Question

The following truss is subjected to vertical loads of 20 KN at joints E and D. In addition to the loads, support A settles by 5 mm and member AB and BC are subjected to a temperature drop of 50°C. Given Young’s modulus, E = 200 GPa, cross sectional area for each member, A = 500 mm² and coefficient of thermal expansion of, α = 1.25 x 10^-5/°C.

Find the internal forces in each member using force method.

20 KN 20 KN V E D 3m A 4m 4m

Answer 1

20KN 20KN J20kW j E 3m 5m A To 20 um um RA as redundant assume Let us Hence, According unit load / virtual wook method So-sui {ui (Pili R= + lidi ATI - Ai E Z uia, li PE i where, so is the support settlement in the direction of reaction uiy member forces due to unit load appuled at the place of redundant ping member forces due to external loads.

biven- So = 5mm = 5x10-3 m A = 500mm 200 kN/mma 1.25x10-5/06 sino = 3/5 5 For Pi 20 KN 20KN D Tore 10 Ayo Ac B calculating member forces using using method of joint- Joint A- {fy=0 ☆ PAE TO PAB=0 PAE Efreno → PAB Joint E 20 KN SPOE PAE E Fx=0 - PRE-PAE Coso=0 PE=0 Efy=0 - PBE +20=0 -> PBE = - 20KN PBE

Joint B PBE Ify =0 PBE+PBD sino =0 ☆ -20+PBOX PBD Depec PAB B PBP 100 KN 3 O Efx=0 PBD Coso + PBC -80 KN - SAB=0 ☆ PBC = - 100 x 4 | PBC = -80 Joint D 20KN sfy=0 => PCD & PBD sino +20=0 PDE SRO => PGD + 100 x 3+20=0 'PCD PBD 40 KN - PCD = RD=0 & Fe=0 PRE t PBB coso 100 x 4 -> Ro= Rp = -80 KN (Reaction at D) For mi o

☆ UAE sino = 1 UAE A UAB ! ų 3 UAE UBE UBET UBD sino =0 Joint A {fy=0 UAE 5 3 sfx=0 => UAE Coso + UAB=0 => UAB = - = 3 x 9 | UAB = Joint Fy=0 UAE cose => UDE → UDE => UDE = Efy=0 + VAE sino =0 -UBE Joint B- efy=0 4BD UBDX 3 = 1 5 T olm - UBE UBE AUBC ЧАР LUBO whe

EFx=0 -> UBOCOS + UBC - UAB = 0 TUBC $=0 wloo Jointo e fy=0 = Yes+ ABD sino =0 UDE RD 2 => UCD - oy - Şx? जय Y CD = -1 UBD LUCO He: biven- SA=5810-3 m ATAS = ATBC = 50°C Eyi RE Pili AEi + lidi ATi) E 4.2, li Ai E AE = constant for all members = 2008 500 = 105 KN

Member	Pi	Ui	Li	PiUiLi	Ui2Li
AB	0	-1.3333	4	0	7.1111111
AE	0	1.66667	5	0	13.888889
BE	-20	-1	3	60	3
DE	0	1.33333	4	0	7.1111111
BD	33.3333	1.66667	5	277.7778	13.888889
BC	-26.667	-2.6667	4	284.4444	28.444444
CD	-40	-1	3	120	3

Hence 0+0+ 60 +0+ 277,776 +284.44 +120 AE РЕ AE R: 5x10-3 AE + (-133X1,25X10°5X50 -2.667x1.25x10*% 50 XY х ч + 7,11 + 13.89 AE AE 3 + Toll + 13189 +28.44 + 3 AE AE AE AE AE > R= 5*10*3- [cox 10-5 +277,778 X105+284.44x10-5+120x105 332.5x10-5 – 666:75x10-5) 10.5 ( 76444) R= gog Hence, Fi= pit ui, R (C) => FAB = 0 - 1.33% 9.9 = -13.167 KN () FAE = 0 + 1,67x9,9 = 16.53 KN FBE = -20-189. g = 29.9 KN (0) FDE = 1.33x go g = 13.167 AN (T) FBD = 33.33 71.678 gig 95 49.86 Andet) FBC = -26.67 - 2167*9.9 = -53.1 (T) Fro= - yorg.g -49, 9 KN (c)