. Theory of structures and strength of materials. Fig. 166. SHEARING FORCES AND BENDING MOMENTS. 107. I inch ; join DA. The ordinate from any point of DA to OAis the bending moment at its foot. For example, at ii:^ O the ordinate is ^, or 300 , and this is equal to80 X 3i, i-e., the bending moment. Ex. 9. A beam OA, Fig. 167, of which the weight may beneglected, rests upon two supports ^at 0 and A, 30 ft. apart, and carriesa uniformly distributed load of 200lbs. per lineal foot, together with asingle weight of 600 lbs. at a pointB dividing the beam into segmentsOB, BA, of which
. Theory of structures and strength of materials. Fig. 166. SHEARING FORCES AND BENDING MOMENTS. 107. I inch ; join DA. The ordinate from any point of DA to OAis the bending moment at its foot. For example, at ii:^ O the ordinate is ^, or 300 , and this is equal to80 X 3i, i-e., the bending moment. Ex. 9. A beam OA, Fig. 167, of which the weight may beneglected, rests upon two supports ^at 0 and A, 30 ft. apart, and carriesa uniformly distributed load of 200lbs. per lineal foot, together with asingle weight of 600 lbs. at a pointB dividing the beam into segmentsOB, BA, of which the lengths are 10and 20 ft. respectively. Determinethe shearing force and bending mo-ment at the points C and D, distant5 ft. from the nearest end. Also,illustrate graphically the shearingforce and bending moment at differ-ent points of the beam. Let R^, R^ be the reactions at Oand A, respectively. Then i?, . 30 = 600 . 20 + 200 . 30. 15 = 102000; .-. R^ = 3400 lbs., and R„ = 200 . 30 + 600 — R^ = 3200 lbs. The Shearing Force at C = 3400 — 200 . 5 = 2400 lbs. Z)=3400—200. 25—600= —22
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