Question

Question 7 Determine the reactions and the force in each member of the following truss using the Force Method. 30k 1609 40k - 2 panels at 8 fft = 16 ft - EA = constant Page 3 of 4

Answer 1

STEPS

1. First, find the static indeterminacy of structure
2. Then find whether it is internal or external indeterminacy
3. Here it was external indeterminacy, so remove the external reaction
4. Here I have removed Cx and found the displacement at C due to given loading
5. Then find Displacement at C due to unit load
6. The horizontal displacement at C should be zero. From this condition, we will get the redundant force
7. Then we can find the member forces

30k D - 60 CM 40k to Degree of Static Indeterminacy, Ds = m + re- 2; m= No. of members = 5 j= Total No. of joints = 4 re= Total enternal reactions = 4 : Ds = 5+ 4 - 2x4 Ds=1 Stoucture is statically indeterminent * It is because of an additional enternal reaction. - For a structure, we have only 3 equilibrium equations (& Fn = 0 , { Fy = 0, CM₂ = 0) But here we have 4 reaction Forces of so we have to take one foce as redundant. - Take Ca as redundant

Resolving - > 40 cos 60 60 81 Anvelt det 40k 40 Sin 60 $40 ces 60 40 Sin 60 * 20k 2013 So Finding An, Ay and Cy EMA=0 30x6+ 2053 x 8 = Cy x16 f alt [28lt fcy 2003 K 28. 57k Cy=28.57k {Fy = 0 . Ay + Cy = 2053 Ay= 2053 - 28.57=6.07k EFx = 0 Ax+20+30=0 An=-50 30k sok A Book oft 811 2013k 602 28.57%

Now we med to find force in each member. Tensile member torces = tre compressive member torces = ove At joint ostant (&) = tan 3 A Star () 50. B 8 B 0 = 36.9° Efn=0 50 = FAB & FAD cos 36.9 - 607 o {fy=0 force) 6.07+ FAQ Sin 36.9=0 FAD = - 10.12 k (compressive Put FAD in equation o 50 - FAB + (-10.12) cos(36.9°) FAB = 58.1k

At joint D FOR 10.12 - Fon 531153) R_>30 531 53.1 For troB Efn=0 FDA sin 53.1= foc sin 53.1 +30 -10.12 sin 53.t! - For Sin 53.1 +30 Foc = – 47.63k A-336-9 – – – – – – 361-C {fy=0 FDA COS 53.1 + FOB & Foc cos 53.1 = 0 -10.12 cos 53:1 + FOB + (-47-63) COS 53.) = 0 FOR = 34.67k

at joint B AFBD. {fy = 0 FB = 34.67 = ABD 2013 - 20x1.734 = 34.68 So it is correct 2013 { Fn = 0 FBA = 20 + FBC 58.1= 20+ Fec Fac = 38.1k So our structure with Member aR30k Forces 47.63 134.67 47.63 10.12 434.67 38 | 50 Ar 581 6.07 Member forces unit load is of 581 B k 38.1 i 2038 28.57 in Structure when a applied in the direction Ca - D E B

© only it is acting horizontally @c. No enternal forces. we need to find Ay, An, and Cy EM=0 / Ik horizontal force has no moment about a So Ay=0 Ef=0 . at joint c A at joint at D joint Á 'B voi FED=0 FAD=0 FCB = elk FAB = 1 u

AE Horizontal Displacement of c due to given loadings - SPKL P= member forces due to given Loading k=member forces due to unit load at C L = Length of the memberi Horizontal displacement at c, due to unit load. = {k²L But Cn is the reactiona acting at C .: Horizontal displacement due to Cn. = { Cek? Since there is no support Settlement. Horz displacement at C = 0 AE AE e, & (PKL) + Cu < (142) = 0 AE = constant. .: EPKL + (n EK²=0 x=-{PKL {K²L

According to consistency conditions Final force in member = Pt Cak K² I forces (k) Final - 10.12 Calculation Table Member Length forces forces Member AB 8 581 1 464.8 10 AD | 10 -10.12 o BD6 34.67 olo o 34.67 BC 8 38.11 304.8 8. -10 CD 10 -97.63 o lo lo - 4763 3PKL = 769:6| Ek?2 = 16 lololololo G = - EPK L = -769.6 = – 48.1k {K ²L 16

30k A 1.9k 481 a 40K 40K 6.07K 28.57k Efn = 0 :: An + 48.1=30+20 An=1.9k found Note In the previous page, Member forces are using S=Ptluk P=member forces due given Loading k=member forces due to unit Load Ca = Redundant force. S=final Force on member Sample Calculation for member AB AB, P = 58.1 k=1 Ca= - 48.1 S = 58.1 + (-48:1) x) = 10k