This solution calculates the bending stress ($\sigma$) in a steel I-beam using the flexure formula.
To use the flexure formula consistently, convert all units to N and mm:
The bending stress is calculated using the formula:
$ \sigma = \frac{M \cdot y}{I} $
Substitute the converted values into the formula:
$ \sigma = \frac{(96 \times 10^6 \text{ N-mm}) \times (100 \text{ mm})}{240 \times 10^6 \text{ mm}^4} $
Simplify the expression:
$ \sigma = \frac{96 \times 100}{240} \text{ N/mm}^2 $
$ \sigma = \frac{9600}{240} \text{ N/mm}^2 $
$ \sigma = 40 \text{ N/mm}^2 $
Since $1 \text{ N/mm}^2 = 1 \text{ MPa}$, the bending stress is:
$ \sigma = 40 \text{ MPa} $
A simply supported RCC beam of cross section $0.4 \text{ m} \times 0.6 \text{ m}$ covers a span of $8 \text{ m}$. It is subjected to a uniformly distributed load of $30 \text{ kN/m}$. If the unit weight of concrete is $24 \text{ kN/m}^3$, the tensile stress (in $N/mm^2$, rounded off to two decimal places) at the bottom of the beam at mid-span is______
A rectangular beam section of size 300 mm (width) X 500 mm (depth) is loaded with a shear force of 600 kN. The maximum shear stress on the section in N/mm² is ___________
A simply supported beam AB has a clear span of 7 meter. The bending moment diagram (BMD) of the beam due to a single concentrated load is shown in the figure below.
The magnitude of the concentrated load in kN is __________.