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norbayamhdsimin

Published on December 4, 2014

Strength Of Materials Simply Supported Beam 9 First integral equation will give, dx dy EI = 1 4 2 C Wx ------------------------------------------------------------------ (1) When x = L / 2, dy/dx = 0. Replace into equation (1) 0 = 1 4 ) 2 / ( 2 C L W 1 C = 16 2 WL Then equation (1) will be dx dy EI = 16 4 2 2 WL Wx Maximum slope occurs at point A, where x = 0 max dx dy EI = 16 2 WL max dx dy = max = EI WL 16 2 The second integral equation will give EIy = 2 16 12 2 3 C x WL Wx ------------------------------------------------------- (2) When x = 0, y = 0, C2 = 0 Maximum deflection, y max occurs at x = L/2 EIy = 16 ) 2 / ( 12 ) 2 / ( 2 3 L WL L W EIy = ) 2 ( 16 ) 8 ( 12 3 3 WL WL = 32 96 3 3 WL WL = 96 3 3 3 WL WL = 96 2 3 WL = 48 3 WL max y = EI WL 48 3 (Negative sign indicates downward deflection)