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Shear force, Bending Moment and Axial Force Diagrams.
For the strip of width (dx) infinitesimal element
V + dv = v + w *dx
W= ------------- 1
Slope of shearing force diagram of any section equal to the intensiting the load
at that section.
By integration:-
] = ∫ ( )
V 2 - V 1 = ∫ ( )
∫ ( )^ --------------
Changing in shearing force between x 1 &x 2 section equals the total applied load
between x 1 &x 2.
M + dM = M + Vdx +
( )
dM = v. dx
Vx = ---------------- 3
Slope of bending moment diagram at any section equals the shear force at that
section.
By integration:-
] = ∫ ( )
M 2 - M 1 = ∫ ( )
∫ ( )^ -------------- 4
Change in bending moment diagram between section x 1 & x 2 equal to the area
under (v – diagram) between x 1 & x 2 plus any additional concentrated moment
applied between x 1 & x 2.
Differentiate Equation:-
With respect to x
W = ------------ 5
EX1:- Draw the axial force; shear force and bending moment diagrams.
3 m 2 m 3 m 2 m 3 m
30 kN/ m 32.5 kN 100 kN
A B^ C^ D^ E
45 25
-7.
2nd degree^45
45 30
105
3rd degree
S.F.D
B.M.D
52.5 kN 120 kN
30 kN/ m
F
A 1 = 1/ (n+1), الكيرف مقعر A 2 = n / (n+1),الكيرف محدب
EX2: Draw the axial force; shear force and bending moment diagrams for the
frame shown below.
10 kN/ m
50 kN (^) 50 kN 20 kN
10 kN 5 m
2 m
1 m
2 m 1 m 2 m 4 m
A
B C E
D
40 kN
105 kN
15 kN
Sol: - SEGMENT (AB)
10 kN/ m
5 m
A
B
A.F.D S.F.D B.M.D
4 m
1 m
Segment BCE
50 kN (^) 50 kN 20 kN
2 m 1 m 2 m 4 m
B C E
15 kN
10 kN
105 kN
10 kN
A.F.D
S.F.D
B.M.D
EX3:- Draw the axial force; shear force and bending moment diagrams for the
frame as shown below.
Sol:
∑
Cy6 – (1202) =0, Cy= 40 kN
∑ (^) , 40 + Ay = 0, Ay= -40 kN = 40 kN
∑ , 120 – Ax = 0, Ax = 120 kN
For segment. AB
∑ , -40 + By = 0, By = 40 kN
∑ (^) , 120 – 120 – Bx = 0, Bx = 0
∑ , MB + 1202 + 403 -120*4 =
MB= 120 kN.m
Segment BC
32 24
24 32
120
120
A.F.D
S.F.D
B.M.D
EX4: Draw the axial force; shear force and bending moment diagrams for the
frame shown below.
Sol:
For segment (DEF)
∑ ,
Fy * 2 – (50*1) =
Fy= 25 kN
For whole structure
∑ => Fx6 + (254) – (50*3) – (10 1) – (403) =
Fx = 30 kN
∑ ,
Ay +25 -50- (5*2) = 0, => Ay= 35 kN
∑ , => -Ax – 30 + 40 = 0, => Ax= 10 kN
Segment ABC
40
10
35
35
30
40
10
35
35
30
1010
30
60
A.F.D S.F.D B.M.D
60
EX5: Draw the axial force; shear force and bending moment diagrams for the
frame shown below.
1 kN/ m
20 kN
10 m
A
B C
D
5 m 5 m
5 kN (^) 15 kN
10 kN
Sol:-
For Member AB:
1 kN/ m
10 m
A
B
5 kN
10 kN
5 kN 50
A.F.D S.F.D B.M.D
10
50
For Member CD:
C
D
15
15
Zero Zero
A.F.D S.F.D B.M.D
For Member BC:
40 kN
B
2 m 3 m
C
4 m
A
3 m
10 kN/ m
