A simply supported beam of width 100 mm, height 200 mm and length 4 m is carrying a uniformly distributed load of intensity 10 kN/m. The maximum bending stress (in MPa) in the beam is __________ (correct to one decimal place).

The correct answer is 30.

Given Data:
Width of the beam, $b=100\text{mm}$
Height of the beam, $h=200\text{mm}$
Length of the beam, $L=4\text{m}$
Uniformly distributed load, $w=10\text{kN/m}$

As shown in the diagram, the beam is simply supported at both ends and carries a uniformly distributed load (UDL) of $10\text{kN/m}$ over its entire length of $4\text{m}$.

Step 1: Calculate the Maximum Bending Moment ($M_{\text{max}}$)
For a simply supported beam subjected to a uniformly distributed load, the maximum bending moment occurs at the mid-span and is given by the formula:

$M_{\text{max}}=\frac{w{L}^2}{8}$

Substituting the given values into the formula:

$M_{\text{max}}=\frac{10\text{kN/m}\times {\left(4\text{m}\right)}^2}{8}=\frac{10\times 16}{8}=20\text{kNm}$

Converting this into Newton-millimeters (N·mm) to match the other units:

$M_{\text{max}}=20\times {10}^6\text{N·mm}$

Step 2: Calculate the Moment of Inertia ($I$) and Maximum Distance from the Neutral Axis ($y_{\text{max}}$)
For a rectangular cross-section, the moment of inertia about the neutral axis is given by:

$I=\frac{b{h}^3}{12}$

Substituting the cross-sectional dimensions:

$I=\frac{100\text{mm}\times {\left(200\text{mm}\right)}^3}{12}=\frac{100\times 8\times {10}^6}{12}=\frac{2}{3}\times {10}^8{\text{mm}}^4$

The maximum distance from the neutral axis to the outer fibers is:

$y_{\text{max}}=\frac{h}{2}=\frac{200\text{mm}}{2}=100\text{mm}$

Step 3: Calculate the Maximum Bending Stress (${\sigma }_{\text{max}}$)
Using the flexure formula, the maximum bending stress is given by:

${\sigma }_{\text{max}}=\frac{M_{\text{max}}\cdot y_{\text{max}}}{I}$

Substituting the calculated values:

${\sigma }_{\text{max}}=\frac{(20\times {10}^6\text{N·mm})\times 100\text{mm}}{\frac{2}{3}\times {10}^8{\text{mm}}^4}$

Simplifying the expression:

${\sigma }_{\text{max}}=\frac{2\times {10}^9}{\frac{2}{3}\times {10}^8}=30{\text{N/mm}}^2=30\text{MPa}$