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MECHANICS OF MATERIALS
CHAPTER
9 Deflection of Beams
MECHANICS OF MATERIALS
9- 2
Deformation of a Beam Under Transverse LoadingRelationship between bending moment and
curvature for pure bending remains valid for general transverse loadings.
EIxM )(1
Cantilever beam subjected to tip load,
EIPx
1
Curvature varies linearly with x
At free end A, AA
ρρ
,01
At support B, PLEI
BB
,01
MECHANICS OF MATERIALS
9 - 3
Deformation of a Beam Under Transverse Loading• Overhanging beam
• Reactions at A and C
• Bending moment diagram
• Curvature is zero at points where bending moment is zero, i.e., at each end and at E.
EIxM )(1
• Beam is concave upwards where bending moment is positive and concave downwards where it is negative.
• Maximum curvature occurs where the moment magnitude is maximum.
• An equation for the beam shape, i.e., elastic curve, is required to determine maximum deflection and slope.
MECHANICS OF MATERIALS
9 - 4
Equation of the Elastic Curve
• Thus,
2
2
232
2
2
1
1dx
yd
dxdy
dxyd
• Substituting and integrating,
21
1
2
21
CxCdxxMdxyEI
CdxxMdxdyEIEI
xMdx
ydEIEI
2/32
2
2
2
2
2
22
2
2
2
1
1
1
1tan1tantan
1
1
tan
dxdydx
yddxdy
dxyd
dxd
dxd
dxdy
dxd
dxd
dd
dxyd
dxd
dxdydx
ddsd
dxdy
MECHANICS OF MATERIALS
9 - 5
Equation of the Elastic Curve
21 CxCdxxMdxyEI
• Constants are determined from boundary conditions
• Three cases for statically determinate beams,
– Simply supported beam0,0 BA yy
– Overhanging beam0,0 BA yy
– Cantilever beam0,0 AAy
• More complicated loadings require multiple integrals and application of requirement for continuity of displacement and slope.
MECHANICS OF MATERIALS
9 - 6
Direct Determination of Elastic Curve from Load Distribution
• For a beam subjected to a distributed load,
xwdxdV
dxMdxV
dxdM
2
2
• Equation for beam displacement becomes
xwdx
ydEIdx
Md 4
4
2
2
432
2213
161 CxCxCxC
dxxwdxdxdxxyEI
• Integrating four times yields
• Constants are determined from boundary conditions.
MECHANICS OF MATERIALS
9- 7
Example 1
For portion AB of the overhanging beam, (a) derive equation for the elastic curve, (b) find maximum deflection, (c) evaluate ymax.
MECHANICS OF MATERIALS
9- 8
Example 1
xLaP
dxydEI 2
2
- Differential equation for the elastic curve,
SOLUTION:
Develop expression for M(x) and derive differential equation for elastic curve.
- Reactions:
LaPR
LPaR BA 1
- From FBD for section AD,
LxxLaPM 0
MECHANICS OF MATERIALS
9- 9
Example 1
PaLCLCLLaPyLx
Cyx
61
610:0,at
0:0,0at
113
2
Integrate differential equation twice and apply boundary conditions to obtain elastic curve.
213
12
61
21
CxCxLaPyEI
CxLaP
dxdyEI
xLaP
dxydEI 2
2
32
6 Lx
Lx
EIPaLy
PaLxxLaPyEI
Lx
EIPaL
dxdyPaLx
LaP
dxdyEI
61
61
3166
121
3
22
Substituting,
MECHANICS OF MATERIALS
9- 10
Example 1Locate point of zero slope i.e., point
of maximum deflection.
32
6 Lx
Lx
EIPaLy
LLxL
xEI
PaLdxdy
mm 577.0
331
60
2
Evaluate maximum deflection.
32
max 577.0577.06
EI
PaLy
EIPaLy
2
max 0642.0
469
23
max m10300Pa10200m5.4m2.1N102000642.0
y
mm2.5max y
MECHANICS OF MATERIALS
9 - 11
Statically Indeterminate Beams• Consider beam fixed at A and roller support at B.
• There are four unknown reaction components.
• Conditions for static equilibrium are000 Ayx MFF
So beam statically indeterminate to degree one. Say RB is redundant.
21 CxCdxxMdxyEI • We also have beam deflection equation,
which introduces two unknowns but provides three additional equations from the boundary conditions (used to solve for C1 , C2 , RB ):
0,At 00,0At yLxyx
MECHANICS OF MATERIALS
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Example 2
For the uniform beam, find reaction at A, derive equation for elastic curve, and find slope at A.
Beam is statically indeterminate to one degree (i.e., one excess reaction which static equilibrium alone cannot solve for).
SOLUTION:
• Develop differential equation for elastic curve (will be functionally dependent on reaction at A).
• Integrate twice, apply boundary conditions, solve for reaction at A and obtain elastic curve.
MECHANICS OF MATERIALS
9 - 13
Example 2• Consider moment acting at section D,
LxwxRM
MxLxwxR
M
A
A
D
6
032
1
0
30
20
LxwxRM
dxydEI A 6
30
2
2
• The differential equation for the elastic curve,
MECHANICS OF MATERIALS
9 - 14
Example 2
LxwxRM
dxydEI A 6
30
2
2
• Integrate twice
215
03
14
02
12061
2421
CxCL
xwxRyEI
CLxwxREI
dxdyEI
A
A
• Apply boundary conditions:
01206
1:0,at
0242
1:0,at
0:0,0at
214
03
13
02
2
CLCLwLRyLx
CLwLRLx
Cyx
A
A
• Solve for reaction at A
0301
31 4
03 LwLRA LwRA 010
1
MECHANICS OF MATERIALS
9 - 15
Example 2
xLwL
xwxLwyEI
3
05
030 120
112010
161
xLxLxEIL
wy 43250 2120
• Substitute for C1, C2, and RA in the elastic curve equation,
42240 65120
LxLxEIL
wdxdy
EILw
A 120
30
• Differentiate once to find the slope,
at x = 0,
MECHANICS OF MATERIALS
9 - 16
Method of Superposition
Principle of Superposition:
• Deformations of beams subjected to combinations of loadings may be obtained as a linear combination of the deformations from the individual loadings
• Procedure is facilitated by tables of solutions for common types of loadings and supports.
MECHANICS OF MATERIALS
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Example 3
For the beam and loading shown, find slope and deflection at point B.
SOLUTION:
Superpose the deformations due to Loading I and Loading II as shown.
MECHANICS OF MATERIALS
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Example 3
Loading I
EI
wLIB 6
3
EIwLy IB 8
4
Loading II
EI
wLIIC 48
3
EIwLy IIC 128
4
In beam segment CB, bending moment is zero and the elastic curve is a straight line.
EI
wLIICIIB 48
3
EI
wLLEI
wLEI
wLy IIB 3847
248128
434
MECHANICS OF MATERIALS
9 - 19
Example 3
EI
wLEI
wLIIBIBB 486
33
EI
wLEI
wLyyy IIBIBB 3847
8
44
EIwL
B 487 3
EIwLyB 384
41 4
Combine the two solutions,
MECHANICS OF MATERIALS
9 - 20
Application of Superposition to Statically Indeterminate Beams
• Method of superposition can be applied to find reactions of statically indeterminate beams.
• Designate one of the reactions as the redundant and eliminate or modify the support.
• Note that you must ensure that redundant chosen does not make structure unstable
• Determine beam deformation without redundant reaction.
• Treat redundant reaction as an unknown load which, together with the other (i.e., applied) loads, must produce deformations compatible with the original supports.
MECHANICS OF MATERIALS
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Example 4
For the uniform beam and loading shown, find reaction at each support and slope at A.
SOLUTION:
• Release “redundant” support/reaction at B, and find deformation.• Apply reaction at B as an unknown load to ensure zero displacement at B.
MECHANICS OF MATERIALS
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Example 4• Distributed Load:
EIwL
LLLLLEI
wy wB
4
334
01132.0
32
322
32
24
• Redundant Reaction Load:
EI
LRLLEILRy BB
RB322
01646.033
23
• For compatibility with original supports, yB = 0
EI
LREI
wLyy BRBwB
3401646.001132.00
wLRB 688.0
• From statics, wLRwLR CA 0413.0271.0
xLLxxEI
wy w334 2
24
EILbPayax
3,At
22
a b
MECHANICS OF MATERIALS
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Example 4
EI
wLEI
wLwA
3304167.0
24
EI
wLLLLEIL
wLRA
322 03398.0
3360688.0
EI
wLEI
wLRAwAA
3303398.004167.0
EIwL
A3
00769.0
Slope at A,