Chapter 5, Problem 5.9.6P

A steel pipe is subjected to a quadratic distributed load over its height with the peak intensity q₀at the base (see figure). Assume the following pipe properties and dimensions: height L, outside diameter d = 200 mm, and wall thickness f = 10 mm. Allowable stresses for flexure and shear are o~_a=125 MPa and T_a= 30 MPa,

1. If L = 2.6 m, Fmd^_0ayM (kN/m), assuming that allowable flexure and shear stresses in the pipe are not to be exceeded.
2. If q₀= 60 kN/m, find the maximum height L_raajl(m) of the pipe if the allowable flexure and shear stresses in the pipe arc not to be exceeded.

  

To determine

The maximum uniform load intensity$q_{\text{max}}$.

Answer to Problem 5.9.6P

The maximum load intensity$q_{\text{max}}$is$-4428\text{kN}/\text{m}$.

Explanation of Solution

Given information:

The allowable stress is$125\text{MPa}$, the allowable shear stress is$30\text{MPa}$, the thickness of the pipe is$10\text{mm}$, the height is$2.6\text{m}$, and the diameter is$200\text{mm}$.

Write the expression intensity.

$q=q_{\text{max}}\left[1-{\left(\frac{x}{L}\right)}^2\right]$…… (I)

Here, the intensity is$q$, the length is$L$, the distance is$x$, and the maximum intensity is$q_{\text{max}}$.

Write the expression for the differential Equation.

$\begin{array}{l}{E}_1\frac{d^{4}y}{d{x}^4}=-qx\\ E_1\frac{d^{4}y}{d{x}^4}=\frac{-q_{\text{max}}}{L}\left[L^2-x^2\right]\end{array}$…… (II)

Here, the modulus of elasticity is$E_1$.

Integrate the Equation (II) with respect to x.

$E_1\frac{d^{3}y}{d{x}^3}=\frac{-q_{\text{max}}}{L}\left[L^{2}x-\frac{x^3}{3}\right]+c_1$…… (III)

Applying, the boundary condition at point$A$,$0$for$x$,and$R_A$for$E_1\frac{d^{3}y}{d{x}^3}$in Equation (III)

$\begin{array}{l}{R}_A=0+c_1\\ R_A=c_1\end{array}$…… (IV)

Here, the reaction force at$A$is$R_A$.

Substitute$R_A$for$c_1$in Equation (III).

$E_1\frac{d^{3}y}{d{x}^3}=\frac{-q_{\text{max}}}{L}\left[L^{2}x-\frac{x^3}{3}\right]+R_A$…… (V)

Applying, the boundary condition at point$B$,$L$for$x$,and$R_B$for$E_1\frac{d^{3}y}{d{x}^3}$in Equation (V)

$\begin{array}{l}{R}_B=\frac{-q_{\text{max}}}{L}\left[L^{2}L-\frac{L^3}{3}\right]+R_A\\ R_A-R_B=\frac{-q_{\text{max}}}{L}\left[L^3-\frac{L^3}{3}\right]\\ R_A-R_B=\frac{2}{3}{q}_{\text{max}}L\end{array}$…… (VI)

Here, the reaction force at the point$B$is$R_B$.

Integrate the Equation (V) with respect to x.

$E_1\frac{d^{2}y}{d{x}^2}=\frac{-q_{\text{max}}}{L}\left[\frac{L^2{x}^2}{2}-\frac{x^4}{4}\right]+R_{A}x+c_2$…… (VIII)

Substitute$M$for$E_1\frac{d^{2}y}{d{x}^2}$in Equation (VI)

$M=\frac{-q_{\text{max}}}{L}\left[\frac{L^2{x}^2}{2}-\frac{x^4}{4}\right]+R_{A}x+c_2$…… (IX)

Here, the bending moment is$M$.

Applying the boundary condition at fixed end$A$,$0$for$x$,and$0$for$M$in Equation (IX).

$\begin{array}{l}0=\frac{-q_{\text{max}}}{L}\left[\frac{L^2\times 0}{2}-\frac{0}{4}\right]+R_A\times 0+c_2\\ 0=0+c_2\\ c_2=0\end{array}$

Applying the boundary condition at proper end$B$,$L$for$x$,$0$for$c_2$,and$0$for$M$in Equation (IX).

$\begin{array}{l}0=\frac{-q_{\text{max}}}{L}\left[\frac{L^2\times L^2}{2}-\frac{L^4}{4}\right]+R_{A}L+0\\ R_{A}L=\frac{5}{12}{q}_{\text{max}}{L}^2\\ R_A=\frac{5}{12}{q}_{\text{max}}L\end{array}$

Substitute$\frac{5}{12}{q}_{\text{max}}L$for$R_B$, in Equation (VI).

$\begin{array}{l}\frac{5}{12}{q}_{\text{max}}L-R_B=\frac{2}{3}{q}_{\text{max}}L\\ R_B=-\frac{2}{3}{q}_{\text{max}}L+\frac{5}{12}{q}_{\text{max}}L\\ R_B=-\frac{1}{4}{q}_{\text{max}}L\end{array}$

$h_1$, the maximum bending moment is$M_{\text{max}}$,

Substitute$\frac{5}{12}{q}_{\text{max}}L$for$R_A$, and$0$for$c_2$in Equation (IX).

$M=\frac{-q_{\text{max}}}{L}\left[\frac{L^2{x}^2}{2}-\frac{x^4}{12}\right]+\frac{5}{12}{q}_{\text{max}}Lx$…… (X)

Differentiate the Equation (X) with respect to x

$\begin{array}{l}\frac{dM}{dx}=\frac{-q_{\text{max}}}{L}\left[\frac{2L^{2}x}{2}-\frac{4x^3}{12}\right]+\frac{5}{12}{q}_{\text{max}}L\\ 0=\frac{-q_{\text{max}}}{L}\left[\frac{2L^{2}x}{2}-\frac{4x^3}{12}\right]+\frac{5}{12}{q}_{\text{max}}L\\ \frac{5}{12}{q}_{\text{max}}L=\frac{q_{\text{max}}}{L}\left[\frac{2L^{2}x}{2}-\frac{4x^3}{12}\right]\\ \frac{5}{12}{L}^3=L^{2}x-\frac{x^3}{3}\end{array}$…… (XI)

Write the expression of the flexible stress.

${\sigma }_a=\frac{Mc}{\frac{\pi }{4}\left(d_0^4-d_1^4\right)}$…… (XII)

Here, the allowable stress is${\sigma }_a$, the outer diameter is$d_0$,and inner diameter is$d_1$.

Write the expression of the allowed shear stress.

$\tau =\frac{4M_{\text{max}}}{3LA}\left(\frac{d_0^2+d_0{d}_1+d_1^2}{d_1^2+d_0^2}\right)$…… (XIII)

Here, the shear stress is$\tau$.

Write the expression for the area.

$A=\frac{\pi }{4}\left[d_0^2-d_1^2\right]$…… (XIV)

Calculation:

Substitute$2.6\text{m}$for$L$in Equation (XI).

$\begin{array}{l}\frac{5}{12}\times {2.6}^3{\text{m}}^3={2.6}^{2}x-\frac{x^3}{3}\\ 7.32{\text{m}}^3=6.76x-\frac{x^3}{3}\\ 20.28{\text{m}}^{3}x-x^3=21.96{\text{m}}^3\\ x=1.08\text{m}\end{array}$

Substitute$1.08\text{m}$for$x$, and$2.6\text{m}$for$L$in Equation (X).

$\begin{array}{l}M=\frac{-q_{\text{max}}}{2.6\text{m}}\left[\frac{{\left(2.6\text{m}\right)}^2\times {\left(1.08\text{m}\right)}^2}{2}-\frac{{\left(1.08\text{m}\right)}^4}{12}\right]+\frac{5}{12}{q}_{\text{max}}\left(2.6\text{m}\times 1.08\text{m}\right)\\ M=\frac{-q_{\text{max}}}{2.6\text{m}}\left[4.056{\text{m}}^2\right]+1.17{\text{m}}^2\\ M=-0.39q_{\text{max}}\end{array}$

Substitute$-0.39q_{\text{max}}$for$M$,$125\text{MPa}$for${\sigma }_a$,$200\text{mm}$for$d_0$,$10\text{mm}$for$d_1$,and$0.1$for$c$in Equation (XII).

$\begin{array}{l}125\text{MPa}=\frac{\left(-0.39q_{\text{max}}\right)\left(0.1\right)}{\frac{\pi }{4}\left[{\left(200\text{mm}\right)}^4-{\left(10\text{mm}\right)}^4\right]}\\ 125\text{MPa}\left(\frac{{10}^3\text{kN}}{1\text{MPa}}\right)=\frac{-0.39q_{\text{max}}}{\left[125600.78\times {10}^4{\text{mm}}^4\right]}\\ -0.39q_{\text{max}}=125\times {10}^3\text{kN}\times \left[\left(125600.78\times {10}^4{\text{mm}}^4\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\right]\\ q_{\text{max}}=-402\text{kN}/\text{m}\end{array}$

Substitute$200\text{mm}$for$d_0$, and$10\text{mm}$for$d_1$in Equation (XIII).

$\begin{array}{c}A=\frac{\pi }{4}\left[{\left(200\text{mm}\right)}^2-{\left(10\text{mm}\right)}^2\right]\\ =\frac{\pi }{4}\left[39900{\text{mm}}^2\right]\left[\frac{1\text{m}}{1000\text{mm}}\right]\\ =0.031{\text{m}}^2\end{array}$

Substitute$0.031{\text{m}}^2$for$A$,$200\text{mm}$for$d_0$,$10\text{mm}$for$d_1$,$-0.39q_{\text{max}}$for$M$,$30\text{MPa}$for$\tau$,and$2.6\text{m}$for$L$in Equation (XIV).

$\begin{array}{l}30\text{MPa}=\frac{4\times \left(-0.39q_{\text{max}}\right)}{3\times 2.6\text{m}\times 0.031{\text{m}}^2}\left[\frac{{\left(200\text{mm}\right)}^2+\left(200\text{mm}\times 10\text{mm}\right)+{\left(10\text{mm}\right)}^2}{{\left(200\text{mm}\right)}^2+{\left(10\text{mm}\right)}^2}\right]\\ 30\text{MPa}\left(\frac{{10}^3\text{kN}/{\text{m}}^2}{1\text{MPa}}\right)=\frac{4\times \left(-0.39q_{\text{max}}\right)}{3\times 2.6\text{m}\times 0.031{\text{m}}^2}\left(1.05\right)\\ -0.39q_{\text{max}}=1727\text{kN}/\text{m}\\ q_{\text{max}}=-4428\text{kN}/\text{m}\end{array}$

Conclusion:

The maximum load intensity$q_{\text{max}}$is$-4428\text{kN}/\text{m}$.

To determine

The maximum height of the pipe.

Answer to Problem 5.9.6P

The maximum height of the pipe is$0.17\text{m}$.

Explanation of Solution

Write the expression of the bending moment in terms of length.

$\frac{{\sigma }_{\text{a}}\times \left(\frac{\pi }{4}\left(d_0^4-d_1^4\right)\right)}{c}=\frac{-q_{\text{max}}}{L}\left[\frac{L^2{x}^2}{2}-\frac{x^4}{12}\right]+\frac{5}{12}{q}_{\text{max}}Lx$…… (XV)

Calculation:

Substitute$125\text{MPa}$for${\sigma }_a$,$-402\text{kN}/\text{m}$for$q_{\text{max}}$,$1.08\text{m}$for$x$,$200\text{mm}$for$d_0$, and$10\text{mm}$for$d_1$in Equation (XV).

$\begin{array}{l}\frac{125\text{MPa}\times \left(\frac{\pi }{4}\left({\left(200\text{mm}\right)}^4-{\left(10\text{mm}\right)}^4\right)\right)}{0.1}=\frac{-402\text{kN}/\text{m}}{L}\left[\begin{array}{l}\left[\begin{array}{l}\frac{L^2{\left(1.08\text{m}\right)}^2}{2}-\\ \frac{{\left(1.08\text{m}\right)}^4}{12}\end{array}\right]\\ +\frac{5}{12}L\left(157\text{kN}/\text{m}\right)1.08\text{m}\end{array}\right]\\ \left[\frac{125\text{MPa}\times 1.25\times {10}^9{\text{mm}}^4}{0.1}\left(\frac{1\text{m}}{1000\text{mm}}\right)\left(\frac{{10}^3\text{kN}}{1\text{MPa}}\right)\right]=\left[\begin{array}{l}\left(-233.16L\text{kN}\cdot \text{m}\right)+\frac{45.42{\text{m}}^3}{L}\\ -180.9\text{kN}\times L\end{array}\right]\\ L=0.17\text{m}\end{array}$

Conclusion:

The maximum height of the pipe is$0.17\text{m}$.