2. Statically Determinate Trusses A truss is a structure consists of a number of straight members Pin-connected together The members at the joints are riveted or welded at their ends. All the truss members will be link members subjected to either axial tension or compression. Members carrying tension are usually called ties and those carrying compression are termed struts .
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4. Stability and Determinancy 1-Assume a truss and let it have (m) members, (j) joints and (r) external reaction components. 2-The number of unknowns is equal to m+r. 3-Each joint is subjected to a system of concurrent forces in equilibrium, the number of the available equations is equal to 2j; two equations for each joint . If 2J = m+r , the truss is stable and statically determinate 2J > m+r, the truss is unstable 2J < m+r, the truss is statically indeterminate
5. Methods of Analysis for Statically Determinate Trusses 1-Method of Joints. 2-Method of Sections. 3-Stress Diagram Method (Graphical Method)
6. Method of Joints Determine the reaction components from the equations of equilibrium for the structure. The analysis should start with a joint where two members only meet Applying the two equations available for a set of concurrent forces in equilibrium, Σx=0 and Σy=0. After finding the forces in these two members, an adjacent joint will have two unknown forces only. Compression Force Tension Force
7. 3 0t Example 5.1 a J= 10 2J=20 m=17 r=3 2J=m+r The truss is stable and statically determinate ƩX=0 Xa=0 7t 8t 2 1 4 5 6
8. F jh =7t in comp . F kj 3 7t 8t 2 1 4 5 6 0t Example 5.1 a 3 F bj F Jc
9. F jc 7t 8t 2 1 4 5 6 F hc F bc F jc 7 8 F hg F cg F cg 3 9 At e, ƩY=0 F ef = 8t in comp. ƩX=0 F de = 0 t 8 At F, ƩY=0 F ef = 4+ F df × 1/ F df = 4 in tension
10. ΣM a = 0 = 2 × 4 - Yb × 2 Y b = 4 t. ↓ ΣY = 0 = 2 + 4 – Ya Y a = 6 t. ↑ ΣX = 0 X b = 0 Example 5.2 b Equilibrium of joint a ΣX F ab =0 d ΣX F cd =0 b ΣX F cb =0 c ΣX F cd =0, F cb =0 F cf =0 g ΣX F hg =0 g ΣY F fg =0 K ΣY F jk =0 J ΣY F je =0 6t 4t J= 11 2J=22 m=19 r=3 2J=m+r The truss is stable and statically determinate 0t
11. F jh = F jh 6t 4t At a ΣY = 0 F ac = 6t in comp. At b ΣY = 0 F bd = 4t in ten. At d ΣY = 0 F df = F bd =4t in ten. At c ΣY = 0 F ce =F ac = 6t in comp. At i ΣY = 0 F ij ×1/ =2 F ij = 2 t ΣX = 0 F ik = F ij ×2/ = 4t 1 2 3 4 5 6 At k ΣX = 0 F ke = F ik =4t in comp. At J F ij = F jh =2 in ten. At e ΣY = 0 F eh = F ce = 6t in comp. ΣX = 0 F ef = F ke =4t in comp. At f ΣY = 0 F df =F hf ×1/ F hf = 4 8 7 9
12. Sheet 5 Prob. 11 From symmetry of loading Ya = Yb = 72/2= 36t ΣMc = 0 for left part 12×4+12 ×8+6 ×12+4.5 X A =12 Y A =12 ×36 X A = 48t ΣX = 0 X B = 48t 2J= 32 M+r =26 +6=32 The truss is stable and statically determinate 8 m Y B Y A X B X A C
13. For Part AC At A ΣX = 0 0.8 F AE cos = X A =48 F AE = 60 t in comp. ΣY = 0 60 0.6 F AE sin + F AD = YA=36t F AD = 0t Ɵ Ɵ 48t 48t 36t 6t 8m
14. At D ΣY = 0 0.781 F DE sin = 6 F DE = 7.68 t in comp . ΣX = 0 0.624 F DF =F DE cos = 4.8t in tens. At E ΣX = 0 0.624 0.8 0.97 F DE cos + F AE cos = F EG cos F EG =54.42t in comp. ΣY = 0 F DE sin +F EG sin + F EF = F AE sin F EF =16.8t in comp. α Ɵ α Ɵ 6t 8m Ɵ 48t 36t 48t
15. At F ΣY = 0 16.81 F FG sin 45+ 12 = F EF F FG = 6.8 t in tens. ΣX = 0 F DF + F FH = F FG cos 45 4.8+ F FH = 4.8 F FH = 0 t At L ΣY = 0 F CL = 6t in comp. ΣX = 0 F HL = 0 t 45 6t 8m C 36t 48t
16. 45 At H ΣX = 0 F HC = 0 t ΣY = 0 F HG = 12t in comp. At G ΣX = 0 0.99 0.97 F GF cos 45 + F GC cos = F GE cos F GC = 48.46 t in comp α 6t C 36t 48t
18. At a ΣX = 0 0.477 Fac cos =6 Fac=13.42t in tens. At b ΣX = 0 Fbc cos =6 Fbc=13.42t in tens. ΣY = 0 0.894 Fbc sin +18 = Fbe Fbe=30t in comp. At c ΣY = 0 Fcd sin =Fcb sin +Fac sin Fcd= 26.84t in tens. ΣX = 0 Fcd cos +Fcb cos = Fac cos +Fce Fce =12t in comp. 1 3 2 Ɵ Ɵ Ɵ Ɵ Ɵ Ɵ Ɵ Ɵ Ɵ Ɵ Ɵ Ɵ Ɵ 18t 12t 6t 6t
19. At d ΣX = 0 Fdf sin = Fcd cos Fdf = 13.42t in tens. ΣY = 0 Fde= Fdf cos +Fcd sin Fde= 30t in comp. At e ΣY= 0 30 30 Fef sin 45+ Fbe = Fde Fef = 0t ΣX = 0 Feg= Fce Feg =12 t in comp 1 3 2 Ɵ Ɵ Ɵ Ɵ Ɵ 4 Ɵ Ɵ Ɵ Ɵ 45 5 Ɵ Ɵ 18t 12t 6t 6t
20. At f ΣX = 0 Ffh = Fdf Ffh= 13.42 in tens. ΣY = 0 Fgf = 0t At g ΣY = 0 fgh = 0t ΣX = 0 Fgj= Feg Fgj= 12t in comp. At J ΣY = 0 Fhj=0t ΣX = 0 Fji=Fgj Fji=12 t in comp. At i ΣY = 0 Fhi cos = 6 Fhi= 13.42 in tens. 1 3 2 Ɵ Ɵ Ɵ Ɵ 4 Ɵ 45 5 Ɵ 6 7 8 9 10 Ɵ Ɵ Ɵ Ɵ 18t 12t 6t 6t
