CHAPTER 7: Forces in Beams
Transcript of CHAPTER 7: Forces in Beams
![Page 1: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/1.jpg)
1
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
![Page 2: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/2.jpg)
2
F
F
A
B
Compression
F
F
A
B
Tension
7.1 Internal Forces in Members
C
C
BF
F
A
C F
F
C
C
BF
F
A
C F
F
![Page 3: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/3.jpg)
3
A
BC
D
EF
W
G
A
B
C
D
Ax
Ay
T Cx
CyFEB
J
D
T V
N
J
M
A
B
C
Ax
Ay
Cx
CyFEB
JVM
N
![Page 4: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/4.jpg)
4
A
B
C
PP
V
NM
A
DP
M
N
V
PD
B
C
D
![Page 5: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/5.jpg)
5
Example 7.1
In the frame shown, determine the internal forces (a) in member ACF atpoint J and in member BCD at point K. (b) determine the equation ofinternal forces of member ACD and draw its diagram. This frame hasbeen previously considered in sample
Guide : Assuming that Framecan be treated as a rigid body.The reactions and the forcesacting on each member of theframe are determined.Member ACF is cut at point Jand member BCD is cut atpoint K, the internal forces arerepresented by an equivalentforce-couple system and canbe determined by consideringthe equilibrium of either part.
2.7 m
4.8 m
3.6 m1.2 m
1.5 m2.7 m
E F
B CD
J
K
A
2400 N
1000 N
![Page 6: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/6.jpg)
6
2.7 m
4.8 m
3.6 m1.2 m
1.5 m2.7 m
E F
B CD
J
K
A
2400 N
• Define the state of the problem
Assuming that Frame can be treated as a rigid body. Member ACF iscut at point J and member BCD is cut at point K, the internal forces arerepresented by an equivalent force-couple system.
(a)
1200 N
![Page 7: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/7.jpg)
7
2.7 m
2.7 m
E
B
A
750 N
2400 N
1200 N
2400 N
450 N
2.7 m
4.8 m
3.6 m1.2 m
1.5 m2.7 m
E F
B CD
A
2400 N
Ex = 1200 N
Ey = 750 N F = 3150 N
2.7 m
4.8 m
2.7 mF
B C
A
2400 N
450 N
3600 N
3150 N
2400 N
• Construct the physical model, construct mathematical model and solve equations
1200 N
1200 N
Ex = 1200 N
1.2 m
CD
2400 N2.4 m
B
2400 N
2400 N3600 N1200 N
![Page 8: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/8.jpg)
8
x
V (N)
-1200
2400
x
M (N•m)
A = -2880
A = +2880
-2880-1800
x
N (N)
1.2 m
CD
2400 N2.4 m
B
3600 N1200 N
x2
x1
Nx1 = +2400 NNx2 = 0
Vx1= -1200 NVx2= 2400 N
Mx1 = -1200x1 N•mMx2 = -2400x2 N•m
• The diagram of internal forces of member BCD
2400 N
2400 N
+2400
![Page 9: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/9.jpg)
9
7.2 Various Types of Loading and SupportBEAMS
P1 P2
A B CD
Concentrated loads
A C
Distributed load
w
B
![Page 10: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/10.jpg)
10
Simply supported beam
L
Overhanging beam
L
Cantilever beam
L
Continuous beam
L2L1
Beam fixed at one end and simplysupported at the other end
L
Fixed beam
L
Hinge
![Page 11: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/11.jpg)
11
7.3 Shear and Bending Moment in a BeamInternal Loadings at a Specified Point
• Sign Convention
N N
V
V
M M
NM
V
V
N
M
Positive Sign Convention
![Page 12: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/12.jpg)
12
Shear and Moment Functions
+ ΣMx1 = 0:Mx1 = (P/2)x1
ΣFy = 0: Vx1 = P/2+
P
A B
L/2 L/2
x1 x2
A
P/2Vx1
Mx1x1
BP/2
x2
Vx2
Mx2
+ ΣMx2 = 0:Mx2 = (P/2)x2
ΣFy = 0: Vx2 = -P/2+
P/2 P/2
V
x
P/2
-P/2M
x
PL/4
![Page 13: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/13.jpg)
13
x1
x2
x3
x1 x2 x3
w
P
AB C
D
w
P
AB C
D
OR
![Page 14: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/14.jpg)
14
. .B C
A D
F1 F3F2
w = w(x)
M1
.M2
.
w
x ∆x
w(x)∆x
ε∆x
∆x
w(x)
O . M + ∆∆∆∆M
V + ∆∆∆∆V
ΣFy = 0:+
0)()( =∆+−∆− VVxxwV
+ ΣMO = 0:
0)()()( =∆++∆∆+−∆− MMxxxwMxV ε2)()( xxwxVM ∆−∆=∆ ε
xxwV ∆−=∆ )(
M
V
7.4 Relations Among Load, Shear, and BendingMoment
![Page 15: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/15.jpg)
15----------(7-4)
----------(7-3)
----------(7-2)
----------(7-1)
2)()( xxwxVM ∆−∆=∆ ε
xxwV ∆−=∆ )(
Dividing by ∆x and taking the limit as ∆x 0, these equation become
Vdx
dM =
Slope of Moment Diagram = Shear
)(xwdxdV −=
Slope of Shear Diagram = -Intensity of Distributed Load
Equations (7-1) and (7-2) can be “integrated” from one point to another between concentrated forces or couples, in which case
dxxwV ∫−=∆ )(Change in Shear = -Area under Distributed Loading Diagram
and
∫=∆ dxxVM )(
Change in Moment = Area under Shear Diagram
![Page 16: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/16.jpg)
16
∆x
M
V
M + ∆∆∆∆M
V + ∆∆∆∆V
.M´
O
F
∆x
M
V
M + ∆∆∆∆M
V + ∆∆∆∆V
+ ΣMO = 0: 'MM =∆
ΣFy = 0:+ FV −=∆
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.
In this case, if an external couple moment M´ is applied clockwise, ∆M is positive, so that themoment diagram jumps upward, and when M acts counterclockwise, the jump (∆M) must bedownward.
![Page 17: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/17.jpg)
17
ML MRM´0
P
VLVR
ML MR
VLVR
ML MRw1
w2
VLVR
ML MRw1 w2
VLVR
ML MRw0
VL VR
Slope = VL
Slope = VR0
0
ML
MR
ML MR
0
0
VL
VR
MR-wo Slope = VL
Slope = VR
ML
VR
VL
ML MR
Slope = -w1
Slope = -w2
Slope = VR
Slope = VL
ML
VL
VR
Slope = w1
Slope = -w2 MR
Slope = VR
Slope = VL
![Page 18: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/18.jpg)
18
Example 7.2
From the structure show. Draw the shear and bending moment diagramfor the beam and loading.(a) use the equations each section.(b) use the relations among load, shear, and bending moment.
2.5 m 3 m 2 m
20 kN 40 kN
A B
CD
![Page 19: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/19.jpg)
19
2.5 m 3 m 2 m
20 kN 40 kN
A B
CD
kNBy 465
)5.7(20)2(40 =+=
Dy = 20 + 40 - 46 = 14 kN
(a) use the equations each section.
x1 x3
x1
20 kN
A
Vx1
Mx1
+ ΣMx1 = 0:Mx1 = -20x1
ΣFy = 0: Vx1 = -20+
x2
+ ΣMx2 = 0: Mx2 = 46x2 - 20(2.5 + x2) Mx2 = 26x2 - 50
ΣFy = 0: Vx2 = 46-20 = 26+x2
2.5 m
20 kN
A B
Vx2
Mx2
46 kN
Dx3
14 kN
Mx3
Vx3+ ΣMx3 = 0:Mx3 = 14x3
ΣFy = 0: Vx3 = -14+
![Page 20: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/20.jpg)
20
(a) use the equations each section.
x1 x3x2
x
V (kN)
-14
26
-20
x
V (kN•m)
-50
28
2.5 m 3 m 2 m
20 kN 40 kN
A B
CD
46 kN14 kN
Vx1 = -20
Mx1 = -20x1
Vx2 = 26
Mx2 = 26x2 - 50
Vx3 = -14
Mx3 = 14x3
![Page 21: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/21.jpg)
21
x
V (kN)
-20
26
-14 -14
26
-20A = -50
A = +78
A = -28
x
M (kN•m)
-50
28
(b) use the relations among load, shear, and bending moment.
2.5 m 3 m 2 m
20 kN 40 kN
A B
CD
46 kN14 kN
20 kN 40 kN
46 kN14 kN
![Page 22: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/22.jpg)
22
Example 7.3
Draw the shear and bending moment diagram for the the mechanical linkAB. 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.
4 mm12 mm 6 mm 10 mm
32 mm
40 N/mm
A BC D
400 N
![Page 23: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/23.jpg)
23
400 N
4 mm12 mm 6 mm 10 mm
32 mm
480 N
A BC D
400 N kN
By
36532
)22(400)6(480
=
+=Ay = 480 + 400 - 365 = 515 kN
480 N
365 N515 N
1600 N•mm
V (N)
x
515
35
-365 -365V (N•mm)
x
A = +3300
A = 5110A = 210
33005110
3510
1600 N•mm
365 N515 N 400 N
480 N
![Page 24: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/24.jpg)
24
Example 7.4
Draw the shear and moment diagrams for the beam shown in the figure.
9 m
20 kN/m
![Page 25: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/25.jpg)
25
9 m
20 kN/m
+
(2/3)9 = 6 m(1/2)(9)(20) = 90 kN
90-60 = 30 kN 90(6)/9 = 60 kN
x
V (kN)
30
+
60
-
x
M (kN•m)
V = 0
M
)9
20)()(21( xx
3x
ΣFy = 0:+
x = 5.20 m
0)9
20)()(21(30 =− xxx
+ ΣMx = 0:
0)2.5(30)32.5)](
92.520)(2.5)(
21[( =−+M
M = 104 kN•m
104
V = 0
= 5.20 m
x
)9
20( x
30 kN
![Page 26: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/26.jpg)
26
Example 7.5
Draw the shear and moment diagrams for the beam shown in the figure.
3 kN5 kN•m
A BC D
3 m 1.5 m 1.5 m
![Page 27: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/27.jpg)
27
3 kN5 kN•m
A BC D
3 m 1.5 m 1.5 m0.67 kN 2.33 kN
V (N)x (m)
0.67+
-2.33
-
M (kN•m)x (m)
2.01+
-1.49
3.52
-+
![Page 28: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/28.jpg)
28
Example 7.6
Draw the shear and moment diagrams for the compound beam shown inthe figure. Assume the supports at A is fix C is roller and B is pinconnections.
12 m12 m 15 m
8 kN 30 kN•m
A B C
hinge
![Page 29: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/29.jpg)
29
8 kN
30 kN•m
Ay
Ax
MA
By
Bx
Bx
By
Cy
= 48 kN•m
= 0
= 6 kN
= 2 kN
= 0
= 2 kN
0 =
= 2 kN
12 m12 m 15 m
8 kN 30 kN•m
A B C
hinge
![Page 30: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/30.jpg)
30
12 m12 m 15 m
8 kN
A B C
V (kN)x (m)
6 6
-2 -2
x (m)M (kN•m)
-48
24
-30
8 m
30 kN•m30 kN•m
6 kN
48 kN•m
2 kN
![Page 31: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/31.jpg)
31
Example 7.7
Draw the shear and moment diagrams for the compound beam shown inthe figure. Assume the supports at A and C are rollers and B and D arepin connections.
5 kN 3 kN/m2 kN/m60 kN • mHinge
10 m 6 m 4 m 6 m 6 m
AB C D
![Page 32: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/32.jpg)
32
By
Bx
Bx
ByCy Dy
Dx
Ay
= 45 kN
= 0 kN
= -6 kN
= 0 kN= 16 kN
= 16 kN
0 =
= 4 kN5 kN 9 kN 9 kN
4 m 4 m
60 kN • m20 kN
5 kN 3 kN/m2 kN/m60 kN • mHinge
10 m 6 m 4 m 6 m 6 m
AB C D
![Page 33: CHAPTER 7: Forces in Beams](https://reader031.fdocuments.us/reader031/viewer/2022021006/620384f6da24ad121e4a5c87/html5/thumbnails/33.jpg)
33
V (kN) x (m)
-16-21 -21
24
6
M (kN • m) x (m)
2 m
6064
-96
-180
4
5 kN2 kN/m60 kN • m
Hinge
10 m 6 m 4 m 6 m 6 mB C D
3 kN/m
A4 kN
45 kN
6 kN