Question

Week 8, Question 1: For the truss shown in the following figure, the temperature of member BC is raised by 60 °C, and member BD is raised by 40 °C. EA= 300000 N for all members and a= 1/75000 18 C. Use the stiffness method to do the following: B 5 m с D Part 1. Calculate the displacements at the joints: a) Ax = mm b) Ay = mm c) B. = mm d) B. = mm e) Cx = mm Part 2 Calculate the force in each member: a) NAB N b) NAC = N c) NBD = N d) NBC = e) Ncp = N f) NAD = N

the required data is there , just need to caslculate it

Answer 1

The solution is provided in the images below. Revert back in case of any query.

solution: 26 ☺ 3 3 4 5m Q Do 27 -Smi Given: LEA: 300000 N 1/25000 Kece at Be : 60°c DI Bo: 40° Element commestiwity matrix: Element Node 13 Nod 2 length. I 1 2 _3 O 3 4 -1 2. 3 4 5 4 5 S 5 552 552 - O 3 1/12 6 W2 -V 4

Elemental stiffness matrix : ke: i lm EA le Im m² - 1² -lm -hm -ľ -lm -lm - m² lw lm m² er -m² Pemex toto ki 60000 . o EX O 1 2 o O O O o lo ke 00009 -1 O ky O 0 OO -o 00 O kg = 60000 252 -) -1 -1 -1 -| - . ko 60000 252 -) -) -

Applying matrix elemental matricel, we get guolial stiffeness R As .per B.C., Q P₂ = 2q = 0. Reduring rows and willmal 2,78 8 of k we. .g.et K : 105 10243 0 -0,424_ -0.424 0 -0.6 O 3 1.024 -0.424 -0.424 1.024 -0.6. 0 -0.424 S 0.424 1.024 0.424 -0.424 1.024 6 Assembling the force orce matrix Thexwal jace..im_BC, The = _EA" DīAL- 240 N TBC: 1% 'Ps Too cosus - 169.70N E = Ibesinys = 169.70N Louny TBC F = F --)69.70N be the force Since OT Be is the il compressive be split into components at the respective can nodes.

For BD TBO EA KAT BO 16ON TGO = F" = TBO 2 IGON - 2 3 QS and Fe = -160N TOO Force matrix, -164.7 169.7 O 169.7 169.7 + 160 O -160 Reducing force matrix woresponding to _93,03498 and applying XQ = F we -get, -169.7 k 23 24 0 O co & ad 169.7 329.7

P3 24 0.002 sm 0.001 m Ay Bal Qs By Q6 0.0035 m 0.0032 m 0.0046 m ca - 0.0002 Ca Aa Ay. 0.0005 0.0002 -0.1953 0.4714 0.1953 0.6667 2.8619 Ps Bx 0.0007 By 0.0029 Stress in element... To El. cm. i m]q .EXOT I de element, Force in Fe I miq EA XOT. EA [-d. le 60000 Fi- FAC O -1 1 -- 0.0002 -0. 2 0.00053 0.0002

A FAC: 12N 5: Eng6०००० - ० । . ० - 0.0005. 0.0002 0.0007 ०.००29 - 12N. F₃ = Foo : 60000 [0 - 0.0007 Ohxh- 0.0024 4N. FFo - 60000 ၀၀၁ ၊ ० । o -यी -0.0002 O Eep, 12N. . 04 60000 -0.0002 09 KH O 0.007 0.0029 -1262 F: hp - 60000 1 म 0.0005 0.0002 . . (80 : -9N