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(Solved): Q.5 A simply supported beam is subjected to a uniformly distributed load (UDL ...



Q.5 A simply supported beam is subjected to a uniformly distributed load (UDL) of \( \boldsymbol{w} \mathrm{kN} / \mathrm{m} ???????

Q.5 A simply supported beam is subjected to a uniformly distributed load (UDL) of \( \boldsymbol{w} \mathrm{kN} / \mathrm{m} \) and a concentrated load of \( 15 \mathrm{kN} \) at mid-span as shown in Figure Q.5(a). Its cross section of the beam is shown in Figure \( Q .5(b) \) below. (a) Determine the moment of inertia \( I_{x x} \) of the beam about the \( \mathrm{x}-\mathrm{x} \) axis. (3 marks) (b) Express the support reactions and the maximum bending moment \( \boldsymbol{M}_{\max } \) in terms of \( w \mathrm{kN} / \mathrm{m} \). Hence, draw the bending moment diagram. (6 marks) (c) If the maximum permissible bending stress of this beam is \( 250 \mathrm{~N} / \mathrm{mm}^{2} \), determine the uniformly distributed load of \( w \mathrm{kN} / \mathrm{m} \) that can be carried by the beam. (5 marks) (d) Sketch the bending stress distribution across the section of the beam under the maximum bending stress. (4 marks) (e) Under the loading \( \boldsymbol{w} \) obtained in (c) and the point load of \( 15 \mathrm{kN} \), determine the maximum shear stress induced in the beam. (7 marks) (Not to scale)


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The solution for the given question is from diagram (a) moment of inertia about x-x axis (b) Calculate reactions Shear
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