Question

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Freeman Part II: Problem 5 (15 points) Draw the shear and moment diagrams for the beam and determine the maximum shear and the maximum moment. 15 kN 5 kN/m 80 kNm B -5 m 5 m (a)

Answer 1

(5x5) KN 13 KM 80 kn.m K 205m LA205 ht -5m 3 5m RA Steba: calculate Reac Hon force at A and C. EMA =0 -80+ RX10 – 15x5 – 25x7.5 = 0 Re = 34.25 kW CH) E Mc = 25x2.5 + 15x5 - RAX10 – 80 = 0 TRA= 5.75 KN Step at Draw Spear force and Bending moment drogoam . Take section between AB and Draw FBD le (0<x<5) =0 80 Knim Mor) El Z Fy A x vex) 5.75 KN 5075 - V(x) = 0 V = 7157 kn. (6) EM=0 M(x) - 80 -5.75x = 0 M(x) 80 +5.750

at x= 5m M(5) = 80 + 5.7505) = 108.75 kv.m For section between BC 5<x<'10 5 (x+5) KN 80 kn.m 15KN 142 El Efy ol garang Art x '5.75 KN 5.75 -15 - VLM) – 5 (405)=0 VI) = 5.75 -15-50+25 V(n) = 5(3-15 - x) teni () ZN=0 M(K) -80 + 5(x+5)(n-5) +15%82-5) = 5.75x = 0 M(x) = 80 +5.75 X -156-5) - 5 (8~5)? 2 2 Od X= t VB) = 5(3415 - 5) = -9.25 KN. M(5) = 80 +575(5) - 17(5-5) - 5/5-55 = 108.75 kaum 2 at x=10m VI 10) = 5(3.15-10) milio) = 80+ 5.75(10) - 15(40-5) - 5(10-52 =-34.25 KN. = Oknom

15 ton 5kN/m 80kN) 5-75 KN .-34.25kN AV (2) 757 KN с. A x (SFD) -9.25 RN - 34.25 kn/ 1 108.75 (Nik)) AMEA) Not stoought line 80 KN x (BMD) maximum sheque force man - -34.25 kN (at c) monimum bending moment 108.75 kw.m (at B) Momax