Chapter 4Pure Bending

4.1 Introduction

- members subjected to bending- M & M' are equal and opposite couples - Said to be pure bending

Bending due to pure bending

Bending due to eccentric loading

Bending due to transverse loading

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 +

x−components :∫ x dA=0moment− y−axis :∫ z x dA=0

moment−z−axis :∫−y x dA=M

4.3 Deformations in a symmetric member in pure bending

-M & M' are positive sign +

There must exist a surface , where normal strain and stress are zero. This surface is called the neutral surface.

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

The maximum value of longitudinal strain

m=c

rewritten;

x=− ycm

4.4 Stress and Deformations in the Elastic Range x=E x

E x=− ycE m

x=− ycm

∫ 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)

I/c dependent only upon the geometry of the cross section, this ratio called the elastic section modulus

elastic modulus=S= Ic

m=MS

Same in area, different in elastic section modulus

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

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

4.5 Deformations in a Transverse Cross Section

y=−x

z=−x

x≠0, y= z=0

x=−y

4.7 Stress Concentrations

max=K nominal nominal=McI

Homeworks & Assignments

-4.1-4.5-4.16-4.65

4.12 Eccentric axial loading in a plane of symmetry

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

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

x= xcentric xbending

x=PA−My

IA=c2=6mm2

I= 14c4

0= PA−

M yo

I

-distance between centroidal and neutral axis

Homeworks:- 4.99- 4.103- 4.121