Question

Using the force method, Determine the reactions at supports. EI is constant.

Determine the reactions at supports. El is constant. 500 lb/ft 3k 10 ft- SA 0 a 6.5 (10) 10 ft - 2.5 uz

Answer 1

PACE NO.: DATE: 1 Solution : - b - a AL oskift loft - MA loft use I lb = 0.001k heb Apply minimum potential (strain energy method (ns in force method) to find out eaction, treating Cy As Redundant they As per minimum potential energy theorem 0 -- - () eguation . where u = Tolay Stooin energy of frame clue to bending = 1 ME LEI . for Span CB8- __ By_ Bac-tyt 3K 11 1. Mx = o osx < 10 ft = {M² dx co . ZEI J UCB =0 for from equilibrium Efyzo , By =ly Efx=0, Bx = 3k Span BA :- -0.5 hlat fmmmmBx=3K loft MA TAY

PAGE !0.: DATE: 1 Mr = Cy (x) - 0.5 (x) (ч.) o3 x < 10 UBA = CM² 2. І єг — JUBA 2LMх а = . 2ЕГ DUBA Әсч 2 vед - Гусе) – без г) юли. — Сусе) раз ) т. эчак - - [ 44 )," ( заре) - slove) This now from eg ①, we get as o = о т а тоев) - oss( Toos) = 4-23 5 К Applying equilibrium eq in the FBD of the span BC we get as 2 B = as ва = 0, 2 = By зк ifx =o, Br = 3K

PAGE NO.: | DATE: 1 1 - Applying equilibrium equation *_F@Dof_BA_spam, weget_4e- By = 1*29Sk MA Hх *= 0 , P tex =e, - (2 м), го By = 6.5 Х 10 - By = 3.425 к He = A = 3K, - мд +5+ (16) (0) – ву (w) -o M9 = ost (w) (s) — (+375 () = » М4 = 6.25 k-fe