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4N Problem 6. The beam shown is loaded with a linear distributed load for the left half and a constant distributed load for t

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Page No. YEN MA * 0.5m 0.5m A = 0 Moment about E) MA - (4X005)-(6X0.5 x 2x0.5) | -(6X0.5x0.75) = 0 MA-2-0.5-2.25=0 Too MA = 4(2 Page No. Date: To for section AC x = RA - Wan E = 8.50 - wa 2 12xxx 2 xx = 8.5 - 6x2 Mx- % = -MA + 8.50 - 12a xxx a 2 3 =Page No. Date: My-y -4.75 +8.5% -1.5%+1.5 -48 to -3 (x²+0.25-x) =-32+(8-5-15-4+3) - 4.75 +1.5 +2 - 0.75 = -3x² +62-3 6 ShearPage No. Date: Bending Moment diagram o Mara - 2x + 8.5a-4.75 ocaso.nl 0.5<x</ Mggr = -3x² + 6x 3 -0075 KNM -4-75 KN-M for o<Page No. Date: by a +8528-47526 141] Foro.&n</ M = EI dy da² dy = 4 SM da es |{=38°462-3) d? -_ *+33°-327627Page No Date: At x=0.5 forsection AC for section CB - dyl dala=0.5 AC dy dx 10.5 CB 1 4551 - 4251 - 5 * 1 - 2 + 3x2 3x + C JPare No Date: D y =?_382=0.468752 453 =% [ 38 41422° +2.38 2°] Ja=0.5 - - 0.5 t = -0.420625 C3 = 0.079 375 By tag 4x?= 1:52-

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