Question

There is a small mistake on drawing. total lenght is 1 m

25% 3. Draw the shear and moment diagrams for the loaded beam. Calculate the deflection at midpoint (E=200GPa, L=lm, 1=100000000 mm, wo=100kN/m, M=10kN.m, P=50kN) using integration method. 100.00 KN 10 kN-m 50.0 KN 0.25m 0.25m 0.25m 0.25m

Answer 1

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x X D y bir 100KN/m B JOKN-m | SON E 20 2 TRE RA 0.25 0.25m 0.25m 0.25m 24315KM -0.625 KM 1-50.625 -50.625 12.8125 KW-M 12-656 N-m 12.968 KN-m 2-8125XN {Fy=0 & Net force iny disections. RATRY - 100x 0.25 +50=75 - EMA=o of Moment about A=oy 160x 0.25 X0-75 +10+50 0.75- Rex1 = 0 RE = 50.625 RN Putting Rein Eyn til RA= 24.375 KN

Taking section x-x in AB - 100 kN/m 1 RA= 24.375 Shear force, V(x) = 24375 - 100 ml. at 250, V = 24.375 at n=0.25, v= -0.625 RN moment, BM= 24.375% - 100XXXX BM: 24.375 % 100xx? atno, B.M=0 at x= 0.25, B M = 2.968 KN-M Bending 2 Taking section x-x in DE TRE-50.625 KN Shear force = -50.625 KM BM = 50 625 XX- at no B.M=0 at - 0.25 , BM: 12.656 KN-M.

Taking Section X- in cp from right L SORM -0.2 TRE=50 0.25 x - 50.625 KN X SF: SO- 50.625 SF= -0.625 KN B-M 50 625 *(1+0:25) - 50 x 21 at x=0, BM = 12.656 KN-. at x1 = 0.25, B.M= 12.8125 RN-m. SOKN 100 KN/m מאסן RE= 50-625 R=24375KW k x Mx = 243752 - 100X-254(-0125) +10-50*/*-0.95) we know du² 2 24 375 2² 2 2 dl + EI dry Mx = 24.375 X - 25(X-0-125) +10-SO (0-0.15) Intego ating both sides - EI dy 25(2-0125)? 10-50(X-0-75)? Again integrating both sides – EI Y = 24.375x2? - 15(X-0125) 4 vox. 5X20-23) c*+62

lor 104 Putting conditions - at n=0, y = 0 = 6=0 at n=1, y=0 0 = 24395 - 251-0,12%)19-501.003) + 6 G=6.14 +10% 2 6 6 Es y = 24.375 M3 - 25 (2-0.125) -50 (X-0.75)3 6.140 6 at x=0.5- y =) 10x.52 EI Y= 6 = 24.375X0.53 25(0.5-0.125) 6 6.14xos. E = 2000 Pa = 200x106 KN I= 108 mm 109 EIY= -1852 y = -1.852 200x106x 16-4 y = 0.0926mm