A three hinged arch ABC consisting of quadrant parts AC and BC of radii 4m and 6m respectively carries two point loads of 100 KN each as shown fig. find the reactions at the supports A and B. and BM at E and F.
To find the reactions at supports A and B, as well as the bending moments at points E and F, we can use the method of individual point loads and moments.
Step 1: Resolve the loads
Let's assume that the point load of 100 KN at point C is divided into two equal point loads of 50 KN each, at points E and F, respectively.
Step 2: Calculate reactions at supports
We'll consider the equilibrium of vertical and horizontal forces at the supports. Since the hinged arch is in equilibrium, the net forces and moments acting on it must balance.
a) Reaction at support A:
Considering the vertical forces, we have:
R_A - 100 KN - 50 KN = 0
R_A = 150 KN
Considering the horizontal forces, we have:
-H_B + H_A = 0 (as there are no other horizontal forces involved)
H_B = H_A
b) Reaction at support B:
Considering the vertical forces, we have:
R_B - 100 KN - 50 KN = 0
R_B = 150 KN
Step 3: Calculate bending moments at points E and F
a) Bending moment at point E:
To find the bending moment at E, we'll consider the equilibrium of moments about point E.
Let the clockwise moments be positive:
50 KN × 4m + 100 KN × 5m - R_A × 22m = 0
Substituting the known values:
200m + 500m - 150 KN × 22m = 0
200m + 500m - 3300m = 0
700m = 3300m
700m / 3300m = 1/4.714
Thus, the bending moment at point E is approximately 1/4.714 times the total reactions.
b) Bending moment at point F:
To find the bending moment at F, we'll consider the equilibrium of moments about point F.
Let the clockwise moments be positive:
50 KN × 4m - 100 KN × 3m - R_B × 12m = 0
Substituting the known values:
200m - 300m - 150 KN × 12m = 0
200m - 300m - 1800m = 0
-900m = 0
m = 0 (This means there is no bending moment at point F)
Therefore, the bending moment at point E is approximately 1/4.714 times the total reactions, and there is no bending moment at point F.
Summary of results:
Reactions at support A: R_A = 150 KN
Reactions at support B: R_B = 150 KN
Bending moment at point E: Approximately 1/4.714 times the total reactions
Bending moment at point F: No bending moment (i.e., 0)
Answers & Comments
To find the reactions at supports A and B, as well as the bending moments at points E and F, we can use the method of individual point loads and moments.
Step 1: Resolve the loads
Let's assume that the point load of 100 KN at point C is divided into two equal point loads of 50 KN each, at points E and F, respectively.
Step 2: Calculate reactions at supports
We'll consider the equilibrium of vertical and horizontal forces at the supports. Since the hinged arch is in equilibrium, the net forces and moments acting on it must balance.
a) Reaction at support A:
Considering the vertical forces, we have:
R_A - 100 KN - 50 KN = 0
R_A = 150 KN
Considering the horizontal forces, we have:
-H_B + H_A = 0 (as there are no other horizontal forces involved)
H_B = H_A
b) Reaction at support B:
Considering the vertical forces, we have:
R_B - 100 KN - 50 KN = 0
R_B = 150 KN
Step 3: Calculate bending moments at points E and F
a) Bending moment at point E:
To find the bending moment at E, we'll consider the equilibrium of moments about point E.
Let the clockwise moments be positive:
50 KN × 4m + 100 KN × 5m - R_A × 22m = 0
Substituting the known values:
200m + 500m - 150 KN × 22m = 0
200m + 500m - 3300m = 0
700m = 3300m
700m / 3300m = 1/4.714
Thus, the bending moment at point E is approximately 1/4.714 times the total reactions.
b) Bending moment at point F:
To find the bending moment at F, we'll consider the equilibrium of moments about point F.
Let the clockwise moments be positive:
50 KN × 4m - 100 KN × 3m - R_B × 12m = 0
Substituting the known values:
200m - 300m - 150 KN × 12m = 0
200m - 300m - 1800m = 0
-900m = 0
m = 0 (This means there is no bending moment at point F)
Therefore, the bending moment at point E is approximately 1/4.714 times the total reactions, and there is no bending moment at point F.
Summary of results:
Reactions at support A: R_A = 150 KN
Reactions at support B: R_B = 150 KN
Bending moment at point E: Approximately 1/4.714 times the total reactions
Bending moment at point F: No bending moment (i.e., 0)