# ﻿ Problem. Determine The Slopes And Deflections At A, B, C And D By Specified Methods. Where E Is Constant. 96 KN 3m— 3m- 3 M 5m- 48 KN-* – 4.5 M – (1) Use The Direct Integration Method (20). (2) Use The Moment-area Method (20). (3) Use The Conjugate-beam Method (20). (4) Use The Method Of Virtual Work (20) (5) Use The Castigliano’s Theorem (20).

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﻿ Question: Problem. Determine the slopes and deflections at A, B, C and D… (1 bookmark) Problem. Determine the slopes and deflections at A, B, C and D by specified methods. Where is constant. 96 kN 3m- 3m в -3m 5 m 48 KN-1 4.5 m – (1) Use the Direct integration Method (20). (2) Use the Moment-area Method (20). (3) Use the Conjugate-beam method (20). (4) Use the method of Virtual work (20). (5) Use the Castigliano’s theorem (20). help me plz…. Show transcribed image text Expert Answer Sai Arpan Joshi answered this 621 answers Was this answer helpful? Bi p ol Hello!! All the methods that you have mentioned, cannot be attempted to obtain what is required, because the time required for all the methods included will be more than 2-4 hours which completely exceeds the time limit given to us to answer. Thus I have managed to solve it using one method, i.e. Method of Virtual Work. Next time before posting such problems ensure you give a condition such that the problem will be solved within 2 hours. 96kN 21 IF 3m 13m 5m 4.5m 24-0=> Ay+ Dy = 96kN X=03 Ax = 48 KN MA=0-> 48×4,5 +96×3 = Dyx6 Dy= 84 kN → Ay= 12 KN 4) By Virtual work method Bending Moment Equations due to loading only A Mx=48x EB Mx = 48x-48(x-4,5) = 216 kNl-m BF→ Mx = 12% -4843 +48×7.5 = 12m +216 FC Mx=842 CO Mx=0 1) Deflections i @A> a) Horizontal Consider aunit force @ A (leftwardy Bending Moment Equations – AE, EB, BF, FC, CD >> m-0 • SHA=0 b) Vertical Consider a unit load @A (Downwards Bending Moment Equations AE, EB, BE, FC, CD m=0 → SVA-O ii) @ D-> a) Horizontal -> Consider a unit force @D(lightwarols) Bending Moment Equations – as AE AEB m=x BF) Sym–5 +7,5 EC m= 5+ 560 CD m=x BF-7–7*47.5 Thr S2 4.5 SHD = 16263 (Rightwards) b) Verticals, Consider a unit force @D (Downwards) Similar to Sva, Suo=o. a) Horizontal-Consider unit force @ P (Rightwarde) Bending nomont Equations AE&EB > m= x BF >> M = 7.5 -1.25% FC 2) m = 1.250 CD m=0 1.25 1.25 4.5 SHISE 4842 216(x), dx + 1.5-1.25*)C12*+216) 2EL 84xxl.sxda 261 EL EL to SETTE (Kightwarde) 5) Vertical Consider unit force @B (Downwarde). Bending foment Equations AE, EB, BE, FC, CD mzo SUB=0 iv) @ c a) Abrizontals Consider unit forre @c (lightwards) >> This is same as the horizontal deffection @B because of the solup obtain ned and sway principals. → SHB = SHC = 7716 (lightwards ] 1.25 1.25 EL h Vertical Same as vertical deflection @B SUR = Sveto 6) Rotations >> i) @ A Consider unit moment @ A (Clockwise ) = Bending moment Equations AE%EE mee m = 1 BE m= -1/6*+ 1 Ef m=%u CD > m=0 ts 4,5 7.5 4.5 a OA = 1458 (Clockwise) ET *) 20Consider a ma unit moment @ (Clockwise) → Bending Moment Equations – AE & EB »m=o BE m=-1/6* RC m= 1/64 -1 CD = m = -1 2 E1 EL & Oo – \$ 0+0+ ((12x+216) (-1/2) dx [(x – 1) (84x) dx T2EL2EL – 06–216 ( Ahofie okurce] i) @B >Consider a unit moment @ B C Clockwise) — Bending Moment Equations AE & EB m=0 Bram=1-1/4 FC m= %* CQ = mzo Alto & OB= 84** /6x dx 1 (12n+216)(1. 261 2GE – OB = 324 (Clockwise) ET i) @ Consider a unit moment @c (Clockwise) – Bending Moment Equations – AE & EB on=0 BE» m 20-%” FC ma /6 – 1 CD » mzo T (1x) (k2x +216) dx ( (Ex-1) (84 x) dx a + 281 = ZEL ZEL De=-216 (Anticlockusíce)