Shear Force & Bending Moment Diagram Lecture
Short Description
Shear Force & Bending Moment Diagram Lecture...
Description
SHEAR FORCE FORCE & BENDING MOMENT DIAGRAM
Various Types Types of Beam Loading & Suppor Supportt •
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Beam Beam - structural member member designed to support loads applied at various points along its length. Beam can be subjected to concentrated loads loads or distributed loads loads or combination of both. Beam design is design is two-step process: 1) determine determine sheari shearing ng forces forces and bending bending moments produced by applied loads 2) select select cross-sectio cross-section n best suited suited to resist resist shearing forces and bending moments
Beer, Ferdinand & Johnston, Russel E. “Vecto “Vectors rs
Various Types of Beam Loading and Support
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Beams are classified according to way in which they are supported. Reactions at beam supports are determinate if they involve only three unknowns. Otherwise, they are statically indeterminate.
Beer, Ferdinand & Johnston, Russel E. “Vectors
Types of Beams •
Depends on the support configuration FH Pin
FH
FV
FV
Fixed
Roller
M Fv
Roller
FV
Pin
FV
FH
Statically Indeterminate Beams Continuous Beam
Propped Cantilever Beam
Types of Beams
SHEAR FORCES AND BENDING MOMENTS
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When a beam is loaded by forces or couples, stresses and strains are created throughout the interior of the beam. To determine these stresses and strains, the internal forces and internal couples that act on the cross sections of the beam must be found.
Shear and Bending Moment in a Beam •
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Wish to determine bending moment and shearing force at any point in a beam subjected to concentrated and distributed loads. Determine reactions at supports by treating whole beam as free-body. Cut beam at C and draw free-body diagrams for AC and CB. By definition, positive sense for internal force-couple systems are as shown. From equilibrium considerations, determine M and V or M’ and V’ . Beer, Ferdinand & Johnston, Russel E. “Vectors
Shear and Bending Moment Diagrams •
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Variation of shear and bending moment along beam may be plotted. Determine reactions at supports. Cut beam at C and consider member AC , V P 2 M Px 2 Cut beam at E and consider member EB, V P 2 M P L x 2 For a beam subjected to concentrated loads, shear is constant between loading points and moment varies linearly.
Beer, Ferdinand & Johnston, Russel E. “Vectors
Sample Problem SOLUTION: •
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Draw the shear and bending moment diagrams for the beam and loading shown.
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Taking entire beam as a free-body, calculate reactions at B and D. Find equivalent internal force-couple systems for free-bodies formed by cutting beam on either side of load application points. Plot results.
Beer, Ferdinand & Johnston, Russel E. “Vectors
Sample Problem SOLUTION: •
•
Taking entire beam as a free-body, calculate reactions at B and D. Find equivalent internal force-couple systems at sections on either side of load application points. F y 0 : 20 kN V 1 0 V 1 20 kN
M 2 0 : 20 kN 0 m M 1 0
M 1 0
Similarly, V 3 26 kN M 3 50 kN m V 4 26 kN M 4 50 kN m V 5 26 kN M 5 50 kN m V 6 26 kN M 6 50 kN m Beer, Ferdinand & Johnston, Russel E. “Vectors
Sample Problem •
Plot results. Note that shear is of constant value between concentrated loads and bending moment varies linearly.
Beer, Ferdinand & Johnston, Russel E. “Vectors
Sample Problem 2 SOLUTION: •
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Draw the shear and bending moment diagrams for the beam AB. The distributed load of 40 lb/in. extends over 12 in. of the beam, from A to C , and the 400 lb load is applied at E .
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Taking entire beam as free-body, calculate reactions at A and B. Determine equivalent internal forcecouple systems at sections cut within segments AC , CD, and DB. Plot results.
Beer, Ferdinand & Johnston, Russel E. “Vectors
Sample Problem SOLUTION: •
Taking entire beam as a free-body, calculate reactions at A and B.
M A 0 : B y 32 in. 480 lb6 in. 400 lb22 in. 0 B y 365 lb
M B 0 : 480 lb26 in. 400 lb10 in. A32 in. 0 A 515 lb
F x 0 : •
B x 0
Note: The 400 lb load at E may be replaced by a 400 lb force and 1600 lb-in. couple at D. Beer, Ferdinand & Johnston, Russel E. “Vectors
Sample Problem •
Evaluate equivalent internal force-couple systems at sections cut within segments AC , CD, and DB.
From A to C : F y 0 : 515 40 x V 0 V 515 40 x
515 x 40 x 12 x M 0
M 1 0 :
M 515 x 20 x 2
From C to D:
F y 0 :
515 480 V 0
V 35 lb
M 2 0 : 515 x 480 x 6 M 0 M 2880 35 x lb in. Beer, Ferdinand & Johnston, Russel E. “Vectors
Sample Problem •
Evaluate equivalent internal force-couple systems at sections cut within segments AC , CD, and DB. From D to B:
F y 0 :
515 480 400 V 0
V 365 lb
M 2 0 : 515 x 480 x 6 1600 400 x 18 M 0 M 11,680 365 x lb in.
Beer, Ferdinand & Johnston, Russel E. “Vectors
Sample Problem •
Plot results. From A to C : V 515 40 x M 515 x 20 x
2
From C to D:
V 35 lb M 2880 35 x lb in.
From D to B:
V 365 lb M 11,680 365 x lb in.
Beer, Ferdinand & Johnston, Russel E. “Vectors
Relations Among Load, Shear, and Bending Moment •
Relations between load and shear: V V V w x 0 dV dx
V w x 0 x
lim
x D
V D V C w dx area under load curve xC •
Relations between shear and bending moment:
M M M V x w x dM dx
x 2
0
M lim V 12 w x V x 0 x x0
lim
M D M C
x D
V dx area under shear curve
xC
Beer, Ferdinand & Johnston, Russel E. “Vectors
Relations Among Load, Shear, and Bending Moment •
Reactions at supports,
•
Shear curve,
R A R B
wL 2
x
V V A w dx wx 0
V V A wx •
L wx w x 2 2
wL
Moment curve, x
M M A
Vdx 0
x
M
L w x dx 2 0 2
w 2
L x x 2
dM M max V 0 M at 8 dx Beer, Ferdinand & Johnston, Russel E. “Vectors wL
Sample Problem 3 SOLUTION: •
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Draw the shear and bendingmoment diagrams for the beam and loading shown.
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Taking entire beam as a free-body, determine reactions at supports. Between concentrated load application points, dV dx w 0 and shear is constant. With uniform loading between D and E , the shear variation is linear. Between concentrated load application points, dM dx V constant . The change in moment between load application points is equal to area under shear curve between points. With a linear shear variation between D and E , the bending moment diagram is a parabola. Beer, Ferdinand & Johnston, Russel E. “Vectors
Sample Problem 3 SOLUTION: •
Taking entire beam as a free-body, determine reactions at supports.
M A 0 : D24 ft 20 kips6 ft 12 kips14 ft
12 kips28 ft 0 D 26 kips
F y 0 : A y 20 kips 12 kips 26 kips 12 kips 0 A y 18 kips •
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Between concentrated load application points, dV dx w 0 and shear is constant. With uniform loading between D and E , the shear variation is linear.
Beer, Ferdinand & Johnston, Russel E. “Vectors
Sample Problem •
Between concentrated load application points, dM dx V constant . The change in moment between load application points is equal to area under the shear curve between points. M B M A 108
M B 108 kip ft
M C M B 16
M C 92 kip ft
M D M C 140 M D 48 kip ft M E M D 48 •
M E 0
With a linear shear variation between D and E , the bending moment diagram is a parabola.
Beer, Ferdinand & Johnston, Russel E. “Vectors
Common Relationships 0
Constant
Linear
Constant
Linear
Parabolic
Linear
Parabolic
Cubic
Load
Shear
Moment
Common Relationships
Load
0
0
Constant
Constant
Constant
Linear
Linear
Linear
Parabolic
M
Shear
Moment
Seatwork: Draw Shear & Moment diagrams for the following beam 12 kN
8 kN
A
C
D
B
1m
3m
1m
Seatwork: Draw Shear & Moment diagrams for the following beam 12 kN
8 kN
A
C
D
B
1m RA = 7 kN
3m
1m RC = 13 kN
12 kN
8 kN
A
C
D
B
1m
3m 8
7 V
1m
8
7 -15
(kN)
-5 7 M (kN-m)
2.4 m
-8
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