Fourth Edition MECHANICS OF - mazzaro.dk OF MATERIALS Fourth Edition Ferdinand P. Beer, E. Russell...
Transcript of Fourth Edition MECHANICS OF - mazzaro.dk OF MATERIALS Fourth Edition Ferdinand P. Beer, E. Russell...
MECHANICS OF MATERIALS
Fourth Edition
Ferdinand P. Beer, E. Russell Johnston, Jr., John T. DeWolf
CHAPTER
Analysis and Designof Beams for Bending
Lecturer: Nazarena Mazzaro, Ph.D.
Aalborg University
Denmark
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
5 - 2Nazarena Mazzaro, AAU
Analysis and Design of Beams for Bending
IntroductionShear and Bending Moment
DiagramsExample 5.01Sample Problem 5.1Sample Problem 5.2Relations Among Load, Shear, and
Bending MomentExample 5.03Sample Problem 5.3Sample Problem 5.5
Part 1: 45 min
Design of Prismatic Beams for BendingSample Problem 5.8Singularity FunctionsExample 5.05
Part 2: 30 min
Exercises
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
5 - 3Nazarena Mazzaro, AAU
Introduction
• Beams - structural members supporting loads at various points along the member
• Objective - Analysis and design of beams
• Transverse loadings of beams are classified as concentrated loads or distributed loads
• Applied 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 is often the critical design criteria
SM
IcM
IMy
mx ==−= σσ
Requires determination of the location and magnitude of largest bending moment
Elastic Flexure Formulas
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
5 - 4Nazarena Mazzaro, AAU
Introduction
Classification of Beam Supports
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
5 - 5Nazarena Mazzaro, AAU
Shear and Bending Moment Diagrams
• Determination of maximum normal and shearing stresses requires identification of maximum internal shear force V and bending couple M.
• Shear force and bending couple at a point are determined by passing a section through the beam and applying an equilibrium analysis on the beam portions on either side of the section.
• Sign conventions for shear forces V and V’and bending couples M and M’ -> positive
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
5 - 6Nazarena Mazzaro, AAU
Example 5.01
Draw the shear and bending-moment diagrams for a simply supported beam AB of span L subjected to a single concentrated load P at it midpoint C
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
5 - 7Nazarena Mazzaro, AAU
Example 5.01
•Determination of the reactions RA=RB=1/2P
•Cut the beam at D and plot free body diagrams with positive V and M. Equilibrium equations:
2/)(;0)(
2/1;0;0
2/;0;0
2/1;0;0
xLPMMxLRM
PRVVRFy
PxxRMMxRM
PRVVRFy
BE
BB
AAD
AA
−==−−=
−=−==+=
===+−=
===−=
∑∑
∑∑
V is constant between concentrated loads and M varies linearly
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
5 - 8Nazarena Mazzaro, AAU
Sample Problem 5.1
For the timber beam and loading shown, draw the shear and bend-moment diagrams and determine the maximum normal stress due to bending.
SOLUTION:
• Treating the entire beam as a rigid body, determine the reaction forces
• Identify the maximum shear and bending-moment from plots of their distributions.
• Apply the elastic flexure formulas to determine the corresponding maximum normal stress.
• Section the beam at points near supports and load application points. Apply equilibrium analyses on resulting free-bodies to determine internal shear forces and bending couples
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
5 - 9Nazarena Mazzaro, AAU
Sample Problem 5.1SOLUTION:
• Treating the entire beam as a rigid body
• Section the beam and apply equilibrium analyses on resulting free-bodies
( )( ) 00m0kN200
kN200kN200
111
11
==+∑ =
−==−−∑ =
MMM
VVFy
( )( ) mkN500m5.2kN200
kN200kN200
222
22
⋅−==+∑ =
−==−−∑ =
MMM
VVFy
0kN14
mkN28kN14
mkN28kN26
mkN50kN26
66
55
44
33
=−=
⋅+=−=
⋅+=+=
⋅−=+=
MV
MV
MV
MV
46;141453405,2200
6040200
==
=⇒×+×−×==
−=⇒+−+−==
∑∑
BD
DDB
BDDB
RRRRM
RRRRFy
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
5 - 10Nazarena Mazzaro, AAU
Sample Problem 5.1• Identify the maximum shear and bending-
moment from plots of their distributions.
mkN50kN26 ⋅=== Bmm MMV
• Apply the elastic flexure formulas to determine the corresponding maximum normal stress.
( )( )
36
3
36
2612
61
m1033.833mN1050
m1033.833
m250.0m080.0
−
−
×
⋅×==
×=
==
SM
hbS
Bmσ
Pa100.60 6×=mσV1=-20, M1=0; V2=-20, M2=-50; V3= 26, M3=-50; V4=26, M4= 28; V5=-14, M5=28, V6=-14, M6=0
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
5 - 11Nazarena Mazzaro, AAU
Sample Problem 5.2
The structure shown is constructed of a W10x112 rolled-steel beam. (a) Draw the shear and bending-moment diagrams for the beam and the given loading. (b) determine normal stress in sections just to the right and left of point D.
SOLUTION:
• 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.
• Section the beam at points near the support and load application points. Apply equilibrium analyses on resulting free-bodies to determine internal shear forces and bending couples.
• Apply the elastic flexure formulas to determine the maximum normal stress to the left and right of point D.
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
5 - 12Nazarena Mazzaro, AAU
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-bodies.
( )( ) ftkip5.1030
kips3030:
221
1 ⋅−==+∑ =
−==−−∑ =
xMMxxM
xVVxFCtoAFrom
y
( ) ( ) ftkip249604240
kips240240:
2 ⋅−==+−∑ =
−==−−∑ =
xMMxM
VVFDtoCFrom
y
( ) ftkip34226kips34:
⋅−=−= xMVBtoDFrom
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
5 - 13Nazarena Mazzaro, AAU
Sample Problem 5.2• Apply the elastic flexure formulas to
determine the maximum normal stress to the left and right of point D.
From Appendix C for a W10x112 rolled steel shape, S = 126 in3 about the X-X axis.
3
3
in126inkip1776
:in126
inkip2016:
⋅==
⋅==
SM
DofrighttheTo
SM
DoflefttheTo
m
m
σ
σ ksi0.16=mσ
ksi1.14=mσ
• Concentrated loads: V constantM varies linearly
• Distributed load: V varies linearlyM parabola
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
5 - 14Nazarena Mazzaro, AAU
Relations Among Load, Shear, and Bending Moment
( )xwV
xwVVVFy
∆−=∆
=∆−∆+−=∑ 0:0
∫−=−
−=
D
C
x
xCD dxwVV
wdxdV
• Relationship between load and shear:
( )
( ) xwVx
MxwxVM
xxwxVMMMM C
∆−=∆∆
⇒∆−∆=∆
=∆
∆+∆−−∆+=∑ ′
21
02
:0
221
∫=−
=→∆
D
C
x
xCD dxVMM
Vdx
dMx ;0
• Relationship between shear and bending moment:
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
5 - 15Nazarena Mazzaro, AAU
Example 5.03
Draw the shear and bending moment diagrams for the simply supported beam and determine the maximum value of the bending moment.
SOLUTION:
•RA=RB=wL/2
•Determination of V and M at any distance from A:
8)(
21)
21(
0;.
)21(
21
.
2
max2
0
0
0
wLMxLxwdxxLwM
MdxVMM
xLwwLwLwxVV
wxdxwVV
x
A
x
A
A
x
A
⇒−=−=
==−
−=−=−=
−=−=−
∫
∫
∫
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
5 - 16Nazarena Mazzaro, AAU
Sample Problem 5.3
Draw the shear and bending moment diagrams for the beam and loading shown.
SOLUTION:
• 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.
• Apply the relationship between bending moment and shear to develop the bending moment diagram.
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
5 - 17Nazarena Mazzaro, AAU
Sample Problem 5.3SOLUTION:
• Taking the entire beam as a free body, determine the reactions at A and D.
( ) ( )( ) ( )( ) ( )( )
kips18
kips12kips26kips12kips200
0Fkips26
ft28kips12ft14kips12ft6kips20ft2400
y
=
−+−−=
=∑
=−−−=
=∑
y
y
A
A
A
DD
M
• Apply the relationship between shear and load to develop the shear diagram.
dxwdVwdxdV
−=−=
- zero slope between concentrated loads
- linear variation over uniform load segment
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
5 - 18Nazarena Mazzaro, AAU
Sample Problem 5.3• Apply the relationship between bending
moment and shear to develop the bending moment diagram.
dxVdMVdx
dM==
- bending moment at A and E is zero
- total of all bending moment changes across the beam should be zero (108-16-140+48=0)
- net change in bending moment is equal to areas under shear distribution segments
- bending moment variation between D and E is quadratic
- bending moment variation between A, B, C and D is linear
dxVMMD
CCD .∫=−
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
5 - 19Nazarena Mazzaro, AAU
Sample Problem 5.5
Draw the shear and bending moment diagrams for the beam and loading shown.
SOLUTION:
• 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.
• Apply the relationship between bending moment and shear to develop the bending moment diagram.
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
5 - 20Nazarena Mazzaro, AAU
Sample Problem 5.5SOLUTION:
• Taking the entire beam as a free body, determine the reactions at C.
⎟⎠⎞
⎜⎝⎛ −−=+⎟
⎠⎞
⎜⎝⎛ −==∑
=+−==∑
330
0
021
021
021
021
aLawMMaLawM
awRRawF
CCC
CCy
Results from integration of the load and shear distributions should be equivalent.
• Apply the relationship between shear and load to develop the shear diagram.
( )curveloadunderareaawV
axxwdx
axwVV
B
aa
AB
−=−=
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−−=∫ ⎟
⎠⎞
⎜⎝⎛ −−=−
021
0
20
00 2
1
- No change in shear between B and C.- Compatible with free body analysis
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
5 - 21Nazarena Mazzaro, AAU
Sample Problem 5.5• Apply the relationship between bending moment
and shear to develop the bending moment diagram.
203
10
320
0
20 622
awM
axxwdx
axxwMM
B
aa
AB
−=
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−−=∫ ⎟
⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛−−=−
( ) ( )
( ) ⎟⎠⎞
⎜⎝⎛ −=−−=
−−=∫ −=−
323 0
061
021
021
aLwaaLawM
aLawdxawMM
C
L
aCB
Results at C are compatible with free-body analysis
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
5 - 22Nazarena Mazzaro, AAU
Design of Prismatic Beams for Bending
• Among beam section choices which have an acceptable section modulus, the one with the smallest weight per unit length or cross sectional area will be the least expensive and the best choice.
• The largest normal stress is found at the surface where the maximum bending moment occurs.
SM
IcM
mmaxmax ==σ
• A safe design requires that the maximum normal stress be less than the allowable stress for the material used. This criteria leads to the determination of the minimum acceptable section modulus.
all
allmM
Sσ
σσ
maxmin =
≤
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
5 - 23Nazarena Mazzaro, AAU
Design of Prismatic Beams for Bending
• Determine σall (If σall is the same for tension and compresion thenfollow 1,2,3 – Otherwise consider 4 in step 2)
• 1-Draw shear and bending-moment diagrams and determine |M|max
• 2-Determine Smin
• 3-Find the dimentions of the beam to use: b, h for S>Smin
• 4-Select the beam section so that σm<= σall for tensile and compressive stresses.
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
5 - 24Nazarena Mazzaro, AAU
Sample Problem 5.7
A 3.6 m-long overhanging timber beam AC is to be designed to support the distributed and concentrated loads shown. Knowing that timber of 100-mm nominal width (90-mm actual width) with a 12 MPa allowable stress is to be used, determine the minimum required depth h of the beam.
SOLUTION:
• Considering the entire beam as a free-body, determine the reactions at A and B.
• Develop the shear diagram for the beam and load distribution. From the diagram, determine the maximum bending moment.
• Determine the minimum acceptable beam section modulus. Find the value for the h.
60 kN/m 20 kN/m 90 mm
2.4 m 1.2 m
MECHANICS OF MATERIALS Beer • Johnston • DeWolf
5 - 25Nazarena Mazzaro, AAU
Sample Problem 5.7• Considering the entire beam as a free-body
( ) ( )( ) ( )( )
AVkN8.2
kN20kN4.14kN2.370VckN2.37
m6.3kN20m2.1kN4.14m4.20
=−=
−−+==
==
−−==
∑
∑
y
yy
A
A
AFB
BM
• Plot shear diagram and determine Mmax.( )
kN2.172.174.148.2414
4.14)4.2)(/6(
−=−=−−=−−=
−=−=−=−
B
AB
AB
VkNkNkNVV
kNmmkNcurveloadunderareaVV
• MA=Mc=0, Between A and B, M decreases an amount equal to area VAB, and between B and C in increases the same amount. Thus the |M|max= 24 kN.m
mmhmmhmmSbh
mmMPa
mkNMS
all
2.365102)90(61
61
10212
.24
362min
2
36maxmin
≥⇒×≥⇒≥
×===σ
14.4 kN 20 kN
2.4 m 1.2m
20 kN
(+24)
(+24)
-17.2 kN
-2.8 kN