A simply supported beam A of lengthl, breadthband depthdcarries a central loadW. Another beam B of the same dimensions carries a central load equal to 2W. The deflection of beam B will be __________ as that of beam A.a)one-fourthb)doublec)one-halfd)four timesCorrect answer is option 'B'. Can you explain this answer?

Question

Answer

Solution:Given, Length of the beam, lBreadth of the beam, bDepth of the beam, dCentral load on beam A, WCentral load on beam B, 2WTo find: Deflection of beam B in terms of beam AAssumptions made:The beam is made of homogenous material and has a uniform cross-section throughout its length.Formula used:The deflection of a simply supported beam with a central load can be calculated using the formula: δ = (Wl^3)/(48EI)where, δ = deflectionW = loadl = length of the beamE = Young's modulus of elasticityI = moment of inertia of the beam's cross-sectionExplanation:The deflection of a beam depends on its length, breadth, depth, material, and the load acting on it. When the load on the beam is increased, its deflection also increases. In this problem, we have two beams A and B of the same dimensions, but the load on beam B is twice that of beam A. Let us calculate the deflection of beam A first. Deflection of beam A:δ_A = (Wl^3)/(48EI_A)Deflection of beam B:δ_B = (2Wl^3)/(48EI_B)We know that the Young's modulus of elasticity (E) and the moment of inertia (I) are the same for both beams A and B, as they have the same dimensions and are made of the same material. Therefore, we can say that: E_A = E_B = EI_A = I_B = ISubstituting these values in the above equations, we get: δ_A = (Wl^3)/(48EI) δ_B = (2Wl^3)/(48EI)Dividing δ_B by δ_A, we get: δ_B/δ_A = (2Wl^3)/(48EI)/(Wl^3)/(48EI) δ_B/δ_A = 2Therefore, the deflection of beam B will be double that of beam A. Answer: b) double

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