b$, $12000\text{ksi}$for $E_c$, $3.42{\text{in}}^{\text{2}}$for $A_p$, $0.19635{\text{in}}^{\text{2}}$for $A_r$in Equation (IV).

  $\begin{array}{c}{\left({\delta }_{net}\right)}_2=F\left(\frac{4\text{ft}+2\times 1\text{in}}{14000\text{ksi}\times 0.19635{\text{in}}^{\text{2}}}+\frac{4\text{ft}}{12000\text{ksi}\times 3.42{\text{in}}^{\text{2}}}\right)\\ =F\left(\frac{\left(4\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)+2\times 1\text{in}}{14000\text{ksi}\times 0.19635{\text{in}}^{\text{2}}}+\frac{\left(4\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)}{12000\text{ksi}\times 3.42{\text{in}}^{\text{2}}}\right)\\ =1.9357\times {10}^{-5}F\end{array}$

Substitute $1.9357\times {10}^{-5}F$for ${\left({\delta }_{net}\right)}_2$and $-13\times {10}^{-3}\text{in}$for ${\delta }_{net1}$in Equation (V).

  $\begin{array}{l}-13\times {10}^{-3}\text{in}+1.9357\times {10}^{-5}F=0\\ F=\frac{-13\times {10}^{-3}}{1.9357\times {10}^{-5}}\\ F=671.6\text{lb}\end{array}$

Substitute $671.6\text{lb}$for $F$in Equation (VI).

  $F_r=671.6\text{lb}$

Substitute $671.6\text{lb}$for $F$in Equation (VII).

  $F_p=-671.6\text{lb}$

Substitute $-671.6\text{lb}$for $F_p$and $3.42{\text{in}}^{\text{2}}$for $A_p$in Equation (VIII).

  $\begin{array}{c}{\sigma }_p=\frac{-671.6\text{lb}}{3.42{\text{in}}^{\text{2}}}\\ =\left(-196.145\text{lb}/{\text{in}}^{\text{2}}\right)\left(\frac{1\text{ksi}}{1000\text{lb}/{\text{in}}^{\text{2}}}\right)\\ =-0.196\text{ksi}\end{array}$

Substitute $671.6\text{lb}$for $F_r$and $0.19635{\text{in}}^{\text{2}}$for $A_r$in Equation (IX).

  $\begin{array}{c}{\sigma }_r=\frac{671.6\text{lb}}{0.19635{\text{in}}^{\text{2}}}\\ =\left(3420.4227\text{lb}/{\text{in}}^{\text{2}}\right)\left(\frac{1\text{ksi}}{1000\text{lb}/{\text{in}}^{\text{2}}}\right)\\ =3.4204\text{ksi}\end{array}$

Conclusion:

The stress produced in pipe is = $-0.196\text{ksi}$.

The stress produced in bar is = $3.4204\text{ksi}$.

To determine

The bearing stress in cap.

The shearing stress in the cap.

Answer to Problem 2.5.25P

The bearing stress in cap is = $2.736\text{ksi}$.

The shearing stress in the cap is = $0.285\text{ksi}$.

Explanation of Solution

Given information:

The pitch of the bolt is $2\text{m}\cdot \text{ins}$, diameter of washer is $\frac{3}{4}\text{in}$, thickness of cap is $1\text{in}$, outer diameter of cast iron pipe is $6\text{in}$, inner diameter of cast iron pipe is $5.625\text{in}$, length of cast iron is $4\text{ft}$, diameter of brass rod is $\frac{1}{2}\text{in}$, modulus of elasticity of steel is $30000\text{ksi}$, modulus of elasticity of brass is $14000\text{ksi}$, modulus of elasticity of cast iron is $12000\text{ksi}$and the number of turns are $\frac{1}{4}$.

Write the expression for area of cap.

  $A_{cap}=\frac{\pi }{4}\left(d_w^2-d_r^2\right)$........ (X)

Here, area of cap is $A_{cap}$, diameter of washer is $d_w$and the diameter of the rod is $d_r$.

Write the expression for bearing stress on cap.

  ${\sigma }_{cap}=\frac{F_r}{A_{cap}}$......... (XI)

Here, bearing stress on cap is ${\sigma }_{cap}$.

Write the expression for shearing stress on the cap.

  ${\tau }_{cap}=\frac{F_r}{\pi d_w{t}_c}$......... (XII)

Here, shearing stress on the cap is ${\tau }_{cap}$.

Calculation:

Substitute $\frac{3}{4}\text{in}$for $d_w$and $\frac{1}{2}\text{in}$for $d_r$in Equation (X).

  $\begin{array}{c}{A}_{cap}=\frac{\pi }{4}\left({\left(\frac{3}{4}\text{in}\right)}^2-{\left(\frac{1}{2}\text{in}\right)}^2\right)\\ =\frac{\pi }{4}\left(0.3125{\text{in}}^2\right)\\ =0.2454{\text{in}}^2\end{array}$

Substitute $0.2454{\text{in}}^2$for $A_{cap}$and $671.6\text{lb}$for $F_r$in Equation (XI).

  $\begin{array}{c}{\sigma }_{cap}=\frac{671.6\text{lb}}{0.2454{\text{in}}^2}\\ =\left(2736.75\text{lb}/{\text{in}}^{\text{2}}\right)\left(\frac{1\text{ksi}}{1000\text{lb}/{\text{in}}^{\text{2}}}\right)\\ =2.736\text{ksi}\end{array}$

Substitute $671.6\text{lb}$for $F_r$, $\frac{3}{4}\text{in}$for $d_w$and $1\text{in}$for $t_c$in Equation (XII).

  $\begin{array}{c}{\tau }_{cap}=\frac{671.6\text{lb}}{\pi \left(\frac{3}{4}\text{in}\right)\left(1\text{in}\right)}\\ =\frac{671.6\text{lb}}{2.3561{\text{in}}^{\text{2}}}\\ =\left(285.036\text{lb}/{\text{in}}^{\text{2}}\right)\left(\frac{1\text{ksi}}{1000\text{lb}/{\text{in}}^{\text{2}}}\right)\\ =0.285\text{ksi}\end{array}$

Conclusion:

The bearing stress in cap is = $2.736\text{ksi}$.

The shearing stress in the cap is = $0.285\text{ksi}$.