E252-Chapter5
Transcript of E252-Chapter5
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5-1Internal Forces in Members
Chapter 5Analysis and Design of Beams for Bending
INTRODUCTION
Internal Forces in Members
Shear and Bending-Moment Diagrams
Design of Prismatic Beams for Bending
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B
A
B
C
P
A E
F
DB
C
5'
10'
4'
8'
80' 40'40'
BA
P P
C D
80' 40'40'
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P P
C D
Cross-section
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5-2Internal Forces in Members
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7 14 7
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10
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B
D
C
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F
A
B
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P
A E
F
DB
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5'
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INTERNAL FORCES IN MEMBERS
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5-3
A
B
ExampleDetermine the internal forcesat point J. P= 5000 lb.Units: Lb, ft.
P
A E
F
DB
C
5
10
4
8
J
1.5
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5-4 Shear and Bending Moment Diagrams
Shear and Bending Moment Diagrams
M
V
Sign Convention
M V
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5-5Shear and Bending Moment Diagrams
Why do we Need to Draw Shear and Bending Diagrams?
80' 40'40'
BA
P P
C D
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5-6
Support Reactions
Segment
A to C
C to D
D to B
Shear Moment
ExampleDraw the shear and bending-moment diagrams for the beam and loadingshown. Label all points of change, maximums and minimums, and theaxes. Units: Lb, ft.
8 44
BA
P P
C D
x
A
P
x
A
P
x
A
P
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5-7
Support Reactions
Segment Shear Moment
ExampleDraw the shear and bending-moment diagrams for the beam andloading shown. Label all points of change, maximums andminimums, and the axes.
L
BA
W
x
A
W
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Caution:
L
BA
W
L
BA
When drawing FBDs, always use the original loading and not theequivalent.
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5-9Relations Among Load, Shear, and Bending Moment
RELATIONS AMONG LOAD, SHEAR, AND BENDING-MOMENT
L
BA
w w
dVw
dx=
dMV
dx=
The change in shear isequal to the area under theload curve.
The change in moment isequal to the area under the
shear curve.
The slope of the sheardiagram is equal to thevalue of the w load.
The slope of the momentdiagram is equal to thevalue of the shear.
M V x =
V w x =
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5-10 Relationship Between Load, Shear, and Bending Moment
A B
Observations about the Shape of Shear/ Moment DiagramsShear Diagrams:
-Are a plot of forces (note the units).
-Discontinuities occur at concentrated forces.
Moment Diagrams:-Are a plot of moments (note the units).
-Discontinuities occur at concentrated moments.
Miscellaneous:-Check your work by noting that you always start and end at zero.
-Always use the original FBD and not the equivalent.
15 kN 25 kN 20 kN
30 kN
4 kips/ft
M
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5-11
Support Reactions
ExampleDraw the shear and bending-moment diagrams for the beam andloading shown. Label all points of change, maximums andminimums, and the axes. Units: kN, m
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BA
8 8
C D
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3 3 3
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5-12
Support Reactions
ExampleDraw the shear and bending-moment diagrams for the beam andloading shown. Label all points of change, maximums andminimums, and the axes. Units: kN, m
3.2
BA
2 kN/m
4
0.8
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5-13
ExampleDraw the shear and bending-moment diagrams for the beamand loading shown. Label all points of change, maximums andminimums, and the axes. Units: Lb, ft.
8 44
BA
30 lb/ft20 lb/ft
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5-14
ExampleSketch the shear and bending-moment diagrams for the beamand loading shown. Label all points of change, maximums andminimums, and the axes. Units: Lb, ft.
8 4
BA
30 lb/ft
4
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5-15
ExampleSketch the shear and bending-moment diagrams for the beamand loading shown. Label all points of change, maximums andminimums, and the axes. Units: Lb, ft.
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BA
30 lb/ft
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5-16
ExampleSketch the shear and bending-moment diagrams for the beamand loading shown. Label all points of change, maximums andminimums, and the axes. Units: Lb, ft.
5 434
BA
30 lb/ft
20 lb/ft
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ExampleHere is an example of how the shape of the girder reflects theshear and bending diagrams.
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B
L
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So why did they put that gap in the bridge?
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BA
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Pins
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ExampleDraw the shear and bending-moment diagrams for the beam andloading shown. Label all points of change, maximums andminimums, and the axes. The addition of the internal pin at thecenter of the beam allows additional head room because rather
than the moment being a maximum in the center it becomes zero.This design is used at Wings Air West in SLO. Total span= 75 ft.
500 lb/ft
A B
500 lb/ft
AB
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ExampleDraw the shear and bending-moment diagrams for the beam andloading shown. Label all points of change, maximums and minimums,and the axes. This example demonstrates that with the addition of aninternal pin we get an additional equation, otherwise we would have too
many unknowns.
Support ReactionsC
A B
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3.5 1 2
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B
3
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5-22 Design of Prismatic Beams for Bending
DESIGN OF PRISMATIC BEAMS FOR BENDING
ExampleDesign the cross sections minimum height of the beam, knowing thatthe grade of wood used has an allowable bending stress of 12 MPa.
Units: N, m.
6
BA
3000
1.5 1.5
3000
h
100
V(N)
M(Nm)
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Cross-section
ExampleKnowing that the grade of copper used has an allowable bendingstress of 15 ksi, determine the minimum wall thickness for the 6"diameter pipe. Units: lb, ft.
3
700500 lb/ft
V(lb)
M
(kin)
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5-24
ExampleKnowing that the allowable bending stress for steel is 160 MPa,determine the most economical W410-shape to support the load.Units: kN, m.
20 kN/m
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BA
100
1.5 1.5
100
Cross-section
X X
V
(kN)
M(kNm)
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Cross-section
AB
6.5
50050 lb/ft
4
ExampleKnowing that the allowable bending stress for steel is 24 ksi, determinethe most economical C7-shape to support the load.Units: lb, ft.
Y Y
V(lb)
M(lbft)
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BA
30 kN/m
ExampleKnowing that the allowable bending stress for steel is 160 MPa,determine the most economical W310-shape to support the load.Units: kN, m.
80
3Cross-section
X X
V(kN)
M(kNm)
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ExampleTwo rolled-steel C150 channels are welded back to back. Knowingthat the allowable bending stress for steel is 160 MPa, determine themost economical channels to support the load.Units: kN, m.
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BA
8
1.5
5 4
1.5 1.5
Cross-section
X X
V
(kN)
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(kNm)
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SUMMARY
Internal Forces in Members
Shear and Bending-Moment Diagrams
Design of Prismatic Beams for Bending
A
B
A
B
C
P
A E
F
DB
C
5'
10'
4'
8'
80' 40'40'
BA
P P
C D
80' 40'40'
BA
P P
C D
Cross-section
A
Sign Convention
V
M
M V x = V w x =