Struc lecture

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Struc lecture

  1. 1. Theory of Structures(1) Lecture No. 5
  2. 2. Statically Determinate Arches An arch : is a curved beam supported at its two ends. Most known types are: fixed- two hinged- three hinged. It can be analyzed by either analytical or graphical methods
  3. 3. Analytical Method Three Hinged Arch 1-The reactions of the arch can be calculated from the three equations of equilibrium and conditional equation at the intermediate hinge. 2- The bending moment at any section determined by calculating the moments of all the forces to the right or to the left of the section. 3- Determining normal and shearing force requires resolving resultant of all forces to the left or to the right of section along the tangent and the normal to it.
  4. 4. 3-The slope of the tangent can be defined if the equation of all the arch is defined first. 4-The slope of the tangent is equal to the first derivative of the equation. 5-Substitute by the coordinates of the specified point in the first derivative to calculate the slope.
  5. 5. Determine the N.F, S.F., and M at points d and e of the three-hinged parabolic arch shown The arch equation is Y= x – 0.025 x 2 Solution Σ Ma= 0 = Yb × 40 – 6 (10 + 20 + 30) Yb = 9 t ↑ Σ Y= 0 = Ya + 9 – 3 × 6 Ya = 9 t ↑ = 0 = 9 × 20 – 6 × 10 – 10 Xa Xa = 12 t -> ΣX = 0 = 12 – Xb Xb = 12 t ←
  6. 6. A section through point d At x = 5, y = Yd = 5 - 0.025 × 5 2 = 4.375 m. = 1-0.05x=1 – 0.05 * 5 = 0.75 = 0.6 and = 0.8 Σ X = 0 = 12 – X X= 12t ← Σ Y = 0 = 9 – Y Y= 9t ↓ ΣMd = 0 = 12 × 4.375 – 9 × 5 + M M = - 7.5 m.t., i.e M= 7.5m.t. (clockwise) Ɵ1 Ysin Ɵ1 Ycos Ɵ1 Xcos Ɵ1 Xsin Ɵ1
  7. 7. Ɵ1 Ysin Ɵ1 Ycos Ɵ1 Xcos Ɵ1 Xsin Ɵ1
  8. 8. At x = 30 y = y e = 30 – 0.025 × 30 2 = 7.5 m . At Point e 180-Ɵ2=153.43 Ɵ2=26.57 = 1 – 0.05 × 30 = = 0.50, = 0.446 = 0.893
  9. 9. ΣMe = 0 = M + 12 × 7.5 – 9 × 10 M is the bending moment at point e = 0 m.t Ne= - (X cos Ɵ2+ Y sin Ɵ2 ) Ne = - (12 × 0.893 + 9 × 0.446) = - 14.74 t Qe=X sin Ɵ2 - Y cos Ɵ2 Qe = 12 × 0.446 – 9 × 0.893 = - 2.685 t For a section just to the right of e ΣX = 0 = 12 – X X = 12 t -> ΣY = 0 = 9 – Y Y = 9 t ↓ 7.5m Ɵ2 X cos Ɵ2 X sin Ɵ2 Y cos Ɵ2 Y sin Ɵ2
  10. 10. For a section just to the left of e ΣX = 0 = 12 – X X = 12 t -> ΣY = 0 = 9 – 6 – Y Y= 3 t ↓ Ne= - (X cos Ɵ2+ Y sin Ɵ2 ) Ne = - (12 × 0.893 + 3 × 0.446) = - 12.05t Qe=X sin Ɵ2 - Y cos Ɵ2 Qe = 12 × 0.446 – 3 × 0.893 = 2.673 t Y cos Ɵ2 X sin Ɵ2 Y sin Ɵ2 X cos Ɵ2 Ɵ2
  11. 11. Example 4.2 : Determine the N.F, S.F., and M at point d of the three-hinged circular arch shown The equation of the arch is Y=ax 2 +bx+c At A x=0 y=0 c=0 At x=8 y=4 4=a(8) 2 + 8b 1=16a+2b (1) A t x=16 y=0 0=a(16) 2 + 16b -b = 16a (2) Substitute from 2 into 1 1=-b+2b b=1 a=-1/16 Y=-x 2 /16 +x
  12. 12. Example 4.2 : Determine the N.F, S.F., and M at point d of the three-hinged circular arch shown The equation of the arch is Y=-x 2 /16+x ΣMb= 0 = Ya × 16+ 4 ×2 – 6 × 10 - 8 × 4- 2 ×2 Ya = 5.5t ΣY= 0 Yb=10.5t ΣMc= 0 for left part 6 ×2+4 ×2 + 4 Xa - 5.5 ×8 = 0 Xa=6t ΣX= 0 Xb=10t Y a = 5.5t Yb =10.5t Xa = 6t Xb = 10t
  13. 13. Y=-X 2 /16+X At X=11 Y = 3.44m dy /dx = -X / 8+1 dy/dx= tan( 180- Ɵ) = -0.375 Ɵ=20.56 SinƟ= 0.351 CosƟ = 0.936 A section through point d Ɵ 10.5 t 10t 10t 0.5t 10 cos Ɵ 0.5 sin Ɵ 10 sin Ɵ 0.5 cos Ɵ Ɵ
  14. 14. Nd= - (10 cos Ɵ+ 0.5 sin Ɵ ) Ne = - (10 ×0.936 + 0.5 × 0.351 ) = - 9.53 t Qd=10 sin Ɵ – 0.5 cos Ɵ Qe = 10 × 0.351 – 0.5 × 0. 936 = + 3.04 t Md + 8 ×1+ 2 ×3+ 10 ×3.44 = 10.5 ×5 Md = 4.1 clockwise Ɵ 10.5 t 10t 10t 0.5t 10 cos Ɵ 0.5 sin Ɵ 10 sin Ɵ 0.5 cos Ɵ 3.44 m 5 m 3 m
  15. 15. Problem 4.2 : Determine the N.F, S.F., and M at point S of the three-hinged circular arch shown Solution Σ MC= 0 = Yb × 6 – 3.6 Xb-2*6*3=0 YB = 0.6 XB + 6 Σ MA= 0 2.8 XB+14YB = 2*14*7 XB= 10 t YB=12 t S 4.8m
  16. 16. S 4.8m 10t 8t 60 60 8sin60 8cos60 10cos 60 10sin60 8t 16t 10t 4.8m Σ X= 0 XA=10t Σ Y= 0 YA=16 t Σ MS =MS-16*4+ 10*4.8+ 8*2=0 MS=0 Ns= - (10 sin 60+ 8 cos60 )=-12.66t Qs= 8 sin 60 -10 cos 60 = - 1.93t 4m 60

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