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MECHANICS OFMATERIALS
Third Edition
Ferdinand P. Beer
E. Russell Johnston, Jr.
John T. DeWolf
Lecture Notes:
J. Walt Oler
Texas Tech University
CHAPTER
2002 The McGraw-Hill Com anies, Inc. All ri hts reserved.
Deflection of Beams
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Deflection of Beams
Deformation of a Beam Under TransverseLoading
Equation of the Elastic Curve
Direct Determination of the Elastic CurveFrom the Load Di...
Statically Indeterminate Beams
Sample Problem 9.1
Sample Problem 9.3
Method of Superposition
Sample Problem 9.7Application of Superposition to Statically
Indeterminate ...
Sample Problem 9.8
Moment-Area Theorems
Application to Cantilever Beams and
Beams With Symmetric ...
Bending Moment Diagrams by Parts
Sample Problem 9.11
Application of Moment-Area Theorems toBeams With Unsymme...
Maximum Deflection
Use of Moment-Area Theorems WithStatically Indeterminate...
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Deformation of a Beam Under Transverse Loading
Relationship between bending moment andcurvature for pure bending remains valid for
general transverse loadings.
EI
x)(1=
Cantilever beam subjected to concentrated
load at the free end,
EIPx=
1
Curvature varies linearly withx
At the free endA, == A
,01
At the supportB, PL
EI
BB
=
,01
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Deformation of a Beam Under Transverse Loading
Overhanging beam
Reactions atA and C
Bending moment diagram
Curvature is zero at points where the bendingmoment is zero, i.e., at each end and atE.
EI
x)(1=
Beam is concave upwards where the bending
moment is positive and concave downwards
where it is negative.
Maximum curvature occurs where the moment
magnitude is a maximum.
An equation for the beam shape or elastic curve
is required to determine maximum deflectionand slope.
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Equation of the Elastic Curve
From elementary calculus, simplified for beamparameters,
2
2
232
2
2
1
1dx
yd
dx
dydx
yd
+
=
Substituting and integrating,
( )
( )
( ) 2100
1
0
2
21
CxCdxxMdxyEI
CdxxMdx
dyEIEI
xMdx
ydEIEI
xx
x
++=
+=
==
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Equation of the Elastic Curve
( ) 21
00
CxCdxxMdxyEI
xx
++=
Constants are determined from boundaryconditions
Three cases for statically determinant beams,
Simply supported beam
0,0 ==
Byy
Overhanging beam0,0 == Byy
Cantilever beam0,0 ==y
More complicated loadings require multiple
integrals and application of requirement forcontinuity of displacement and slope.
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Direct Determination of the Elastic Curve From the
Load Distribution
For a beam subjected to a distributed load,
( ) ( )xwdx
dV
dx
MdxV
dx
dM===
2
2
Equation for beam displacement becomes
( )xw
dx
ydEI
dx
Md==
4
4
2
2
( ) ( )
432
2213
161 CxCxCxC
dxxwdxdxdxxyEI
++++
=
Integrating four times yields
Constants are determined from boundary
conditions.
h
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Statically Indeterminate Beams
Consider beam with fixed support atA and roller
support atB.
From free-body diagram, note that there are four
unknown reaction components.
Conditions for static equilibrium yield000 === AyxF
The beam is statically indeterminate.
( ) 2100
CxCdxxMdxyEI
xx
++=
Also have the beam deflection equation,
which introduces two unknowns but providesthree additional equations from the boundary
conditions:
0,At00,0At ===== yLxyx
h
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Sample Problem 9.1
ft4ft15kips50
psi1029in7236814 64
===
==
aLP
EIW
For portionAB of the overhanging beam,
(a) derive the equation for the elastic curve,
(b) determine the maximum deflection,
(c) evaluateymax.
SOLUTION:
Develop an expression for M(x)
and derive differential equation forelastic curve.
Integrate differential equation twice
and apply boundary conditions toobtain elastic curve.
Locate point of zero slope or point
of maximum deflection.
Evaluate corresponding maximum
deflection.
h
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hird Beer Johnston DeWolf
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Sample Problem 9.1
SOLUTION: Develop an expression for M(x) and derive
differential equation for elastic curve.
- Reactions:
+==L
aPR
L
PaR BA 1
- From the free-body diagram for sectionAD,
( )LxxL
aPM
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ird Beer Johnston DeWolf
9 - 11
Sample Problem 9.1
PaLCLCLLaPyLx
Cyx
61
610:0,at
0:0,0at
113
2
=+===
===
Integrate differential equation twice and applyboundary conditions to obtain elastic curve.
213
12
6
1
2
1
CxCxL
aPyEI
CxL
aP
dx
dyEI
++=
+=
xL
a
Pdx
yd
EI =2
2
=
32
6 Lx
Lx
EIPaLy
PaLxx
L
aPyEI
Lx
EIPaL
dxdyPaLx
LaP
dxdyEI
6
1
6
1
3166
121
3
22
+=
=+=
Substituting,
h
f
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ird Beer Johnston DeWolf
9 - 12
Sample Problem 9.1
Locate point of zero slope or pointof maximum deflection.
=32
6 L
x
L
x
EI
PaLy
LL
xL
x
EI
PaL
dx
dym
m 577.03
316
02
==
==
Evaluate corresponding maximum
deflection.
( )[ ]32max 577.0577.06
=EI
PaLy
EI
PaLy
6
0642.02
max =
( )( )( )
( )( )462
maxin723psi10296
in180in48kips500642.0
=y
in238.0max =y
hi
B J h t D W lf
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ird Beer Johnston DeWolf
9 - 13
Sample Problem 9.3
For the uniform beam, determine the
reaction atA, derive the equation forthe elastic curve, and determine the
slope atA. (Note that the beam is
statically indeterminate to the first
degree)
SOLUTION:
Develop the differential equation for
the elastic curve (will be functionally
dependent on the reaction atA).
Integrate twice and apply boundary
conditions to solve for reaction atA
and to obtain the elastic curve.
Evaluate the slope atA.
hi
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rd Beer Johnston DeWolf
9 - 14
Sample Problem 9.3
Consider moment acting at sectionD,
L
xwxRM
Mx
L
xwxR
M
A
A
D
6
032
1
0
30
20
=
=
=
L
xwxRM
dx
ydEI
A 6
30
2
2
==
The differential equation for the elastic
curve,
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rd Beer Johnston DeWolf
9 - 15
Sample Problem 9.3
L
xwxRM
dx
ydEI A
6
30
2
2
==
Integrate twice
21
503
1
402
1206
1
242
1
CxCL
xw
xRyEI
CL
xwxREI
dx
dyEI
A
A
++=
+==
Apply boundary conditions:
01206
1:0,at
0242
1:0,at
0:0,0at
21
4
03
1
302
2
=++==
=+==
===
CLCLw
LRyLx
CLw
LRLx
Cyx
A
A
Solve for reaction atA
0
30
1
3
1 40
3 = LwLRA = LwRA 010
1
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rd Beer Johnston DeWolf
9 - 16
Sample Problem 9.3
xLwL
xwxLwyEI
= 30
503
0120
1
12010
1
6
1
( )xLxLxEIL
wy 43250 2
120+=
Substitute for C1, C2, and RA in theelastic curve equation,
( )42240 65120
LxLx
EIL
w
dx
dy+==
EI
LwA
120
30=
Differentiate once to find the slope,
atx = 0,
hir Beer Johnston DeWolf
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rd Beer Johnston DeWolf
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Method of Superposition
Principle of Superposition:
Deformations of beams subjected to
combinations of loadings may be
obtained as the linear combination of
the deformations from the individualloadings
Procedure is facilitated by tables of
solutions for common types of
loadings and supports.
hir Beer Johnston DeWolf
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d Beer Johnston DeWolf
9 - 18
Sample Problem 9.7
For the beam and loading shown,
determine the slope and deflection at
pointB.
SOLUTION:
Superpose the deformations due toLoading IandLoading II as shown.
hird Beer Johnston DeWolf
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d
9 - 19
Sample Problem 9.7
Loading I
( )EI
wLIB 6
3
= ( )EI
wLy IB 8
4
=
Loading II
( )
EI
wLIIC
48
3
= ( )
EI
wLy IIC
128
4
=
In beam segment CB, the bending moment is
zero and the elastic curve is a straight line.
( ) ( )EI
wLIICIIB 48
3==
( )EI
wLL
EI
wL
EI
wLy IIB
384
7
248128
434
=
+=
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d
9 - 20
Sample Problem 9.7
Combine the two solutions,
EIwL
B487 3=( ) ( )
EIwL
EIwL
IIBIBB 48633 +=+=
EI
wL
yB 384
41 4
=( ) ( ) EIwL
EI
wL
yyy IIBIBB 384
7
8
44
+=+=
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d
9 - 22
Sample Problem 9.8
For the uniform beam and loading shown,
determine the reaction at each support and
the slope at endA.
SOLUTION:
Release the redundant support at B, and find deformation. Apply reaction atB as an unknown load to force zero displacement atB.
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Sample Problem 9.8
Distributed Loading:
( )
EIwL
LLLLLEI
wy wB
4
334
01132.0
3
2
3
22
3
2
24
=
+
=
Redundant Reaction Loading:
( ) EILRL
LEIL
Ry BBRB
322
01646.033
2
3 =
=
For compatibility with original supports,yB = 0
( ) ( ) EILR
EI
wLyy BRBwB
34
01646.001132.00 +=+=
= wLRB 688.0
From statics,
== wLRwLR C 0413.0271.0
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Sample Problem 9.8
Slope at endA,
( )EI
wL
EI
wLwA
33
04167.024
==
( )EIwLLLL
EILwLRA
32
2 03398.0336
0688.0 =
=
EI
wLA
3
00769.0=( ) ( )EI
wL
EI
wL
RA
wAA
33
03398.004167.0 +=+=
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Moment-Area Theorems
Geometric properties of the elastic curve canbe used to determine deflection and slope.
Consider a beam subjected to arbitrary loading,
First Moment-Area Theorem:
area under (M/EI) diagram between
CandD.
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Moment-Area Theorems
Second Moment-Area Theorem:
The tangential deviation of Cwith respect toD
is equal to the first moment with respect to avertical axis through Cof the area under the
(M/EI) diagram between CandD.
Tangents to the elastic curve atPandPintercepta segment of length dton the vertical through C.
= tangential deviation of C
with respect toD
hird Beer Johnston DeWolf
A li i C il B d B Wi h
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Application to Cantilever Beams and Beams With
Symmetric Loadings
Cantilever beam - Select tangent atA as the
reference.
Simply supported, symmetrically loaded
beam - select tangent at Cas the reference.
hird Beer Johnston DeWolf
B di M t Di b P t
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Bending Moment Diagrams by Parts
Determination of the change of slope and thetangential deviation is simplified if the effect of
each load is evaluated separately.
Construct a separate (M/EI) diagram for eachload.
- The change of slope, D/C, is obtained by
adding the areas under the diagrams.
- The tangential deviation, tD/Cis obtained by
adding the first moments of the areas with
respect to a vertical axis through D.
Bending moment diagram constructed from
individual loads is said to be drawn by parts.
hird Beer Johnston DeWolf
S l P bl 9 11
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Sample Problem 9.11
For the prismatic beam shown, determine
the slope and deflection atE.
SOLUTION:
Determine the reactions at supports.
Construct shear, bending moment and(M/EI) diagrams.
Taking the tangent at Cas the
reference, evaluate the slope andtangential deviations atE.
hird Beer Johnston DeWolf
S l P bl 9 11
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Sample Problem 9.11
SOLUTION:
Determine the reactions at supports.
waRRDB
==
Construct shear, bending moment and
(M/EI) diagrams.
( )EI
waa
EI
waA
EI
LwaL
EI
waA
623
1
422
32
2
22
1
=
=
=
=
hird Beer Johnston DeWolf
S l P bl 9 11
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Sample Problem 9.11
Slope at E:
EI
wa
EI
LwaAA
CECECE
64
32
21 =+=
=+=
( )aLEI
waE 23
12
2
+=
=
+
+=
=
EI
Lwa
EI
wa
EI
Lwa
EI
Lwa
LA
aA
LaA
tty CDCEE
168164
44
3
4
224223
121
( )aLEI
wayE += 28
3
Deflection at E:
hird Beer Johnston DeWolf
A li ti f M t A Th t B
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Application of Moment-Area Theorems to Beams
With Unsymmetric Loadings
Define reference tangent at supportA. Evaluate Aby determining the tangential deviation atB with
respect toA.
The slope at other points is found with respect to
reference tangent.
ADAD +=
The deflection atD is found from the tangential
deviation atD.
hird Beer Johnston DeWolf
Maximum Deflection
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Maximum Deflection
Maximum deflection occurs at pointKwhere the tangent is horizontal.
PointKmay be determined by measuring
an area under the (M/EI) diagram equal
to -A .
Obtainymaxby computing the first
moment with respect to the vertical axis
throughA of the area betweenA andK.
hird Beer Johnston DeWolf
Use of Moment Area Theorems With Statically
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Use of Moment-Area Theorems With Statically
Indeterminate Beams
Reactions at supports of statically indeterminate
beams are found by designating a redundant
constraint and treating it as an unknown load which
satisfies a displacement compatibility requirement.
The (M/EI) diagram is drawn by parts. The
resulting tangential deviations are superposed and
related by the compatibility requirement. With reactions determined, the slope and deflection
are found from the moment-area method.