Question

Problem 7-86 (15 points) 6 IN E Draw the shear and moment diagrams for the beam. SIN

Answer 1

# Solution - General sign conventions - - Upwand force = 9 = Downward force -- Positive (tve) Negative (-ve) . * Shear force Diagram Conventions di e section 8500 tve. sf: -ve. Sito Diagram Conventions - - À Bending moment ---- section Hogg Sagging B.m tve Bim. Glo in ta i.e. section Hogging B.m. - ve B.M.

of Sifis # Variation & B.M'D on s Reana ③ Type of loading . Variation of sif.D. I u cariation of Bomo to point Looed. straight line Inclined line I!!!! UDL Linear parabolic I UUL UUL I Parabolic parabolic 1 Cubic parabola force (@f) ☆ General Reation bet Shear { Bending Moment (m) af = w әх am af aa

blom Now , - -3 kN/m The above diagram is equivalent to kelm 11 3 kr/m F ism — -.5m Draw F.B.P. PBK pm Rat GIA lism -1.5mm A RAV Efx=0 ... Ray + ( x 3x1.5) - (£*3*1.5)=0 ... | Rav=o] EFH=0 RAH = 0

EMA = 0 w I-h- we know, Intensity of load = Area of triangle. = 1/xwon from This will act at distance 2h aper . or at distance ha from Base

EMA = 0 =--($*3*15) ('55+(5^3*1*5) (1.57 3.73) Im=4.5 kN/m - Assumed din is correct as ans positive. is 3kr/m 4.5kom 3 kolm Shear force calculation tve -ve. Shear Force aut A= 0 sif at B = Hi3x1.5 = 2.25 kn sif at c" = 3x3x1.5- £x3x1.5 -0

Draw S. Fo. +2. 25kN parabala. B tve. b x 15m - force 15m - Diagram. Shear 22 As af=w .: Slope at B'shoud be 0. CAs shown in 01.) Tangent (Bero slope ) - ve. Bending moment . Calcutations - +ve Bending moment at A =-415 knom DX Bim at B = -4.5+ (x3) 3 - 2.25 katm B.m at CE-45 +(343x1.5) (2x15+1.5) - (£x3x1.5) (9) = 0 X .5

* Daw Bimio. B _ve X2.25 lcp-mg 4.5 km Bending momat Diagram. as - am ax = FF δα cas slope at a &c should be zero F CS.F) is zero at Afc.

ЗЕРЈm) 3 kplog 3m Loading Diagram 2.25kn +ve Sifo. = 0 .vel 12.25-kom < Bm. 4.5 kp-m B.MiD