Question:
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).
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).
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Find the reaction that carries the shear over the segment containing C, then use the rectangular-section formula \(\tau_{max}=1.5V/A\).
Updated On: Jul 27, 2026
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Correct Answer: 3
Solution and Explanation
Step 1: Read the geometry and locate C.
The beam spans 5 m, with pin support A at one end and roller support B at the other. Point C sits on the neutral axis at 1 m from A, and the load \(P\) is applied 2 m from A, so C lies in the segment between A and the load.
Step 2: Find the reaction that governs the shear at C.
Taking moments about A: \(R_B(5) = P(2)\), so \(R_B = 0.4P\). From vertical equilibrium, \(R_A = P - R_B = 0.6P\). Since C is between A and the load, the shear force throughout that stretch of the beam equals \(R_A = 0.6P\).
Step 3: Use the shear stress formula for a rectangular section.
On the neutral axis of a rectangular cross-section, the shear stress is at its peak value, given by \(\tau_{max} = \dfrac{3V}{2A}\), where \(V\) is the shear force at that section and \(A = 20\times45 = 900\) mm\(^2\) \(= 9\times10^{-4}\) m\(^2\) is the cross-section area.
Step 4: Solve for V, then for P.
\[ V = \frac{2\tau_{max}A}{3} = \frac{2(3\times10^{6})(9\times10^{-4})}{3} = 1800\ \text{N} \]
Since \(V = R_A = 0.6P\), \(P = 1800/0.6 = 3000\) N \(= 3.0\) kN.
Final Answer:
Matching the shear stress at C to the reaction it carries gives an applied load of 3.0 kN. \[ \boxed{P = 3.0\ \text{kN}} \]
The beam spans 5 m, with pin support A at one end and roller support B at the other. Point C sits on the neutral axis at 1 m from A, and the load \(P\) is applied 2 m from A, so C lies in the segment between A and the load.
Step 2: Find the reaction that governs the shear at C.
Taking moments about A: \(R_B(5) = P(2)\), so \(R_B = 0.4P\). From vertical equilibrium, \(R_A = P - R_B = 0.6P\). Since C is between A and the load, the shear force throughout that stretch of the beam equals \(R_A = 0.6P\).
Step 3: Use the shear stress formula for a rectangular section.
On the neutral axis of a rectangular cross-section, the shear stress is at its peak value, given by \(\tau_{max} = \dfrac{3V}{2A}\), where \(V\) is the shear force at that section and \(A = 20\times45 = 900\) mm\(^2\) \(= 9\times10^{-4}\) m\(^2\) is the cross-section area.
Step 4: Solve for V, then for P.
\[ V = \frac{2\tau_{max}A}{3} = \frac{2(3\times10^{6})(9\times10^{-4})}{3} = 1800\ \text{N} \]
Since \(V = R_A = 0.6P\), \(P = 1800/0.6 = 3000\) N \(= 3.0\) kN.
Final Answer:
Matching the shear stress at C to the reaction it carries gives an applied load of 3.0 kN. \[ \boxed{P = 3.0\ \text{kN}} \]
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