Chapter 5.5, Problem 129P

5.128 and 5.129 The beam AB, consisting of a cast-iron plate of uniform thickness b and length L, is to support the distributed load w(x) shown. (a) Knowing that the beam is to be of constant strength, express h in terms of x, L, and h₀. (b) Determine the smallest value of h₀ if L = 750 mm, b = 30 mm, w₀ = 300 kN/m, and σ_all = 200 MPa.

Fig. P5.129

To determine

Express h in terms of x, L, and $h_0$.

Answer to Problem 129P

The expression for h in terms of x, L, and $h_0$ is $\underset{\_}{1.659h_0{\left[\left(\frac{x}{L}-\frac{2}{\pi }\sin \frac{\pi x}{2L}\right)\right]}^{1/2}}$.

Explanation of Solution

Given information:

The uniform thickness of the beam is b.

The length of the beam is L.

The rise of the beam is $h_0$.

Calculation:

By definition;

$\frac{dV}{dx}=-w$

Substitute $w_0\sin \frac{\pi x}{2L}$ for w.

$\frac{dV}{dx}=-w_0\sin \frac{\pi x}{2L}$

Integrate the equation to find V.

$V=-\frac{2w_{0}L}{\pi }\cos \frac{\pi x}{2L}+C_1$ (1)

Apply the boundary condition; $V=0\text{at}x=0$.

Substitute 0 for V and 0 for x in Equation (1).

$\begin{array}{l}0=-\frac{2w_{0}L}{\pi }\cos \frac{\pi \left(0\right)}{2L}+C_1\\ 0=-\frac{2w_{0}L}{\pi }+C_1\\ C_1=\frac{2w_{0}L}{\pi }\end{array}$

Substitute $\frac{2w_{0}L}{\pi }$ for $C_1$ in Equation (1).

$\begin{array}{c}V=-\frac{2w_{0}L}{\pi }\cos \frac{\pi x}{2L}+\frac{2w_{0}L}{\pi }\\ =-\frac{2w_{0}L}{\pi }\left(1-\cos \frac{\pi x}{2L}\right)\end{array}$

By definition;

$\frac{dM}{dx}=V$

Substitute $-\frac{2w_{0}L}{\pi }\left(1-\cos \frac{\pi x}{2L}\right)$ for V and integrate.

$M=-\frac{2w_{0}L}{\pi }\left(x-\frac{2L}{\pi }\sin \frac{\pi x}{2L}\right)+C_2$ (2)

Apply the boundary condition; $M=0\text{at}x=0$.

Substitute 0 for M and 0 for x in Equation (2).

$\begin{array}{l}0=-\frac{2w_{0}L}{\pi }\left(0-\frac{2L}{\pi }\sin \frac{\pi \left(0\right)}{2L}\right)+C_2\\ C_2=0\end{array}$

Substitute 0 for $C_2$ in Equation (2).

$\begin{array}{c}M=-\frac{2w_{0}L}{\pi }\left(x-\frac{2L}{\pi }\sin \frac{\pi x}{2L}\right)+0\\ =-\frac{2w_{0}L}{\pi }\left(x-\frac{2L}{\pi }\sin \frac{\pi x}{2L}\right)\end{array}$

Determine the section modulus (S) of the beam using the relation.

$S=\frac{\left|M\right|}{{\sigma }_{all}}$

Here, the allowable stress in the beam is ${\sigma }_{all}$.

Substitute $\frac{2w_{0}L}{\pi }\left(x-\frac{2L}{\pi }\sin \frac{\pi x}{2L}\right)$ for $\left|M\right|$.

$\begin{array}{c}S=\frac{\frac{2w_{0}L}{\pi }\left(x-\frac{2L}{\pi }\sin \frac{\pi x}{2L}\right)}{{\sigma }_{all}}\\ =\frac{2w_{0}L}{\pi {\sigma }_{all}}\left(x-\frac{2L}{\pi }\sin \frac{\pi x}{2L}\right)\end{array}$

Determine the section modulus (S) of the rectangular cross section using the relation.

$S=\frac{1}{6}b{h}^2$

Here, the width of the beam is b and the depth of the beam is h.

Substitute $\frac{2w_{0}L}{\pi {\sigma }_{all}}\left(x-\frac{2L}{\pi }\sin \frac{\pi x}{2L}\right)$ for S.

$\begin{array}{l}\frac{2w_{0}L}{\pi {\sigma }_{all}}\left(x-\frac{2L}{\pi }\sin \frac{\pi x}{2L}\right)=\frac{1}{6}b{h}^2\\ h^2=\frac{12w_{0}L}{\pi b{\sigma }_{all}}\left(x-\frac{2L}{\pi }\sin \frac{\pi x}{2L}\right)\\ h={\left[\frac{12w_{0}L}{\pi b{\sigma }_{all}}\left(x-\frac{2L}{\pi }\sin \frac{\pi x}{2L}\right)\right]}^{1/2}\end{array}$ (3)

When $x=L;h=h_0$.

Substitute $h_0$ for h and L for x.

$\begin{array}{c}{h}_0={\left[\frac{12w_{0}L}{\pi b{\sigma }_{all}}\left(L-\frac{2L}{\pi }\sin \frac{\pi L}{2L}\right)\right]}^{1/2}\\ {\left(h_0\right)}^2=\frac{12w_{0}L}{\pi b{\sigma }_{all}}\left(L-\frac{2L}{\pi }\right)\\ =\frac{12w_{0}L}{\pi b{\sigma }_{all}}\left(L-\frac{2L}{\pi }\right)\\ \frac{12w_{0}L}{\pi b{\sigma }_{all}}=\frac{{\left(h_0\right)}^2}{L\left(1-\frac{2}{\pi }\right)}\end{array}$ (4)

Substitute $\frac{{\left(h_0\right)}^2}{L\left(1-\frac{2}{\pi }\right)}$ for $\frac{12w_{0}L}{\pi b{\sigma }_{all}}$ in Equation (3).

$\begin{array}{c}h={\left[\frac{{\left(h_0\right)}^2}{L\left(1-\frac{2}{\pi }\right)}\left(x-\frac{2L}{\pi }\sin \frac{\pi x}{2L}\right)\right]}^{1/2}\\ =h_0{\left[\frac{1}{L\left(1-\frac{2}{\pi }\right)}\left(x-\frac{2L}{\pi }\sin \frac{\pi x}{2L}\right)\right]}^{1/2}\\ =1.659h_0{\left[\left(\frac{x}{L}-\frac{2L}{\pi L}\sin \frac{\pi x}{2L}\right)\right]}^{1/2}\\ =1.659h_0{\left[\left(\frac{x}{L}-\frac{2}{\pi }\sin \frac{\pi x}{2L}\right)\right]}^{1/2}\end{array}$

Therefore, the expression for h in terms of x, L, and $h_0$ is $\underset{\_}{1.659h_0{\left[\left(\frac{x}{L}-\frac{2}{\pi }\sin \frac{\pi x}{2L}\right)\right]}^{1/2}}$.

To determine

The smallest value of $h_0$.

Answer to Problem 129P

The smallest value of $h_0$ is $\underset{\_}{197.6\text{mm}}$.

Explanation of Solution

Given information:

The maximum allowable stress is ${\sigma }_{all}=200\text{MPa}$.

The length of the beam is $L=750\text{mm}$.

The distributed load is $w_0=300\text{kN/m}$.

The width of the beam is $b=30\text{mm}$.

Calculation:

Substitute $300\text{kN/m}$ for $w_0$, 30 mm for b, 750 mm for L, and 200 MPa for ${\sigma }_{all}$ in Equation (4).

$\begin{array}{c}\frac{\left(\begin{array}{l}12\times 300\text{kN/m}\times \frac{\text{1000}\text{N}}{\text{1}\text{kN}}\\ \times 750\text{mm}\times \frac{\text{1}\text{m}}{\text{1000}\text{mm}}\end{array}\right)}{\left(\begin{array}{l}\pi \times 30\text{mm}\times \frac{\text{1}\text{m}}{\text{1000}\text{mm}}\\ \times 200\text{MPa}\times \frac{{\text{10}}^{\text{6}}\text{Pa}}{\text{1}\text{MPa}}\end{array}\right)}=\frac{{\left(h_0\right)}^2}{750\text{mm}\times \frac{\text{1}\text{m}}{\text{1000}\text{mm}}\left(1-\frac{2}{\pi }\right)}\\ {\left(h_0\right)}^2=\frac{12\times 300\times 1000\times 750\text{mm}\times 1000\times 750\times \left(1-\frac{2}{\pi }\right)}{\pi \times 30\times 1000\times 200\times {\text{10}}^6\times 1000}\\ h_0=0.1976\text{m}\times \frac{\text{1000}\text{mm}}{\text{1}\text{m}}\\ =197.6\text{mm}\end{array}$

Therefore, the smallest value of $h_0$ is $\underset{\_}{197.6\text{mm}}$.