Question

Safari . 7:08 PM 1. The truss shown is used to support the floor deck. The uniform load on the deck is 3 kit. This load is transferred from the deck to the floor beams, which rest on the top joints of the truss. (Figure is not to scale) a. Determine whether the truss is statically determinant/indeterminant unstable. ( Points) b. Convert the uniformly distributed loading on the truss to point loads (10 Points) c. Calculate the support reactions for the truss (10 Points) d. Can the truss achieve greatest economy? Explain your answer (5 Points) e. Calculate the forces in all the members (1745-85) 3 k/ft 10 ft - 12 ft 14 ft 16 ft-

Answer 1

13K/ft 10 fit a soft * 12ft *14ft * soft The An). Degree of static indeterminacy of a truss is given by Ds = m + R-2j where, m= Number of members = 17 R= Total no. of external reactions = 3 {RA. HA REJ j= total no. of joints = 10 {A, B, C, D, E, F, G, H, I, J} ". Ds = m+ R-zj = 17+3 - (2x16) - 20 - 20 = 0 since, Ds=0 Therefore the given buss is statically determinate fly 6 Ans since the span length is varying for each span. So, we will calculate total load in each span and then we will distribute them equally on the two corresponding joints of that particular span. The distribution of UOL into point load is done on the joints only due to the fact that tresses are not allowed for moments & loads are only applied at joints, otherwise moment will be generated which is not allowed in a tress structure.

total load in JI = 34 10 = 30k i load on joint j = 30 = 15k & » » » I = 30 = 15k total load on TH = 3x12 = 36k i load on joint I = 36 = 18k & load on joint H = 36 = 18k a load on HG = 3x 14 = 42K : load on joint H&G = 4 total load - 42 - 24K total load on GF = 3x16 = 48K i load on joint G&F = 48 = 24k The common joint loads will be superimposed & truss becomes like following 15k (15+18)= 334 (18+21) = 39k 124 +24) = ASK 24K 1040 * soft x 12ft x 14 ft x 16ft total load on truss → Before distribution = 3x (10+12+14+16) = 156k ? → After distribution = (15+ 37+39+45+24) = 156k 5 Equal Menace verified.

thy Refer the figure in part 6 solution. Efv=0 > RATRE = (15 + 33 +39 +45 +24) « » RATRE = 156K 0 E FH=0 - CHA=0 Aul EM @ A=0 > Rex (40+12 +44 +16) = 24x16) + (45 x 36) + (39 x 22) + (33X 10) +(1580) » REX 5 = 3192 [RE = 61.38 A Put this RE in equation ④ RA = 156k - 61.386 = 94.62k . TRA= 94.62K) all @ As). Consider the equilibrium of joint c. As there is no any external force in vertical direction FCH to balance FCH Therefore, for the equilibrium of joint c, FeH=0 All Refer fig in part ① solution o= tant ( 10 ) = 45° B = tant ( 10 ) = 35.54° 2 = tant ( 10) = 32° &= tant (10) = 39.8° per so ER

Consider joint A Y FAT A $94.62K FAB E fv = 0 [FAJ = 94.62 k (compressive) And EFH=0 = [FAB = 0 Consider Joint I = 45 Efu zo Vask sho E - Fjg sino +(15 k) = 94.62k - Frex sin 45' = 79.62k > FR = 79.62 k = 112.6K sints' For = 112.6K (Tensile) Stud EFH=0 → FJI = FR Coso = 112.6 X COS 45° = 79.62 K : For = 79.62 K (compressive) and Consider Joint I 1334 και εκ τ τ. TFB EF=0 => [FIB = 33k (compressive) And Στη ο = [TH = 79.6 2K (Compressive) A

Consider Joint B {Fr=0 =) FBH sind + 112.6 sino = 33K = FRH (Sin 39-8) +(112.6 x sin 45") = 33K FBH BH BATO . Fou = -72.83K or [FBH = 72.83 K (compressive) skal ) - - | FH 0 FBH Cosopt FBC = 142.6 cose -12 83 x Co 5 39 g) + Fax - 44 6 x Co 545 FBC = 135.57K (Tensile) And Consider joint c A FCHO FrH=0 E FH=0 =) Fco = FEB = 135.57K Fco= 135.57K (Tensile) a c Fco o : Fo= 135.57K Consider Joint E EFHTO YFEF { Fu=0 61.38K (compress TRE = 61.38K

Consider joint F x=32° V/ 24K a f 1 61.38k Efuzo => Fig sina +(24K ) = 61-38 W) = FFS = 37.384 = 70.54k Ffp= 70.54k (Tensile) A 37.38K - F 70.54K FES ↑ 61-38K Sin 32" E Fu=0 3 Fe = FFA COSX = 70.54 x Cos 32° = 59-82K Fra = 59.82% comprevine) = 59.82K (Comprewire) consider Joint G - 45k 59.82K ΣΕ Ο >> Fad = 45k (Compressive) EFH o FGH = 59.824 (compressive) FGH As AFGD consider joint H V 39K Y 59.82K 79.62K H VB I X FHO Efuzo =) FHD sinß + 72.83 sind = 39k & Fuo = 39K - (72.83 k) (sin 39.8°) Sin 35-54 =) Fro - 13.1k or Fho = 13.1K (Tensile) A 72.8K FHD