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: \[ \sigma_{xx} = 200y + 500 (N/mm^2), y \in [-15, +15] \, mm \]

Which of the following statements is/are CORRECT?

• 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:

Net force = integrate stress over area. Net moment = integrate stress \(\times\) distance. For linear variation, odd terms cancel symmetrically.

Answer 1

The Correct Option is A, C

Step 1: Net axial force.
\[ F = \int_{-15}^{15} \sigma_{xx} \, dA, dA = 10 \, dy \] \[ F = 10 \int_{-15}^{15} (200y + 500) \, dy \] \[ = 10 \left[ 100y^2 + 500y \right]_{-15}^{15} \] At \(y=15\): \(22500 + 7500 = 30000\). At \(y=-15\): \(22500 - 7500 = 15000\). Difference = 15000. \[ F = 10 \times 15000 = 150000 \, N = 150 \, kN \] So (A) is correct.

Step 2: Net moment about z.
\[ M_z = \int_{-15}^{15} \sigma_{xx} \cdot y \cdot dA = 10 \int_{-15}^{15} (200y^2 + 500y) \, dy \] \[ = 10 \left[ \tfrac{200}{3} y^3 + 250 y^2 \right]_{-15}^{15} \] Evaluate: At \(y=15\): \(225000 + 56250 = 281250\). At \(y=-15\): \(-225000 + 56250 = -168750\). Difference = 450000. \[ M_z = 10 \times 450000 = 4.5 \times 10^6 \, Nmm = 4500 \, Nm \] So (C) is correct. Final Answer:
\[ \boxed{(A) \; \text{and} \; (C)} \]