8.3 kg/cm2
7.6 kg/cm2
beam of 25 cm width and 50 cm effective depth has a clear span of 6 meters and carries a U.
11.4 kg/cm2
Answer is Right!
Answer is Wrong!
The correct answer is $\boxed{\text{B) 7.6 kg/cm}^2}$.
The maximum shear stress in a beam is given by the following equation:
$$\tau_{max} = \frac{VQ}{Ib}$$
where:
- $\tau_{max}$ is the maximum shear stress,
- $V$ is the shear force,
- $Q$ is the first moment of area of the cross-section about the neutral axis,
- $I$ is the second moment of area of the cross-section, and
- $b$ is the width of the cross-section.
In this case, we are given that:
- $V = 3000 \times 6 = 18000 \text{ N}$
- $Q = \frac{1}{2}bh^2 = \frac{1}{2} \times 0.25 \times 0.5^2 = 0.0625 \text{ m}^3$
- $I = \frac{bh^3}{12} = \frac{0.25 \times 0.5^3}{12} = 0.015625 \text{ m}^4$
- $b = 0.25 \text{ m}$
Substituting these values into the equation for $\tau_{max}$, we get:
$$\tau_{max} = \frac{VQ}{Ib} = \frac{18000 \times 0.0625}{0.015625 \times 0.25} = 7.6 \text{ kg/cm}^2$$
Therefore, the maximum intensity of shear stress is $\boxed{\text{7.6 kg/cm}^2}$.