Minggu 7-2
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Transcript of Minggu 7-2
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Minggu 7-2
Internal Forces
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Objectives
1. To show how to use the method of sections for
determining the internal loadings in a member
2. To generalize the procedure by formulating
equations that can be plotted so they describethe internal shear and moment throughout a
member
3. To analyze the forces and study the geometry
of cables supporting a load
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Shear Force and Bending
Moment DiagramsA BEAM is a long, slender, structural
member designed to support transverse
loadings.
A transverse loading is applied
perpendicular to the axis of the beam.
Beams classified by their supports.
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Need to know the internal shear
force and the internal bending
moment at all locations in thebeam. This information is usually
presented as a graph or diagram of
these values vs. position.
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Pw
a
L
b
x1
x2
x3
Simply Supported Beam
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Pw
a
L
b
x1
x2
x3
Cantilever Beam
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Sign Convention
1. Positive Shear Force will cause a
clockwise rotation of the member on
which it acts.
2. Positive Bending Moment will cause
compression on the top surface of the
beam.
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VV
VV
Positive Shear
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M M
MM
Positive Moment
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MM
Compression
Tension
Positive Moment
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Procedure
1. Find the support reactions.
2. Identify critical points.
3. At representative locations betweencritical points take a section cut.
4. Find V and M and representative locationsbetween critical points.
5. Use proper sign conventions for V and M.6. Plot results.
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Critical Points
1. End of Beam
2. Support
3. Point of application of concentrated
load4. Point of application of concentrated
couple moment.
5. Beginning of distributed load.6. End of distributed load.
7. Change in distributed load definition.
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3 m6 m
6 kN
Determine the internal forces just to the left and the right of the
external force
9 kN m
Critical Points
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4 ft 4 ft
50 lb/ft
A C
Critical Points
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Pw
a
L
b
x1
x2
x3
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Pw
a
L
b
x1x2
x3
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4 ft 4 ft
50 lb/ft
A C
Critical Points
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3 m6 m
6 kN
9 kN m
Shear Force and Bending
Moment Diagram
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FBD of Beam
3 m6 m
6 kN
9 kN m
AyDy
Dx
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x
x
y
y y
D
y
y
y
F 0
D 0
F 0
A 6 D 0
M 0
9 6(6) - A (9) 0
A 5 kN
D 1 kN
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FBD of Beam
3 m 6 m6 kN 9 kN m
5 kN 1 kN
x
x
9x3
3x0
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5 kN
x1
y
0 x 3
F 05 V 0
V 5 kN
M 0
M 5 kN x 0
M 5x kN m
V
M
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3 m 6 kN
5 kN
x
x
V
M
y
3 x 9
F 0
5 kN 6 kN V 0
V 1 kN
M 0
M 5 x 6 x 3 0
M 18 x kN m
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V (kN)
M (kN m)
V = -1 kN
V = 5 kN
M = 5x kN m M = (18-x) kN m
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4 ft 4 ft
50 lb/ft
A C
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4 ft 4 ft
50 lb/ft
Ax
Cx
Cy
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4 ft 4 ft
50 lb/ft
Ax
Cx
Cy
400 lb
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x
x
y
y y
A
y
y
y
F 0
C 0F 0
A C 50(8) 0
M 0
50(8)(4) - C (8) 0
A 200 lb
C 200 lb
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x
50 lb/ft
200 lb V
M
y
2
0 x 8
F 0V 200 lb 50x 0
V 50x 200 lb
M 0x
M 200x 50(x) 02
M 200x 25x lb ft
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x
50 lb/ft50 lb/ft
200 lbV
M
0Mlb200-V8,xat
0Mlb200V0,xat
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V (lb)
M (lb ft)
x50200V
2x25x200M
L
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Pw
a b
w 100 lb / ft
L 15 ft
a 5 ft
b 10 ft
P 1000 lb
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Pw
a
L
b
Ay
Ax
Cy
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x
x
y
y y
y y
A
y
y
y
F 0
A 0
F 0
A C 100(5) 1000 0
A C 1500
M 0
100(5)(2.5) -1000(10) C 15 0
A 750 lbC 750 lb
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x
x750 lb
100 lb/ft
V
M
y
2
0 x 5
F 0
V 750 lb 100x 0
V 750 100x lb
M 0x
M 750x 100(x) 02
M 750x 50x lb ft
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x
x750 lb
100 lb/ft
V
M
f tlb002Mlb250V5,xat
0Mlb750V0,xat
5
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x
750 lb
100 lb/ft
V
M
5 ft
y
5 x 10
F 0
V 750 lb 100(5) 0
V 250 lb
M 0
M 750x 100(5) x 2.5 0M 750x 500x 1250 lb ft
M 250x 1250 lb ft
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x
750 lb
100 lb/ft
V
M
5 ft
f tlb3750Mlb250V,10xat
f tlb2500Mlb250V5,xat
-
10 ft
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1000 lb
x
750 lb
100 lb/ft
V
M
10 ft
5 ft
y
10 x 15
F 0
V 750 lb 100(5) 1000 0
V 750 lb
M 0
M 750x 100(5) x 2.5 1000 (x 10) 0
M 750x 11250 lb ft
10 ft
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1000 lb
x
750 lb
100 lb/ft
V
M
10 ft
5 ft
f tlb0Mlb750-V,15xat
flb3750Mlb750-V,10xat
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2
0 x 5
V 750 lb 100 x
M 750 x 50 x lb ft
5 x 10
V 250 lbM 250 x 1250 lb ft
10 x 15
V 750 lb
M 750 x 11250 lb ft
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V (lb)
M (lb ft)
x100lb750V
lb750V
ftlbx50x750M2
lb250V
ftlb1250x250M
ftlb11250x750M
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V
-1000
-500
0
500
1000
0.0 5.0 10.0 15.0
V
M
0
500
1000
1500
2000
2500
3000
3500
4000
0.0 5.0 10.0 15.0
M
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