ch4 presentation
DESCRIPTION
The slide of Mechanics of Materials ITRANSCRIPT
![Page 1: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/1.jpg)
Chapter 4Pure Bending
![Page 2: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/2.jpg)
![Page 3: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/3.jpg)
![Page 4: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/4.jpg)
![Page 5: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/5.jpg)
![Page 6: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/6.jpg)
4.1 Introduction
- members subjected to bending- M & M' are equal and opposite couples - Said to be pure bending
![Page 7: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/7.jpg)
![Page 8: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/8.jpg)
Bending due to pure bending
![Page 9: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/9.jpg)
Bending due to eccentric loading
![Page 10: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/10.jpg)
Bending due to transverse loading
![Page 11: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/11.jpg)
4.2 Symmetric member in pure bending
-At arbitrary section c, the internal forces in the section be equivalent to couple M
-M & M' are positive sign +
![Page 12: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/12.jpg)
x−components :∫ x dA=0moment− y−axis :∫ z x dA=0
moment−z−axis :∫−y x dA=M
![Page 13: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/13.jpg)
4.3 Deformations in a symmetric member in pure bending
-M & M' are positive sign +
![Page 14: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/14.jpg)
![Page 15: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/15.jpg)
There must exist a surface , where normal strain and stress are zero. This surface is called the neutral surface.
![Page 16: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/16.jpg)
The deformation of JK is=L '−L
=− y −=−y
Then the longitudinal strain of JK is
x=L=−y
=− y
Arc DE L= Arc JK L '=− y
![Page 17: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/17.jpg)
The maximum value of longitudinal strain
m=c
rewritten;
x=− ycm
![Page 18: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/18.jpg)
4.4 Stress and Deformations in the Elastic Range x=E x
E x=− ycE m
x=− ycm
![Page 19: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/19.jpg)
∫ x dA=∫ − ycmdA=−
m
c ∫ y dA=0
∫ y dA=0
∫ −y xdA=M
∫ −y− ycmdA=M
m
c ∫ y2dA=M m=McI
x=−My
IFlexural stress
From equilibrium conditions on page 212 (Beer & Johnston)
![Page 20: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/20.jpg)
I/c dependent only upon the geometry of the cross section, this ratio called the elastic section modulus
elastic modulus=S= Ic
m=MS
![Page 21: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/21.jpg)
Same in area, different in elastic section modulus
![Page 22: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/22.jpg)
![Page 23: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/23.jpg)
The deformation of the member caused by the bending moment M is measured by the curvature of neutral surface.
curvature= 1=m
c=m
Ec= 1
EcMcI
1=
MEI
![Page 24: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/24.jpg)
Example 4.01 A steel bar of 20x 60-mm rectangular cross section is subjected to two equal and opposite couples acting in the vertical plane of symmetry of the bar. Determine the value of the bending moment M that causes the bar to yield. Assume yield strength = 250 MPa
=MyI
y=c=0.03mm
I= 112
bh3= 112
20mm60mm3
=250MPa
M = Ic
=3kN.m
![Page 25: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/25.jpg)
4.5 Deformations in a Transverse Cross Section
y=−x
z=−x
x≠0, y= z=0
x=−y
![Page 26: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/26.jpg)
![Page 27: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/27.jpg)
4.7 Stress Concentrations
max=K nominal nominal=McI
![Page 28: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/28.jpg)
![Page 29: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/29.jpg)
Homeworks & Assignments
-4.1-4.5-4.16-4.65
![Page 30: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/30.jpg)
4.12 Eccentric axial loading in a plane of symmetry
![Page 31: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/31.jpg)
![Page 32: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/32.jpg)
x= xcentric xbending
x=PA−My
I
The neutral axis does not coincide with the centroidal axis since stress-x isn't zero at this point
![Page 33: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/33.jpg)
Example 4.07 : Chain made by steel rod dia=12 mm.Determine the largest tensile and compressive stresses in the straight portion of a link, the distance between the centroidal and the neutral axis of a cross section
P=700 NM=Pd=11.2 N.m
![Page 34: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/34.jpg)
x= xcentric xbending
x=PA−My
IA=c2=6mm2
I= 14c4
0= PA−
M yo
I
-distance between centroidal and neutral axis
![Page 35: Ch4 presentation](https://reader031.vdocuments.mx/reader031/viewer/2022013102/54be7e644a79596b288b4676/html5/thumbnails/35.jpg)
Homeworks:- 4.99- 4.103- 4.121