Question

The figure shows the stress distribution across an internal surface of a rectangular beam of height 30 mm and depth 10 mm. The normal stress distribution is given by the expression \( \sigma_{xx} = 200y + 500 \, {N/mm}^2 \), where \( y \) is the distance in mm from the centroidal axis of the beam. Assume that there is no variation in the stress distribution along the z-direction.

• A. The net force in the x direction is 150 kN.
• B. The net force in the x direction is 75 kN.
• C. The net moment about the z axis is 4500 Nm.
• D. The net moment about the z axis is 2250 Nm.

Hint:

For problems involving stress distribution across a beam, always calculate the net force by integrating the stress function over the area. For the moment, multiply the stress by the distance from the neutral axis.

Answer 1

The Correct Option is A, C

We are given the stress distribution along the height of the beam with the expression \( \sigma_{xx} = 200y + 500 \), where \( y \) is the distance from the centroidal axis of the beam. The height of the beam is 30 mm, and the depth is 10 mm.
Step 1: Net Force in the x direction
The net force in the x direction is the integral of the stress distribution across the height of the beam. The expression for the force is: \[ F_x = \int_{-15}^{15} \sigma_{xx} \, b \, dy, \] where \( b = 10 \, {mm} \) is the width of the beam. Substituting the given stress distribution: \[ F_x = \int_{-15}^{15} (200y + 500) \times 10 \, dy. \] This can be split into two integrals: \[ F_x = 10 \times \left[ \int_{-15}^{15} 200y \, dy + \int_{-15}^{15} 500 \, dy \right]. \] For the first integral: \[ \int_{-15}^{15} 200y \, dy = 0 \quad \text{(since it's an odd function over a symmetric interval)}. \] For the second integral: \[ \int_{-15}^{15} 500 \, dy = 500 \times 30 = 15000 \, {N}. \] Thus, the net force in the x direction is: \[ F_x = 10 \times 15000 = 150000 \, {N} = 150 \, {kN}. \] Thus, option (A) is correct.
Step 2: Net Moment about the z axis
The net moment about the z-axis is calculated by integrating the moment arm times the stress across the beam height. The moment about the z-axis is given by: \[ M_z = \int_{-15}^{15} \sigma_{xx} \, y \, b \, dy. \] Substituting the expression for \( \sigma_{xx} \): \[ M_z = \int_{-15}^{15} (200y + 500) \times y \times 10 \, dy. \] This can be split into two integrals: \[ M_z = 10 \times \left[ \int_{-15}^{15} 200y^2 \, dy + \int_{-15}^{15} 500y \, dy \right]. \] The second integral is zero because it's an odd function over a symmetric interval: \[ \int_{-15}^{15} 500y \, dy = 0. \] For the first integral: \[ \int_{-15}^{15} 200y^2 \, dy = 200 \times 2 \times \int_0^{15} y^2 \, dy = 200 \times 2 \times \left[ \frac{y^3}{3} \right]_0^{15} = 200 \times 2 \times \frac{3375}{3} = 450000 \, {N mm}. \] Thus, the net moment about the z axis is: \[ M_z = 10 \times 450000 = 4500000 \, {N mm} = 4500 \, {Nm}. \] Thus, option (C) is correct.