Uniform beam AB has a weight of 0.5kN/m and it is 16m long. It is subjected to a

Question

Uniform beam AB has a weight of 0.5kN/m and it is 16m long. It is subjected to a loading system as shown. a) Replace the loading on the beam by a single resultant force b)Specify the location of this resultant on the beam from point A c) Find the support reactions at A and B Image can be found at (Question 7) https://docs.google.com/viewer?a=v&q;=cache:AkJhFPRrEfkJ:www.coursehero.com/load_question_attachment.php?q_att_id=275417+CIV1501+assignment+3&hl;=en≷=au&pid;=bl&srcid;=ADGEESgotGhwVi-wmLtI1irerjMM8_5kpV6xPtnvXhyATYG4veEgbTHzlpAwonS1Y1drG_0qpEIR89-JmcEPSPm4Q-zgAU1aZLujxM2lSg2z1oeLej2Qvj-byUHclzbuyrJ6sjV8QkGy&sig;=AHIEtbS4jI4x3kQ5TCiZ5Hs6w8xWXORrGg

Explanation / Answer

) the various forces which are applicable are as follows: - force due to the weight of the beam = 0.5 * 16 = 8 KN force due to the linear force of 3 KN/m = 3 * 5 = 15 KN force due to the inclined wedge = 6 * 5 = 30 KN thus the resultant force would be = 30 - 8 - 15 = 7 KN b)since the bar resting on A & B thus the net torque should be zero taking torque along the point a R X 7 = 20 thus R = 20/7 = 2.86 m from the point A c)take torque at point c Ra*6+20+6*6*6/2=Rb*6+3*6*6/2 6Ra+128=6Rb+54 Ra+21.33=Rb+9 Ra-Rb=-12.33 and Ra+Rb=132.5 Ra=60.085 Rb=72.415 N

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