# Structural Analysis Example: Arch Bridges

For the given arch bridge loaded, as shown above, determine the following;

(a) The support reactions
(b) The bending moment diagram of the girder
(c) The shear force diagram of the girder
(d) The axial force diagram of the girder

## Support Reactions

âˆ‘MB = 0
30Ay â€“ (4 Ã— 15) Ã— (15/2 + 15) = 0
30Ay â€“ 1350 = 0
Therefore; Ay = 1350/30 = 45 kN

âˆ‘MGL= 0
15Ay â€“ (4 Ã— 152)/2 + 5H = 0
15(45) + 5H = 0
Therefore; H = -225/5 = -45 kN

âˆ‘MA = 0
30By â€“ (4 Ã— 152)/2 = 0
30By â€“ 450 = 0
Therefore; By = 450/30 = 15 kN

âˆ‘MGR= 0
15By + 5H = 0
15(15) + 5H = 0
Therefore; H = -225/5 = -45 kN

## Analysis of the Joints

Joint 2

Ï† = tan-1(1/6) = 9.46Â°

âˆ‘FX = 0
-F2-1cosÏ† + H = 0
-F2-1cos9.46 â€“ 45 = 0
-F2-1 = 45/cos 9.46 = 45.62 kN
F2-1 = -45.62 kN

âˆ‘FY = 0
-F2-1sinÏ† â€“ F2-6 = 0
-F2-6 = F2-1sinÏ†
-F2-6 = -45.62 sin9.46 = 7.498 kN
F2-6 = 7.498 kN

Joint 1

Î± = tan-1(4/6) = 33.69Â°

âˆ‘FX = 0
-F1-AcosÎ± + F1-2sinÏ† = 0
-F1-AcosÎ± â€“ 45.62sin(9.46) = 0
-F1-A = 45.62sin(9.46)/cos(33.69) = 54.082 kN
F1-A = -54.082 kN

âˆ‘Fy = 0
-F1-AsinÎ± + F1-2sinÏ† â€“ F1-5 = 0
54.082sin(33.69) â€“ 45.62sin9.46 â€“ F1-5 = 0
22.501 â€“ F1-5 = 0
F1-5 = 22.501 kN

Therefore, the loading on the girder can be summarised as follows;

When the inclined forces are resolved into their vertical and horizontal components;

Fy = 54.082sinÎ± = 54.082sin(33.69) = 30 kN
Fx = 54.082cosÎ± = 54.082cos(33.69) = 45 kN

Check;
âˆ‘Fy â†‘ = 45 + 15 + 2(22.501)+ 2(7.498) = 120 kN
âˆ‘Fy â†“ = (4 Ã— 15) + 2(30) = 120 kN

## Bending Moment

(Coming from the left)
MA = 0
M5 = (45 Ã— 6) â€“ (30 Ã— 6) â€“ (4 Ã— 62)/2 = 18 kNm
M6 = (45 Ã— 12) â€“ (30 Ã— 12) â€“ (4 Ã— 122)/2 + (22.501 Ã— 6) = 27 kNm
MGL = (45 Ã— 15) â€“ (30 Ã— 15) â€“ (4 Ã— 152)/2 + (22.501 Ã— 9) + (7.498 Ã— 3) = 0
(Coming from the right)
MB = 0
M8 = (15 Ã— 6) â€“ (30 Ã— 6) = -90 kNm
M7 = (15 Ã— 12) â€“ (30 Ã— 12) + (22.501 Ã— 6) = -45 kNm
MGR = (15 Ã— 15) â€“ (30 Ã— 15) + (22.501 Ã— 9) + (7.498 Ã— 3) = 0

## Shear Force

VA = 45 â€“ 30 = 15 kN
V5L = 45 â€“ 30 â€“ (4 Ã— 6) = -9 kN
V5R = 45 â€“ 30 â€“ (4 Ã— 6) + 22.501 = 13.501 kN
V6L = 45 â€“ 30 â€“ (4 Ã— 12) + 22.501 = -10.499 kN
V6R = 45 â€“ 30 â€“ (4 Ã— 12) + 22.501 + 7.498 = -3 kN
VG = 45 â€“ 30 â€“ (4 Ã— 15) + 22.501 + 7.498 = -15 kN
V7L = -15 kN
V7R = -15 + 7.498 = -7.502 kN
V8L = -7.502 kN
V8R = -7.502 + 22.510 = 15 kN
VB = 15 kN

## Axial Force

NA-B = H = Ax = 45 kN (Tension)