Struc lec. no.444

Theory of Structures(1) Lecture No. 4

Statically Determinate Rigid Frames A Frame : is a structure composed of a number of members connected together by joints all or some are rigid. Hinged or Pin-connection allows relative rotation Between the ends of the connected members. Rigid connection does not allow relative rotation between the ends of the connected members.

Internal Stability and Determinacy If No. of members= m No. of External reaction components = r No. of unknowns = 3m + r. No. of joints= j No. of equilibrium Eq.(s) = 3j; No. of condition Eq.(s) =S Total No. of available Eq.= 3j+S IF 3j + S > 3m + r, the frame is unstable. 3j + S = 3m + r, the frame is statically determinate. 3j + S < 3m + r, the frame is statically indeterminate. Equations Unknowns

Draw N.F, S.F. and B.M.D (s) for the shown frame 1) ∑ X = 0 X A = 3t 2) ∑ M B = 0 10 Y A + 3 × 2 – 8 × 7= 0 Y A = 5t 3) ∑ Y = 0 Y B = 3t Solution of Example 3.1 A B r = 3 J=4 S=0 m=3 3J+S=12 3m+r=12 Stable & Determinate

∑ X = 0 X A = 4t ∑ M B = 0 12 Y A – 24 × 6 – 4 × 6= 0 Y A = 14t ∑ Y = 0 Y B = 10t Solution of Example 3.2

Solution of Example 3.3 1) ∑ X = 0 X B = 4t 2) ∑ M B = 0 16 Y A + 4 × 6 – 32 × 8– 4 × 18 = 0 Y A = 19t 3) ∑ Y = 0 Y B = 17t r = 3 J=4 S=0 m=4 3J+S=15 3m+r=15 Stable & Determinate

Solution of Example 3.4 1) ∑ X = 0 X B = 2.5t 2) ∑ M A = 0 8 Y B + 2.5 × 2.5+ 3 × 2 – 3×0.75–1×10×3 = 0 Y B = 2.5t 3) ∑ Y = 0 Y A +Y B = 16t Y A = 13.5t 2.5 t 3m 10t B A

2.5 t 3.125 m 3.125 m WL 2 / 8 = 8 11.75 t.m

Solution of Example 3.5 1) ∑ X = 0 X A = 0 2) Y A = Y B =6t A B

6 sin α =5.7 α α 6 cos α =1. 9 sin α =0.948 cos α =0.316 α α 6 cos α =1. 9

Solution of External Example ∑ X = 0 X a =12 t ∑ M a = 0 12 × 2+ 16 × 3 –6Yb=0 Yb=12 t ∑ Y = 0 Y a +Y b =16 Ya= 4t 3m 3m 3 t/m 12 t 4m 12 t 12 t 4 t

36 4t 4t 12t 12t 12t 12 t 4 t 3m 3m 3 t/m 12 t 4m 12 t 12 t 4 t

Solution of External Example ∑ X = 0 Xa =5t ∑ M b = 0 8Y a + 15×2–4 × 8×4 –5 × 6 =0 Y a =16t ∑ Y = 0 Y A +Y b =32+15= 47 Y b = 31t 4t/m 5t 16t 31t 5t

31t 15t 16t 16t 16t 16t 16t 5t 32t 5t Wwwwwww

Solution of External Example ∑ M b = 0 for part ab 2×1.5+4 ×4.5 –Ya × 6 =0 Y a =3.5t ∑ Me = 0 Ya 15 ×3+ 3.5 × 8 + 3 × 4 = 2 ×3.5+4× 6.5 + 8Yd Y d = 6.5t ∑ X = 0 Xe =3t ∑ Y = 0 Y a +Y e + Yd=4+2+15 Ye= 11t 3t 4t 3t 11t 6.5t 3.5t 15t 5m

4 1 2.5 3 1.5 4 4m F g F g c c MF=3.5 ×1.5= 5.25t.m Mg=3.5 ×4.5-4 × 3= 3.75t.m Mc=3.5 ×8-4 × 6.5 -2 × 3.5-3 × 1= -8t.m 5.25 3.75 B.M.D 14.1 8 12

Struc lec. no.444

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Struc lec. no.444