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Question 2 (10 Marks) The area of the homogeneous plate is 10 ft as shown in Figure 2. The vertical reactions on the plate at A and B are 80 lb and 84 lb, respectively. Suppose that you want to equalize the reactions at A and B by drilling a l-ft-diameter hole in the plate. (a) Calculate the horizontal distance from A should the centre of the hole will be drilled. (b) Determine the resulting reactions at A and B 1 SA Figure 2

Answer 1

Sel:- Given required data: B Homogeneous plate area A) = loft² A 5ft vertical reaction plate @ A = 801b vertical reaction on plato @ B - 84 db Diameter of drilled hole, Ift (a) weight of homogeneour plate = 80+44 = 164 16 for the centroid, moments about A = 0 ie 2 MA 84 Wix - 19*5 = 0 8445 84+5 164 2.56 ft => x = Location of centroid 2.56ft from A. 164 10 16.4.14 A weight density (W) = W Weight of doilled hole Id w 4 T & P + 16.4 4 W = 12.88 lb

Weight of the plate after drilling (Wed W - W₂ = 164-12.88 = 151.1216 The new centroid for the reautions to be equal es x - 5ft - 2.5 ft the center i The B-coordinate drilled hole. of of y = W, a-Weal W2 164* 2.56- IS1:127 2:5 12.88 Y = 3.26 ft The distane from A to center of drilled hole Reaction at A = ISIR 2 R. 75.56 lb WN 2 Resulting reactions a A and B are eqreal je Reaction at B= 75.5616

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