CHAPTER 7:
Forces in Beams
„
Beams
– Various types of loading and support
– Shear and bending moment in a beam
– Shear and bending moment diagrams
– Relations among Load, Shear, and
Bending Moment
1
7.1 Internal Forces in Members
F
B
B
F
F
B
C
B
F
C
A
A
C
F
F
C
A
F
C
F
F
C
F
A
F
Tension
F
Compression
2
D
E
F
C
B
W
A
G
D
J
T
N
D
C
Cy
B
Ax
Cx
FEB
J V
T
N
V
M
J
C
Cy
B
M
Cx
FEB
Ax
A
Ay
A
Ay
3
B
B
M
D
N
D
P
P
A
V
C
P
M
N
A
V
P
D
C
4
Example 7.1
In the frame shown, determine the internal forces (a) in member ACF at
point J and in member BCD at point K. (b) determine the equation of
internal forces of member ACD and draw its diagram. This frame has
been previously considered in sample
3.6 m
1000 N
A
1.2 m
2400 N
J
2.7 m
B
K
2.7 m
1.5 m
C
E
4.8 m
D
F
Guide : Assuming that Frame
can be treated as a rigid body.
The reactions and the forces
acting on each member of the
frame are determined.
Member ACF is cut at point J
and member BCD is cut at
point K, the internal forces are
represented by an equivalent
force-couple system and can
be determined by considering
the equilibrium of either part.
5
(a)
• Define the state of the problem
3.6 m
1200 N
A
1.2 m
2400 N
J
2.7 m
B
K
2.7 m
1.5 m
C
E
D
F
4.8 m
Assuming that Frame can be treated as a rigid body. Member ACF is
cut at point J and member BCD is cut at point K, the internal forces are
represented by an equivalent force-couple system.
6
• Construct the physical model, construct mathematical model and
3.6 m
solve equations
1200 N
1.2 m
A
2400 N
2.7 m
B
C
2.7 m
Ex = 1200 N
450 N
1200 N
2.7 m
Ex = 1200 N
E
Ey = 750 N
A
2.7 m
B
1.5 m
4.8 m
D
F
A
2400 N
2400 N
1200 N
E
450 N
F = 3150 N
2.4 m 1.2 m 2400 N
2400 N
D
B
C
2400 N
1200 N 3600 N
2.7 m
B
2400 N
3600 N
C
2.7 m
F
4.8 m
750 N
2400 N
3150 N
7
• The diagram of internal forces of member BCD
Nx1 = +2400 N
Nx2 = 0
Vx1= -1200 N
Vx2= 2400 N
Mx1 = -1200x1 N•m
Mx2 = -2400x2 N•m
2.4 m 1.2 m 2400 N
2400 N
B
x1
D
C 2400 N
1200 N 3600 N
x2
N (N)
+2400
x
2400
V (N)
A = +2880
x
A = -2880
-1200
M (N•m)
x
-1800
-2880
8
BEAMS
7.2 Various Types of Loading and Support
P1
A
B
P2
C
D
Concentrated loads
w
A
B
Distributed load
C
9
L
L1
Simply supported beam
L2
Continuous beam
L
L
Overhanging beam
Beam fixed at one end and simply
supported at the other end
L
L
Cantilever beam
Fixed beam
Hinge
10
7.3 Shear and Bending Moment in a Beam
Internal Loadings at a Specified Point
• Sign Convention
M
M
N
V
N
V
N
N
V
M
M
V
Positive Sign Convention
11
Shear and Moment Functions
P
x1
x2
A
B
L/2
P/2
x1
L/2
P/2
Mx1
A
P/2
+
+
+ ΣMx1 = 0: Mx1 = (P/2)x1
Vx1
Mx2
x2
ΣFy = 0: Vx2 = -P/2
+ ΣMx2 = 0: Mx2 = (P/2)x2
V
ΣFy = 0: Vx1 = P/2
B
Vx2
P/2
P/2
x
M
PL/4
-P/2
x
12
P
w
A
B
x1
D
C
x2
x3
OR
P
w
A
x1
B
x2
C
D
x3
13
7.4 Relations Among Load, Shear, and Bending
F1
F2
F3
Moment
w = w(x)
A
.
.
.
.
M1
B
C
M2
x
w(x)∆x
D
∆x
w
+
w(x)
ΣFy = 0:
V − w( x) ∆x − (V + ∆V ) = 0
ε∆x
V
M
O
∆x
.
M + ∆M
V + ∆V
∆V = − w( x)∆x
+ ΣMO = 0:
− V∆x − M + w( x) ∆xε (∆x) + ( M + ∆M ) = 0
∆M = V∆x − w( x)ε (∆x) 2
14
∆V = − w( x)∆x
∆M = V∆x − w( x)ε ( ∆x) 2
Dividing by ∆x and taking the limit as ∆x
0, these equation become
dV
= − w(x)
dx
----------(7-1)
Slope of Shear Diagram = -Intensity of Distributed Load
dM
=V
dx
----------(7-2)
Slope of Moment Diagram = Shear
Equations (7-1) and (7-2) can be “integrated” from one point to another between
concentrated forces or couples, in which case
∆V = − ∫ w( x)dx
Change in Shear = -Area under Distributed Loading Diagram
----------(7-3)
and
∆M = ∫ V ( x)dx
Change in Moment = Area under Shear Diagram
----------(7-4)
15
F
V
V
M
M + ∆M
∆x
M
V + ∆V
+
ΣFy = 0:
M´
.
O
∆x
M + ∆M
V + ∆V
∆V = − F
Thus, when F acts downward on the beam, ∆V is negative so that the shear diagram shows
a “jump” downward. Likewise, if F acts upward, the jump (∆V) is upward.
+ ΣMO = 0:
∆M = M '
In this case, if an external couple moment M´ is applied clockwise, ∆M is positive, so that the
moment diagram jumps upward, and when M acts counterclockwise, the jump (∆M) must be
downward.
16
P
ML
0
MR
0
VL
M´
w0
ML
VL
ML
w1
0
MR
ML
0
-wo
MR
VR
VL
w1
ML
w2 M
R
VL
VR
VR
VL
0
MR
Slope = VR
VL
VR
w2
MR
MR
ML
VR
VL
ML
VR
Slope = VL
Slope = VR
Slope = VL
ML
Slope = -w1
Slope = VR
Slope = -w2
VR
VL
MR
Slope = VL
ML
MR
Slope = VR
Slope = w1
Slope = -w2
VR
Slope = VL
ML
MR
17
Example 7.2
From the structure show. Draw the shear and bending moment diagram
for the beam and loading.
(a) use the equations each section.
(b) use the relations among load, shear, and bending moment.
20 kN
40 kN
B
A
2.5 m
3 m
C
D
2 m
18
(a) use the equations each section.
20 kN
x1
2.5 m
By =
20 kN
A
x2
B
A
3 m
D
C
2 m
40( 2) + 20(7.5)
= 46 kN
5
x1 Mx1
+
Dy = 20 + 40 - 46 = 14 kN
ΣFy = 0: Vx1 = -20
+ ΣMx1 = 0: Mx1 = -20x1
20 kN
Vx1
B
A
2.5 m
46 kN
+
40 kN
x3
ΣFy = 0: Vx3 = -14
+ ΣMx3 = 0: Mx3 = 14x3
x2 Mx2
Vx2
Mx3
+
ΣFy = 0: Vx2 = 46-20 = 26
+ ΣMx2 = 0: Mx2 = 46x2 - 20(2.5 + x2)
Mx2 = 26x2 - 50
x3
D
Vx3
14 kN
19
(a) use the equations each section.
20 kN
x1
A
x2
B
2.5 m
3 m
40 kN
x3
C
D
2 m
46 kN
Vx1 = -20
V (kN)
Vx2 = 26
14 kN
Vx3 = -14
Mx1 = -20x1 Mx2 = 26x2 - 50 Mx3 = 14x3
26
x
-14
-20
28
V (kN•m)
x
-50
20
(b) use the relations among load, shear, and bending moment.
20 kN
40 kN
B
A
2.5 m
3 m
C
D
2 m
46 kN
V (kN)
26
14 kN
26
A = +78
A = -28
A = -50
-20
-20
M (kN•m)
-14
x
-14
28
x
-50
21
Example 7.3
Draw the shear and bending moment diagram for the the mechanical link
AB. The distributed load of 40 N/mm extends over 12 mm of the beam,
from A to C, and 400 N load is applied at E.
40 N/mm
B
A
C
12 mm
D
6 mm 4 mm 10 mm
400 N
32 mm
22
480 N
B
A
Ay = 480 + 400 - 365
= 515 kN
D
C
6 mm 4 mm 10 mm
400 N
32 mm
12 mm
480 N
1600 N•mm
515 N
V (N)
400 N
365 N
515
35
A = +3300
x
A = 210
V (N•mm)
480(6) + 400(22)
32
= 365 kN
By =
3300
3510
A = 5110
-365
5110
-365
x
23
Example 7.4
Draw the shear and moment diagrams for the beam shown in the figure.
20 kN/m
9m
24
x
3
1
x
( )( x )(20 )
2
9
(1/2)(9)(20) = 90 kN
20 kN/m
(2/3)9 = 6 m
x
(20 )
9 M
90-60 = 30 kN
90(6)/9
= 60 kN
9m
V (kN)
30 kN
V=0
x
30
+
V=0
+
x
x = 5.20 m
-
ΣFy = 0:
1
x
30 − ( )( x)(20 ) = 0
2
9
x = 5.20 m
60
M (kN•m)
104
+ ΣMx = 0:
1
5. 2 5. 2
M + [( )(5.2)(20
)]( ) − 30(5.2) = 0
2
9
3
+
x
M = 104 kN•m
25
Example 7.5
Draw the shear and moment diagrams for the beam shown in the figure.
3 kN
5 kN•m
A
B
C
3m
D
1.5 m
1.5 m
26
3 kN
5 kN•m
A
B
C
0.67 kN
3m
V (N)
0.67
D
1.5 m
1.5 m
+
2.33 kN
x (m)
-2.33
2.01
M (kN•m)
+
3.52
-
+
x (m)
-1.49
27
Example 7.6
Draw the shear and moment diagrams for the compound beam shown in
the figure. Assume the supports at A is fix C is roller and B is pin
connections.
8 kN
A
12 m
12 m
hinge
B
30 kN•m
C
15 m
28
8 kN
A
12 m
12 m
hinge
B
30 kN•m
C
15 m
30 kN•m
0 = Bx
By = 2 kN
Cy = 2 kN
MA = 48 kN•m
8 kN
By = 2 kN
Bx = 0
Ax = 0
Ay = 6 kN
29
8 kN
30 kN•m
48 kN•m
A
6 kN
12 m
12 m
6
B
C
15 m
2 kN
6
V (kN)
x (m)
-2
-2
24
8m
M (kN•m)
x (m)
-30
-48
30
Example 7.7
Draw the shear and moment diagrams for the compound beam shown in
the figure. Assume the supports at A and C are rollers and B and D are
pin connections.
60 kN • m
5 kN
2 kN/m
3 kN/m
Hinge
A
C
B
10 m
6m
4m
D
6m
6m
31
60 kN • m
5 kN
2 kN/m
3 kN/m
Hinge
A
C
B
10 m
6m
4m
D
6m
6m
20 kN
60 kN • m
By = 16 kN
Bx = 0 kN
4m
Ay = 4 kN
5 kN
9 kN
4m
9 kN
Dx = 0 kN
0 = Bx
By = 16 kN
Cy = 45 kN
Dy = -6 kN
32
60 kN • m
5 kN
2 kN/m
3 kN/m
Hinge
A
4 kN
B
10 m
4m
6m
C
6m
D
6m
6 kN
45 kN
24
V (kN)
4
6
x (m)
2m
64
-16
-21
-21
60
M
(kN • m)
x (m)
-96
-180
33