Given: The beam loaded and supported as shown below. The beam has a length, L, e

Question

Given: The beam loaded and supported as shown below. The beam has a length, L, equal to 8 m. The beam is supported by a pin at a distance of L/10 from its left end and a roller at a distance of L/5 from its right end. The left half of the beam is acted upon by an upward directed distributed load that decreases linearly from a value of q0 equal to 6 kN/m at the left end to a value of q0/2 at the midpoint of the beam. The right half of the beam is acted upon by a downward directed distributed load that decreases linearly from a value of q0 equal to 6 kN/m at the right end to a value of q0/2 at the midpoint of the beam. Find: The support reactions.

Explanation / Answer

Okay. So first we want to turn our UDL's into workableconcentrated loads. The magnitude of the force generated by each UDL will be the areaof those UDL's. A = b x [(h1 + h2) / 2]   where b is the base, h1 is thelarger of the max height, and h2 is the min height. In thiscase, h1 = 6, and h2 = 3. A = the magnitude of the force ofthe UDL. This force will act at d, where d = [ b (h1 +2h2) ] / [3 (h1 +h2] **note that d is measured from the large end (h1). So W (concentrated load of UDL) = 4 x [(6 + 3) / 2] =18KN (This same magnitude is used for both UDL's since they are the samein shape and size. W is acting down on the right UDL, and Wis acting up on the left UDL) d = [ 4 (6 +2*3) ] / [3 (6 + 3] = 1.78m This tells us that W will act 1.78m from the big end of eachUDL. We can draw those forces into our diagrams and make surewe dimension correctly. Once the dimensions are labelled, wecan use other given dimensions to determine how far W is from eachsupport (this will help us fill in values for our momentequation). Now do M at B for Ay and M at A for By. M at A = 0 (clockwise positive) -By(5.6) - 18(0.98) + 18(5.42) = 0      By = 14.27KN (upwards) M at B = 0 (clockwise positive) -Ay(5.6) + 18(4.62) = 18(0.18) = 0      Ay = 14.27KN (downwards) I hope this helps. Good luck.

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