The document discusses statically determinate arches and their analysis using analytical methods. It provides examples of calculating reactions, bending moments, normal forces, and shear forces at different points along three-hinged parabolic and circular arches given their equations. Calculations are shown for determining these values at various points along the arches using equations of equilibrium and geometry.

1. Theory of Structures(1) Lecture No. 5

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. 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. 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. 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. 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. Ɵ1 Ysin Ɵ1 Ycos Ɵ1 Xcos Ɵ1 Xsin Ɵ1

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. 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. 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. 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. 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. 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. 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. 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. 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