Question

7–53. Draw the shear and moment diagrams for the beam. 1.5 kN/m - 3 m Prob. 7–53

Answer 1

Here i taking R_A . R_B as vertival forces at point A & B

b=breath, h=height,
first cal culate the total load on beam = xbxh =*3*1.5 = 2.25KN Now, calculate moment at point B taking anti clockwis Clock wise tue - EMB=o G RAX 3 - 2:25x4x2= 0 3RA = 2025 Raz:75 KN forces Efy = 0 (sumation of vertical is zero) 2:25 - RB-RA=0 2-25 - RB-0.75=0 {RB = 1.5KN)

2.25KN SKN/m For SED, cut the section at a distance So, Sfx = PA I Wa xn =RA - IX nxa - Ra=n? Sf at Just right of A (n=0) (1-0) SFA = +0.75KN (12=3m) SfB = RA --*? 3.75-182 - +1.5 KN = +1.5 KM SED clx - dlsx = Wn SED 51-52 = area e under loading 0.75-0 = XXX n asof @ 52 In Jim - SKN BMD BM = Rp.x={{xnxwn)x . RAM - 2 X 2 X u X 2 max BM 20.86 30 Parabola RAX n38 T2 BMn Just right of A >= 0 , BMA = 0.75X0-0 FOKNim 21 = 53m n= 3, BME = 0.75X3-03 - OKN)m BMC = 0.75x53-(53) = 0.BOKNIM

first cal culate the total load on beam = xbxh =*3*1.5 = 2.25KN Now, calculate moment at point B taking anti clockwis Clock wise tue - EMB=o G RAX 3 - 2:25x4x2= 0 3RA = 2025 Raz:75 KN forces Efy = 0 (sumation of vertical is zero) 2:25 - RB-RA=0 2-25 - RB-0.75=0 {RB = 1.5KN)

2.25KN SKN/m For SED, cut the section at a distance So, Sfx = PA I Wa xn =RA - IX nxa - Ra=n? Sf at Just right of A (n=0) (1-0) SFA = +0.75KN (12=3m) SfB = RA --*? 3.75-182 - +1.5 KN = +1.5 KM SED clx - dlsx = Wn SED 51-52 = area e under loading 0.75-0 = XXX n asof @ 52 In Jim - SKN BMD BM = Rp.x={{xnxwn)x . RAM - 2 X 2 X u X 2 max BM 20.86 30 Parabola RAX n38 T2 BMn Just right of A >= 0 , BMA = 0.75X0-0 FOKNim 21 = 53m n= 3, BME = 0.75X3-03 - OKN)m BMC = 0.75x53-(53) = 0.BOKNIM