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Mechanics of Materials Assignment

   

Added on  2020-03-16

4 Pages353 Words33 Views
Solutions Q1)a)Taking ∑MB = 0RA * 3.75 – [17.5*3.75 *3.75/2 + 13 * 1.25]RA = 37.15 KNTotal load = 13 + 17.5 * 3.75 = 78.625 KNRB = 78.625 – 37.15 = 41.48 KNMaximum bending moment, shear force (SF) = 0Therefore, 37.15 – (13 + 17.5 * x) = 017.5x = 24.15X = 1.38 mb)37.15 * 1.38 – 17.5 * 1.3822 = 34.6035 KN/mModulus Z = MPt = 34.6035∗106165= 2.09718 * 105 mm3= 209.718 cm3Try IPN = 200 Having;Z = 214 cm3I = 2140 cm4c)Deflection equation Y = W0x24EI(L3−2Lx2+x3)+Pb6LEI(lb(x−a)3+(l2−b2)x+x3)Y =215.62∗10−18x24∗2.140∗107∗2∗1011(3.753−2∗3.75x2+x3)+13∗1.256∗3.75∗2.140∗107∗2∗1011(l1.25(x−2.5)3+(3.752−1.252)x+x3)Y = 215.62∗10−18x−30.6667∗10−18x3+4.08878∗10−18x4+¿Graphical presentation
Mechanics of Materials Assignment_1
00.511.522.533.54-900-800-700-600-500-400-300-200-1000100x in my*10-18inmMaximum deflection = -800 *10-18 mQ2) from the provided table d = 450 mmb = 170 mmt = 24.3 mmFs = PlengthTotal length = 2*170 + 450*2 * ½ = 790Fs = 340∗103790= 430.38 N/mmFT = P∗e∗d2IxxIxx = 2*225*170 + 450312∗22
Mechanics of Materials Assignment_2

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