Question

Solve the following truss problem. All truss members are ANSI 2x2x0.25 hollow square tubes (with rounded corners) for which the cross-section area is A-1.5891 in2. The material has a modulus of E-29E6 psi. Length of element 1 and 5 is L-20 inches, and length of element 3 and 6 is 2L 40 inches. 7 5 6 P-1000 lb 2. 1. Solve in an Excel spreadsheet using the truss element. Note that there are only four different element stiffness matrices (look for length and orientation). Report the following: (1) Show the stiffness matrices for each of the four different elements. You can write these in terms of spring constant k (for example 0.5ki where ki is the spring constant of element 1) (2) Assemble the constrained global stiffness matrix and load vector using coefficients of each element. Again, write in terms of ki values and Pi, P values (where Pue R are the x and y components of the external applied load at node ) (3) Solve the nodal displacements. (4) Calculate the axial force and normal stress in each element shots showing X displacement

Answer 1

1 2 31 e iguxt esing (an be falulted Lerg th of membe s Gnd 34.64 inches .2 1 (21x10 p)x C-5g91in)46083900 Jbs by 9iven EA Sh ffness msthix r thoss element is Cossing -5h20 -fos Osin 0 (os9sino C7= EA L osbin -3 in in '8 (oss in0 TH i) for element nd LE34 64nches Gind -1 Lak460g 100 -1 34.64 -1330366-628 133 0366.628 C 1330366 628 13303 66. 628 :)for ele ment L2 20 inhes, -90, Sin sin t0- 1, (os0- as 900 l-L 403900 20 1

230 4195 -230419 5 -23 04195 0 2304195 for element (G 0.5 L 40 inihes, $ano o30 Sin0- Sin 30ニ 0.5 (os 30 =0 866 -0.433 075 O.493 ka46083900 0.433 o25 -075 -0:433 -0.25 84073.125 48858 218 8 64073.125 -49%85821? 433 O.25 0.433 75 40 0 25 0.433 0'433 298024.375 -498858-28 288024-379 864073 125 49Y858212 499858-218 -864073.125 -498 8 S8218 -4988 58218 -282024.354988821828024.375 fv) For tlement 2 4 L 40 inche SI9 So C.S 1O.7S 433 75 -0433 0.25 O.433 fk = 460939000.433 0.433 75 -0:433 40 Lo.433-0.25 -0.433 64073-125 48958 218 364073.125 498858.21% -48858218 288024.335 4988 58 218 -288024-375 -86407-3.12 9 498858218 L49 88 518 218 288024 375 -4988 98.218288024-37s 864073. 125-498858 218 m the stchre De qtee of freecum exi hng 2) No dleyree of fredom uil exist at foint haing hringe ty 3since these fn upport dis lecement inX ditechon at int S diplo cemont in Y-ditechon at ont o elo a

The 6 dlegre Figute 1. mc hs if cleno fre by mch term 0f h - an te foundt fro efement shfness metry srgig t 864073-125 2194439.953 o t 4998 58218 = 498 858.218 k12k21 419-31 133 0366.628 14 kai O kisksi 96407 3.125 L kaIミ-49 8958-218 krr 288024.375 0+ 288024.375 423ko2 24- kan O - 4988 58. 218 k52 k62 -288024. 335 (33o 3 66.628+1330366:628 k26 Laうニ O 266 0333256 2304195 k44 ニ0+0 +2304195 - k45 54 O P12304195 k55 = 864033.125ナ0+ 864073.125 t1330365-62? 30 S8 512, 848 498858.21+0-49 %Y58 218 toO sc -288024 375+23041 95 千 288024.375ヤ0 2 28802 43.79 2194139.753 49868-2181390 366-62 0-8640}3:125 -4488 49%89.-218 288024.375 0-498858-21-28824 375 =+3303 66698 Z6694 33 256 O C 2304195 2304195 964093.125 -4989 58218 30 59 ST2.978 497858.218 18g024 375 -2304195 2880243.5

3ince there i oo 1 load -(00 to deg oly cores poning free do m in rep re direchon The goveminy p, ue hon of shess mathk 5el viny On hudal obfain he onhnown ue Can clis piacements Ctowarls hyht 000 26 0.00 P0 5 (do wnwerls S75 Clowards rig ht (downusrls) Oo00 3 0-00080 .O0026 tspaoma SY2 (downwarts) .00080 EA A (4) f, L 20 in For element 1- 4608 3900 20 Co-0) os 90t (o-0)sin t0 La 34-64 In = ele ment 2- 46083 900 34.64 2-133 0366 f, (0-C-0.00026) 628 -345 895 1b Cromhtession

element 3- La 40in -EA 160 83700 0 00026-0) ros 150 40 t 000080-o )sin 1S0 20f.425b Clension) for L-34.641 e lement 4 -EA (s-400 (8y3- Sya)sino 00 013) fus0 +(0-(-0 000 80)sino - 460y3 900 34.64 Co-o 172.948 1b Cnsion For flement S L R0inhes, O-90 12-Sy4sin -EA L fs 460 83900 00026-0.000 12 ) (os T0 000 80-(-0.000 80)) 5in 0 L A0inche, 9-300 or element 6 ATCSx-d) o0tCS2-Sy5)sin0 7 L (-0000 26-0.00026) os 30 40 46083 900 +(-000080-(-0o0205)) 5in 30 201 233 bs (ro mp La 94.64 n -t tud u E A

46083 900 34.64 (0.00013-0.00 0 26) ros o t(-0.000 80- (-0.00 205J) in0 172.948 1by (Fension) NO HMGI Ste ieach ele ment 1s A -2(7667 6si rombiessive) 345 895 Th89-7 126. 754 bsi 201.425 1.5891 172.948 1-5891 (fensile) F4 108.834 psi Cfensile) F5 (-5891 201.233 58 122.948 -126.633 6s (rompsive) (03.834 bs Cfens ile) MRI A Answet