fourth edition mechanics of materials - hacettepeyunus.hacettepe.edu.tr/~boray/5_beams...
TRANSCRIPT
MECHANICS OFFourth Edition
MECHANICS OF MATERIALS
CHAPTER
MATERIALSFerdinand P. BeerE. Russell Johnston, Jr.John T DeWolf
Analysis and DesignJohn T. DeWolf
Lecture Notes:of Beams for Bending
J. Walt OlerTexas Tech University
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Analysis and Design of Beams for Bending
IntroductionShear and Bending Moment DiagramsSample Problem 5.1
l blSample Problem 5.2Relations Among Load, Shear, and Bending MomentSample Problem 5 3Sample Problem 5.3Sample Problem 5.5Design of Prismatic Beams for BendingSample Problem 5.8
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 2
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Introduction
• Beams - structural members supporting loads at
• Objective - Analysis and design of beams
Beams structural members supporting loads at various points along the member
• Transverse loadings of beams are classified asTransverse loadings of beams are classified as concentrated loads or distributed loads
• Applied loads result in internal forces consistingApplied loads result in internal forces consisting of a shear force (from the shear stress distribution) and a bending couple (from the normal stress distribution)normal stress distribution)
• Normal stress is often the critical design criteriaMMMSM
IcM
IMy
mx ==−= σσ
Requires determination of the location and
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 3
magnitude of largest bending moment
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Introduction
Classification of Beam Supports
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 4
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Shear and Bending Moment Diagrams
• Determination of maximum normal and shearing stresses requires identification of maximum internal shear force and bending couple.
• Shear force and bending couple at a point are determined by passing a section through the beam and applying an equilibrium analysis on pp y g q ythe beam portions on either side of the section.
• Sign conventions for shear forces V and V’and bending couples M and M’
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 5
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Sample Problem 5.1SOLUTION:
• Treating the entire beam as a rigid g gbody, determine the reaction forces
• Section the beam at points near psupports and load application points. Apply equilibrium analyses on resulting free-bodies to determine
For the timber beam and loading shown, draw the shear and bend-
t di d d t i th
resulting free bodies to determine internal shear forces and bending couples
moment diagrams and determine the maximum normal stress due to bending.
• Identify the maximum shear and bending-moment from plots of their distributionsdistributions.
• Apply the elastic flexure formulas to determine the corresponding
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 6
determine the corresponding maximum normal stress.
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Sample Problem 5.1SOLUTION:
• Treating the entire beam as a rigid body, determine eat g t e e t e bea as a g d body, dete ethe reaction forces
∑ ∑ ==== kN14kN46:0 from DBBy RRMF
• Section the beam and apply equilibrium analyses on resulting free-bodies
kN200kN200 11 −==−−∑ = VVFy
( )( ) 00m0kN200 111
11
==+∑ =
∑
MMMy
kN200kN200 22 −==−−∑ = VVFy
( )( ) mkN500m5.2kN200 222
22
⋅−==+∑ =
∑
MMMy
mkN50kN26 33 ⋅−=+= MV
mkN28kN14
mkN28kN26
55
44
33
⋅+=−=
⋅+=+=
MV
MV
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 7
0kN14 66 =−= MV
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Sample Problem 5.1• Identify the maximum shear and bending-
moment from plots of their distributions.
mkN50kN26 ⋅=== Bmm MMV
• Appl the elastic fle re form las to• Apply the elastic flexure formulas to determine the corresponding maximum normal stress.
( )( )36
2612
61
m1033.833
m250.0m080.0−×=
== hbS
36
3
1033833mN1050
−⋅×
==S
M Bmσ 36 m1033.833 ×S
Pa100.60 6×=mσ
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 8
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Sample Problem 5.2SOLUTION:
• Replace the 10 kip load with an equivalent force-couple system at D. Find the reactions at B by considering the beam as a rigid body.g y
• Section the beam at points near the support and load application points.
The structure shown is constructed of a W10x112 rolled-steel beam. (a) Draw th h d b di t di
pp pp pApply equilibrium analyses on resulting free-bodies to determine internal shear forces and bendingthe shear and bending-moment diagrams
for the beam and the given loading. (b) determine normal stress in sections just
internal shear forces and bending couples.
• Apply the elastic flexure formulas toto the right and left of point D. • Apply the elastic flexure formulas to determine the maximum normal stress to the left and right of point D.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 9
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Sample Problem 5.2SOLUTION:
• Replace the 10 kip load with equivalent force-couple system at D. Find reactions at B.
• Section the beam and apply equilibrium analyses on resulting free-bodiesanalyses on resulting free bodies.
kips3030:
−==−−∑ = xVVxFCtoAFrom
y
( )( ) ftkip5.1030 221
1 ⋅−==+∑ = xMMxxMy
:DtoCFrom
( ) ( ) ftkip249604240
kips240240
2 ⋅−==+−∑ =
−==−−∑ =
xMMxM
VVFy
( ) ftkip34226kips34:
⋅−=−= xMVBtoDFrom
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 10
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Sample Problem 5.2• Apply the elastic flexure formulas to
determine the maximum normal stress to th l ft d i ht f i t Dthe left and right of point D.
From Appendix C for a W10x112 rolled steel shape S = 126 in3 about the X-X axissteel shape, S = 126 in about the X-X axis.
inkip2016:
⋅MDoflefttheTo
k i0163
iki1776:in126p
==
MDofrighttheTo
Smσ ksi0.16=mσ
3in126inkip1776 ⋅
==SM
mσ ksi1.14=mσ
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 11
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Relations Among Load, Shear, and Bending Moment
( ) xwVVVFy =∆−∆+−=∑ 0:0• Relationship between load and shear:
xwV ∆−=∆
−= wddV
∫−=−D
C
x
xCD dxwVV
dx
x∆
• Relationship between shear and bending moment:
( )
( )221
02
:0
xwxVM
xxwxVMMMMC
∆−∆=∆
=∆
∆+∆−−∆+=∑ ′
dM
∫
=
DxdxVMM
Vdx
dM
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 12
∫=−Cx
CD dxVMM
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Sample Problem 5.3
SOLUTION:
T ki th ti b f b d• Taking the entire beam as a free body, determine the reactions at A and D.
• Apply the relationship between shear and load to develop the shear diagram.
Draw the shear and bending moment diagrams for the beam
• Apply the relationship between bending moment and shear to develop the bending moment diagramand loading shown. moment diagram.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 13
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Sample Problem 5.3SOLUTION:
• Taking the entire beam as a free body, determine the reactions at A and D.
( ) ( )( ) ( )( ) ( )( )ft28kips12ft14kips12ft6kips20ft2400
−−−==∑ A
DM
( ) ( )( ) ( )( ) ( )( )
0Fkips26
ft28kips12ft14kips12ft6kips20ft240
y =∑
==
DD
kips18
kips12kips26kips12kips200
=
−+−−=
y
y
A
A
• Apply the relationship between shear and load to develop the shear diagram.
ddVdV dxwdVwdx
−=−=
- zero slope between concentrated loads
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 14
- linear variation over uniform load segment
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Sample Problem 5.3• Apply the relationship between bending
moment and shear to develop the bending dimoment diagram.
dxVdMVdx
dM==
- bending moment at A and E is zero
- bending moment variation between A, B,
- bending moment variation between D d E i d i
C and D is linear
- net change in bending moment is equal to areas under shear distribution segments
and E is quadratic
- total of all bending moment changes across the beam should be zero
areas under shear distribution segments
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 15
t e bea s ou d be e o
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Sample Problem 5.5
SOLUTION:
T ki h i b f b d• Taking the entire beam as a free body, determine the reactions at C.
• Apply the relationship between shear and load to develop the shear diagram.
Draw the shear and bending moment diagrams for the beam and loading
• Apply the relationship between bending moment and shear to develop the bending moment diagramshown. the bending moment diagram.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 16
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Sample Problem 5.5SOLUTION:
• Taking the entire beam as a free body, determine the reactions at C.
⎞⎛⎞⎛
=+−==∑ 0 021
021
aaawRRawF CCy
⎟⎠⎞
⎜⎝⎛ −−=+⎟
⎠⎞
⎜⎝⎛ −==∑
330 02
102
1 aLawMMaLawM CCC
Results from integration of the load and shear distributions should be equivalent.
• Apply the relationship between shear and load t d l th h dito develop the shear diagram.
axxwdx
axwVV
aa
AB⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎠
⎞⎜⎜⎝
⎛−−=∫ ⎟
⎠⎞
⎜⎝⎛ −−=−
20
00 2
1
( )curveloadunderareaawV
aa
B −=−=
⎥⎦⎢⎣ ⎠⎝⎠⎝
021
00 2
- No change in shear between B and C
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 17
No change in shear between B and C.- Compatible with free body analysis
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Sample Problem 5.5• Apply the relationship between bending moment
and shear to develop the bending moment didiagram.
320
0
20 622 a
xxwdxa
xxwMMa
aAB
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎠
⎞⎜⎜⎝
⎛−−=∫ ⎟
⎠
⎞⎜⎜⎝
⎛⎟⎠
⎞⎜⎜⎝
⎛−−=−
203
100 622
awM
aa
B −=
⎥⎦⎢⎣ ⎠⎝⎠⎜⎝ ⎠⎝
( )L( ) ( )
( ) ⎞⎜⎛ −=−−=
−−=∫ −=−
3 00
1
021
021
aLwaaLawM
aLawdxawMM
C
L
aCB
( )⎠
⎜⎝
==32
306 LaLawMC
Results at C are compatible with free-body anal sisanalysis
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 18