A simply supported beam is subjected to an external point load P as shown in the figure below. The beam has a rectangular cross-section of 20 mm $\times$ 45 mm. A is a pin support and B is a roller support. The shear stress developed at point C, lying on the neutral axis of the beam, is 3 MPa. Neglecting the mass of the beam, the magnitude of the applied load is _______ kN (rounded off to 1 decimal place).

Question

A rectangular section has width $b$ and depth $d$ and a symmetrical triangular section has base $b$ and depth $d$. The ratio of moment of inertia of rectangular section to that of triangular section with respect to their respective centroidal axis will be:

Answer 1

Step 1: Get the average shear stress first.
For a rectangular section the peak (neutral axis) shear stress is 1.5 times the average shear stress, $\tau_{max} = 1.5\tau_{avg}$, where $\tau_{avg} = V/A$. So $\tau_{avg} = 3/1.5 = 2$ MPa, which gives $V = \tau_{avg}A = (2\times10^6)(9\times10^{-4}) = 1800$ N directly.

Step 2: Get the reactions using the ratio of distances.
For a simply supported beam of length 5 m with a single load $P$ at 2 m from A, the reactions split in proportion to the distance from the other support: $R_A = P\times\dfrac{5-2}{5} = 0.6P$ and $R_B = P\times\dfrac{2}{5} = 0.4P$.

Step 3: Match C to the correct segment.
Point C is 1 m from A, which is before the load point at 2 m, so the shear force acting there is the reaction $R_A$ alone, unaffected by the load further down the beam: $V_C = R_A = 0.6P$.

Step 4: Combine and solve for P.
$$0.6P = 1800 \implies P = 3000\ \text{N} = 3.0\ \text{kN}$$

Final Answer:
Working from the average shear stress up to the reaction gives the same 3.0 kN load. \[ \boxed{P = 3.0\ \text{kN}} \]

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Correct Answer: 3

Solution and Explanation

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