analysis of beams and frames theory of structure - i
TRANSCRIPT
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Analysis of Beams and Frames
Theory of Structure - I
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Department of Civil EngineeringUniversity of Engineering and Technology, Taxila, Pakistan
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Lecture Outlines
Shear and Moment Diagrams for Beams Shear and Moment Diagrams for a Frames Moment Diagrams Constructed by the
Method of Superposition Deflected Curves
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. .B C
A D
F1 F3F2
w = w(x)
M1
.M2
.
w
x x
w(x)x
x
x
w(x)
O . M + M
V + V
Fy = 0:+
0)()( VVxxwV
+ MO= 0:
0)()()( MMxxxwMxV
2)()( xxwxVM
xxwV )(
M
V
Shear and Moment Diagrams for a Beam
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----------(4-2)
----------(4-1)
2)()( xxwxVM
xxwV )(
Dividing by x and taking the limit as x 0, these equation become
Vdx
dM
Slope of Moment Diagram = Shear
)(xwdx
dV
Slope of Shear Diagram = -Intensity of Distributed Load
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----------(4-4)
----------(4-3)
Equations (4-1) and (4-2) 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
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x
M
V
M + M
V + V
.M´
O
F
x
M
V
M + M
V + V
+ MO= 0: 'MM
Fy = 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.
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ML MRM´0
P
VLVR
ML MR
VLVR
ML MR
w0
VL VR
Slope = VL
Slope = VR0
0
ML
MR
ML MR
0
0
VL
VR
MR-wo Slope = VL
Slope = VR
ML
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VLVR
ML MR
w1w2
VLVR
ML MR
w1 w2
VR
VL
ML MR
Slope = -w1
Slope = -w2
Slope = VR
Slope = VL
ML
VL
VR
Slope = w1
Slope = -w2 MR
Slope = VR
Slope = VL
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Example 4-7
Draw the shear and moment diagrams for the beam shown in the figure.
9 m
20 kN/m
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9 m
20 kN/m
+
SOLUTION
(2/3)9 = 6 m(1/2)(9)(20) = 90 kN
30 kN 60 kN
x
V (kN)
30
+
60
-
x
M (kN•m)
V = 0
M
)9
20)()(2
1(
xx
3
x
Fy = 0:+
x = 5.20 m
0)9
20)()(2
1(30
xxx
+ Mx = 0:
0)2.5(30)3
2.5)](
9
2.520)(2.5)(
2
1[( M
M = 104 kN•m
104
V = 0
= 5.20 m
x
)9
20(x
30 kN
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Example 4-8
Draw the shear and moment diagrams for each of the beam shown in the figure.
P
L
L
MO
L
wo
L
wo
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+
L
Mo
SOLUTION
P
LPPL
0
0
Mo
0
x
V P
+
x
M
-PL
-
x
V
x
M Mo
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L
wo
wo L
0
wo L2
2L
wo
(wo L)/2
0
wo L2
6
x
V wo L
+
xM
-wo L2
2
-
x
V (wo L)/2
+
xM
-wo L2
6
-
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Example 4-9
Draw the shear and moment diagrams for the beam shown in the figure.
3 kN
5 kN•m
A B
C D
3 m 1.5 m 1.5 m
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3 kN
5 kN•m
A B
C D
3 m 1.5 m 1.5 m
SOLUTION
0.67 kN 2.33 kN
V (N)x (m)
0.67+
-2.33
-
M (kN•m)x (m)
2.01
+
-1.49
3.52
-+
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Example 4-10
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.
12 m12 m 15 m
8 kN 30 kN•m
A B C
hinge
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Ay
Ax
MA
SOLUTION
Cy
By
Bx
Bx
By
0 =
= 2 kN
= 2 kN
= 2 kN
= 0= 0
= 6 kN
8 kN
30 kN•m
= 48 kN•m
12 m12 m 15 m
8 kN 30 kN•m
A B C
hinge
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V (kN)x (m)
6 6
-2 -2
x (m)M (kN•m)
-48
24
-30
8 m
-
+
-
12 m12 m 15 m
8 kN
A B C
30 kN•m30 kN•m
6 kN
48 kN•m
2 kN
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Example 4-11
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.
5 kN 3 kN/m2 kN/m60 kN • m
Hinge
10 m 6 m 4 m 6 m 6 m
AB C D
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By
Bx
Bx
By
Ay
Cy Dy
Dx
= 0 kN
= 16 kN
= 4 kN
= 16 kN
0 == 0 kN
= 45 kN = -6 kN
SOLUTION
5 kN 9 kN 9 kN
4 m 4 m
60 kN • m
20 kN
5 kN 3 kN/m2 kN/m60 kN • m
Hinge
10 m 6 m 4 m 6 m 6 m
AB C D
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V (kN) x (m)
-16-21 -21
24
6
M (kN • m)
x (m)
2 m
60
64
-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
A
4 kN
45 kN
6 kN
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P1
B
P2
Ay
Cx
CyBx
By
MB
Bx
By
MBBx
By
MB
Bx
By
MB
P1
P2
Ay
Cx
Cy
P1
P2
A
BC
Shear and Moment Diagrams for a Frame
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Cx
Cy
Bx
By
MB
MB
P2
P2 = Bx
P2
Ay
Bx
By
MB
By By = Cy
MBMB
MB
A
B C
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Example 4-12
Draw the shear and moment diagrams for the frame shown . Assume A, C and D are pinned and B is a fixed joint.
3 kN/m
15 kN4 m 4 m
12 m
60 kN•mA
B C
D
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Find the Reaction
15 kN
4 m 4 m
B
36 kN
60 kN•m
Ax
Ay
Dx
Dy
= 5 kN
= 42 kN
= 41 kN
= -27 kN
Cx
Cy
Cx
Cy
= 5 kN
= 5 kN
= 42 kN
= 42 kN
3 kN/m
15 kN4 m 4 m
12 m
60 kN•mA
B C
D
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V (kN)
x (m)Bx
By
MB
Member AB
= 276 kN•m
12 m
3 kN
/m
A
B
Ax=41 kN
Ay= 27 kN
= 5 kN
= 27 kN
41
5
M (kN•m)
x (m)
276
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--
V (kN)
x (m)
15 kN
4 m 4 m
B
5 kN
42 kN
C
Member BC
41 kN
27 kN
Bx
By
MB
12 m
3 kN
/m
A
B= 5 kN
= 27 kN
= 276 kN•m
5 kN
27 kN
276 kN•m
-27-42 -42
M (kN•m)
x (m)
276168
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-
V (kN)
x (m)
60 kN•m
5 kN
5 kN
42 kN
D
C
42 kN
12 m
Member CD
-5
-5
M (kN•m)
x (m)
60
+
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Bending moment diagram of frame
276
A
B
+
276168
C
+
60 D
+
3 kN/m
15 kN4 m 4 m
12 m
60 kN•mA
B C
D
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Example 4-13
Draw the moment diagram for the frame shown . Assume A is pin, C is a roller, and B is a fixed joint.
40 kN/m
80 kN
4 m 4 m
2 m3 m
A
B C
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120 kN
1.5 m36.87o
80 kN
4 m 4 m
2 m3 m
B
Ay
Ax
Cy
+ MA = 0;
+ Fx = 0;
+Fy = 0;
82.5 kN =
- (120)(1.5) - (80)(6) + 8Cy = 0 Cy = 82.5 kN, ญ
120 kN =
-Ax + 120 = 0; Ax = 120 kN , ญ
= 2.5 kN
- Ay - 80 + 82.5 = 0; Ay = 2.5 kN , ญ
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Bx
By
MB
Bx
By
MBB
80 kN
82.5 kN
B
C2 m 2 m
2.5 kN
120 kN
120 kN
A
B
36.87o 1.5 m
1.5 m
By´ Bx´
MB´
By´ Bx´
MB´
+ MB = 0:
Member BC
+ Fy = 0:
2.5 kN =
-By - 80 + 82.5 = 0, By = 2.5 kN , ญ
36.87o
By´ cos 36.87By´ sin 36.87
-Bx´cos 36.87 + By´sin 36.87 + 0 = 0 -----(1)
Joint B
+ Fx = 0;
36.87o
Bx´cos 36.87
Bx´sin 36.87
-Bx´sin 36.87 - By´cos 36.87 + 2.5 = 0 -----(2)
+ Fy = 0;
=2.5 kN
= 0 kN
=170 kN•m
170 kN•m=
From eq. (1) and (2): Bx´ = 1.5 kN By´ = 2 kN
1.5 kN = = 2 kN
=1.5 kN
=170 kN•m2 kN=
0 kN =
170 kN•m =
-MB -80(2) + 82.5(4) = 0, MB = 170 kN•m
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1.5 kN
170 kN•m2 kNB
120 kN36.87o
2.5 kN
120 kN
36.87o
53.13
o5 m
120s
in36
.87o
1.5 kN
170 kN•m2 kN
70 kN97.5 kN
120 s
in 36
.87/5
=14
.4 kN
/m
70
-2
x (m)
M (kN•m)
4.86m
+
-
170.1
170
x (m)
V (kN)
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40 kN/m
80 kN
A
B C
-
-
80 kN
82.5 kN
BC170 kN•m
2.5 kN
V (kN) x (m)-2.5
-82.5
M (kN•m) x (m)
170 165
170
+
170 165
+
A
CB
M (kN•m)
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L
P
Mx
-PL
L
wo
Mx
2
2Lwo Parabolic curve
Most loading on beams in structural analysis will be a combination of the loadings shown in the figure below:
Moment Diagrams Constructed by the Method of Superposition
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M
x
Mo
L
Mo
L
wo
Mx
6
2Lwo Cubic curve
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M (kN•m)
x (m)
-20
70
=
M (kN•m)
x (m)
90
M (kN•m)
x (m)
-20
+
M (kN•m)
x (m)
-20
+
12 m
20 kN•m20 kN•m 5 kN/m
12 m
5 kN/m=
12 m
20 kN•m +
12 m
20 kN•m+
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Example 4-14
Draw the moment diagrams for the beam shown at the top of the figure below using the method of superposition. Consider the beam to be cantilevered from the support at B.
6 m
20 kN•m5 kN/m
2 m
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6 m
20 kN•m5 kN/m
2 m
SOLUTION
8.33 kN 6.67 kN
6 m
8.33 kN
+
6 m
5 kN/m+
20 kN•m=
M (kN•m)
x (m)
49.98+
M (kN•m)
x (m)
-20 -20
-20
M (kN•m)
x (m) 4.84
=
M (kN•m)
x (m)
-30
+
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+ M + M
positive moment,concave upward
- M - M
negative moment,concave downward
P1
P2M
x
inflection point
M
x
inflection point
P1
P2
Deflected Curve
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