Chapter 5.4, Problem 110P

(a)

To determine

Write the equations for shear force and bending moments based on singularity function.

Answer to Problem 110P

The equation of shear force as a singularity function is;

$\underset{\_}{V=30-24{\langle x-0.75\rangle }^0-24{\langle x-1.5\rangle }^0-24{\langle x-2.25\rangle }^0+66{\langle x-3\rangle }^0\text{kN}}$.

The equation of bending moment as a singularity function is;

$\underset{\_}{M=30x-24{\langle x-0.75\rangle }^1-24{\langle x-1.5\rangle }^1-24{\langle x-2.25\rangle }^1+66{\langle x-3\rangle }^1\text{kN-m}}$.

Explanation of Solution

Show the free-body diagram of the beam as in Figure 1.

Determine the vertical reaction at point E by taking moment about point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ -24\left(0.75\right)-24\left(1.5\right)-24\left(2.25\right)+E_y\left(3\right)-24\left(3.75\right)=0\\ -18-36-54+3E_y-90=0\\ E_y=66\text{kN}\end{array}$

Determine the vertical reaction at point A by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A_y-24-24-24+E_y-24=0\\ A_y-96+66=0\\ A_y=30\text{kN}\end{array}$

Write the equation of the shear force function as follows;

$\begin{array}{c}V=A_y-24{\langle x-0.75\rangle }^0-24{\langle x-1.5\rangle }^0-24{\langle x-2.25\rangle }^0+E_y{\langle x-3\rangle }^0\\ =30-24{\langle x-0.75\rangle }^0-24{\langle x-1.5\rangle }^0-24{\langle x-2.25\rangle }^0+66{\langle x-3\rangle }^0\text{kN}\end{array}$ (1)

The equation for bending moment as a function of shear force is,

$V\left(x\right)=\frac{dM}{dx}$

Integrate the equation (1) to find M;

$\begin{array}{c}M={\displaystyle \int Vdx}\\ =30x-24{\langle x-0.75\rangle }^1-24{\langle x-1.5\rangle }^1-24{\langle x-2.25\rangle }^1+66{\langle x-3\rangle }^1\text{kN-m}\end{array}$ (2)

Therefore,

The equation of shear force as a singularity function is;

$\underset{\_}{V=30-24{\langle x-0.75\rangle }^0-24{\langle x-1.5\rangle }^0-24{\langle x-2.25\rangle }^0+66{\langle x-3\rangle }^0\text{kN}}$.

The equation of bending moment as a singularity function is;

$\underset{\_}{M=30x-24{\langle x-0.75\rangle }^1-24{\langle x-1.5\rangle }^1-24{\langle x-2.25\rangle }^1+66{\langle x-3\rangle }^1\text{kN-m}}$.

(b)

To determine

The maximum normal stress due to bending using the singularity function.

Answer to Problem 110P

The maximum normal stress due to bending is $\underset{\_}{87.7\text{MPa}}$.

Explanation of Solution

Refer to Equation (2).

Point A $\left(x=0\text{m}\right)$:

Substitute 0 m for x in Equation (2).

$\begin{array}{c}{M}_A=30\left(0\right)-24{\langle 0-0.75\rangle }^1-24{\langle 0-1.5\rangle }^1-24{\langle 0-2.25\rangle }^1+66{\langle 0-3\rangle }^1\\ =0-0-0-0+0\\ =0\end{array}$

Point B $\left(x=0.75\text{m}\right)$:

Substitute 0.75 m for x in Equation (2).

$\begin{array}{c}{M}_B=30\left(0.75\right)-24{\langle 0.75-0.75\rangle }^1-24{\langle 0.75-1.5\rangle }^1-24{\langle 0.75-2.25\rangle }^1+66{\langle 0.75-3\rangle }^1\\ =22.5-0-0-0+0\\ =22.5\text{kN-m}\end{array}$

Point C $\left(x=1.5\text{m}\right)$

Substitute 1.5 m for x in Equation (2).

$\begin{array}{c}{M}_C=30\left(1.5\right)-24{\langle 1.5-0.75\rangle }^1-24{\langle 1.5-1.5\rangle }^1-24{\langle 1.5-2.25\rangle }^1+66{\langle 1.5-3\rangle }^1\\ =45-18-0-0+0\\ =27\text{kN-m}\end{array}$

Point E $\left(x=2.25\text{m}\right)$:

Substitute 2.25 m for x in Equation (2).

$\begin{array}{c}{M}_D=30\left(2.25\right)-24{\langle 2.25-0.75\rangle }^1-24{\langle 2.25-1.5\rangle }^1-24{\langle 2.25-2.25\rangle }^1+66{\langle 2.25-3\rangle }^1\\ =67.5-36-18-0+0\\ =13.5\text{kN-m}\end{array}$

Point E $\left(x=3\text{m}\right)$:

Substitute 3 m for x in Equation (2).

$\begin{array}{c}{M}_E=30\left(3\right)-24{\langle 3-0.75\rangle }^1-24{\langle 3-1.5\rangle }^1-24{\langle 3-2.25\rangle }^1+66{\langle 3-3\rangle }^1\\ =90-54-36-18-0\\ =-18\text{kN-m}\end{array}$

Point F $\left(x=3.75\text{m}\right)$:

Substitute 3.75 m for x in Equation (2).

$\begin{array}{c}{M}_F=30\left(3.75\right)-24{\langle 3.75-0.75\rangle }^1-24{\langle 3.75-1.5\rangle }^1-24{\langle 3.75-2.25\rangle }^1+66{\langle 3.75-3\rangle }^1\\ =112.5-72-54-36+49.5\\ =0\end{array}$

Refer to the calculated bending moment values, the maximum bending moment occurs at point C.

The maximum bending moment in the beam is $\left|M_{\max }\right|=27\text{kN-m}$.

Refer to Appendix C “Properties of Rolled-Steel Sections” in the textbook.

The section modulus (S) for $\text{W}250\times 28.4$ section is $S=308\times {10}^3{\text{mm}}^3$.

Determine the maximum normal stress $\left({\sigma }_m\right)$ using the equation.

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

Substitute $27\text{kN-m}$ for M and $308\times {10}^3{\text{mm}}^3$ for S.

$\begin{array}{c}{\sigma }_m=\frac{27\text{kN-m}\times \frac{\text{1,000}\text{N}}{\text{1}\text{kN}}}{308\times {10}^3{\text{mm}}^3\times {\left(\frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)}^3}\\ =87.7\times {10}^6\text{Pa}\times \frac{\text{1}\text{MPa}}{{\text{10}}^{\text{6}}\text{Pa}}\\ =87.7\text{MPa}\end{array}$

Therefore, the maximum normal stress due to bending is $\underset{\_}{87.7\text{MPa}}$.