Chapter 4 Pure Bending Ch 2 – Axial Loading Ch 3 – Torsion Ch 4 – Bending -- for the designing...

Chapter 4 Pure Bending 

Ch 2 – Axial Loading

Ch 3 – Torsion

Ch 4 – Bending

-- for the designing of beams and girders

4.1 Introduction

A. Eccentric Loading

B. Pure Bending

4.2 Symmetric Member in Pure Bending

M = Bending Moment

Sign Conventions for M:

-- concave upward

⊝ -- concave downward

Force Analysis – Equations of Equilibrium

0xdA

0xz dA ( )xy dA M

Fx = 0

My-axis = 0

Mz-axis = 0

xz = xy = 0

(4.1)

(4.2)

(4.3)

4.3 Deformation in a Symmetric Member in Pure Bending

Assumptions of Beam Theory:

1. Any cross section to the beam axis remains plane

2. The plane of the section passes through the center of curvature (Point C).

Plane CAB is the Plane of Symmetry

The Assumptions Result in the Following Facts:

1. xy = xz = 0 xy = xz = 0

2. y = z = yz = 0

The only non-zero stress: x 0 Uniaxial Stress

The Neutral Axis (surface) : x = 0 & x = 0

L

' ( )L y

' L L

Where = radius of curvature

= the central angle

Line JK (4.5)

Before deformation: DE = JK

Therefore,

Line DE (4.4)

(4.6)

( )y y

The Longitudinal Strain x =

o

x

x

y

L

y

x varies linearly with the distance y from the neutral surface

(4.9)

The max value of x occurs at the top or the bottom fiber:

m

c

(4.8)

Combining Eqs (4.8) & (4.9) yields

x m

yc

(4.10)

4.4 Stresses and Deformation is in the Elastic Range

For elastic response – Hooke’s Law

x xE

( )x m

yE E

c

x m

yc

maxx m

y yc c

Therefore,

(4.11)

(4.10)

(4.12)

Based on Eq. (4.1)

0xdA

maxx m

y yc c

0( ) mx m

ydA dA ydA

c c

(4.1)

(4.12)

(4.13)

Hence,

0ydA first moment of area

Therefore,

Within elastic range, the neutral axis passes through the centroid of the section.

According to Eq. (4.3)

( )xy dA M x m

yc

( )( ) m

yy dA M

c

(4.3)

and (4.12)

It follows

or2 m y dA M

c

(4.14)

Since 2I y dA2 m y dA M

c

m

McI

Eq. (4.24)

can be written as

Elastic Flexure Formula (4.15)

At any distance y from the neutral axis:

x

My

I Flexural Stress (4.16)

If we define

Eq. (4.15) can be expressed as

IElastic section modulus = S =

c

m

M

S (4.18)

(4.17)

Solving Eq. (4.9) m

c

1 m

c

(4.9)

1 1m McEc Ec I

1 MEI

mm E

[ ] m

McI

Finally, we have

(4.21)

3

2

11 1126 6

2

bhI

S bh Ahhc

4.5 Deformation in a Transverse Cross Section

Assumption in Pure Bending of a Beam:

The transverse cross section of a beam remains “plane”.

However, this plane may undergo in-plane deformations.

A. Material above the neutral surface (y>0), ,ⓛ ⓛx x

y x z x

x

y

Since (4.8)

Hence, yy

zy

(4.22)

Therefore,

,y z

B. Material below the neutral surface (y<0), ,x x

,ⓛ ⓛy z

x

y

As a consequence,

Analogous to Eq. (4.8)

For the transverse plane:

x

y

1'

y x x

y y y

= radius of curvature,

1/ = curvature

(4.23)'Anticlastic cur

1 =vature

4.6 Bending of Members Made of Several Materials

x

y

11 1 x

E yE

From Eq. (4.8)

For Material 1:

For Material 2:2

2 2 x

E yE

(Composite Beams)

11 1

E ydF dA dA

22 2

E ydF dA dA

1 12

( )( )

nE y E ydF dA ndA

Designating E2 = nE1

x

MyI

2 xn

Notes:

1. The neutral axis is calculated based on the transformed section.

2.

3. I = the moment of inertia of the transformed section

4. Deformation -- 1

1 ME I

Beam with Reinforced Members:

As = area of steel, Ac = area of concrete

Es = modulus of steel, Ec = modulus of concrete

n= Es/Ec

Beam with Reinforced Members:

As = area of steel, Ac = area of concrete

Es = modulus of steel, Ec = modulus of concrete

n= Es/Ec

( ) ( ) 02

s

xbx nA d x

210

2 s sbx nA x nA d determine the N.A.

4.7 Stress Concentrations

m

McKI

4.12 Eccentric Axial Loading in a Plane of Symmetry

( ) ( ) x x centric x bending

x

P My

A I

4.13 Unsymmetric Bending

-- Two planes of symmetry

y – axis & z-axis

-- Single plane of symmetry – y-axis

--M coincides with the N.A.

For an arbitrary geometry + M applies along the N.A

Fx = 0 0xdA 0xz dA

( )xy dA M My = 0

Mz = 0

(4.1)

(4.2)

(4.3)

Substituting mx

yc

into Eq. (4.2)

(the Centroid = the N.A.)

(moment

equilibrium)

(moment equilibrium)

Plane of symmetry

0 0( ) ( )m myz dA or z y dA

c c

0yzyzdA I

We have

or

Iyz = 0 indicates that y- and z-axes are the principal centroid of the cross section.

Hence, the N.A. coincides with the M-axis.

(knowing m/c = constant)

If the axis of M coincides with the principal centroid axis, the superposition method can be used.

yx

y

M z

I

zx

z

M y

I

= =

+

sinyM M coszM M

Case A

Case B

For Case A

For Case B

For the combined cases : yzx

z y

M zM y

I I

(4.53)

(4.54)

(4.55)

tan tanz

y

I

I

0yz

z y

M zM y

I I

( tan ) z

y

Iy z

I

The N.A. is the surface where x = 0. By setting x = 0 in Eq. (4.55), one has

Solving for y and substituting for Mz and My from Eq. (4.52),

(4.56)

This is equivalent to / tanz

y

Iy z m slope

I

The N.A. is an angle from the z-axis:

(4.57)

4.14 General Case of Eccentric Axial loading

yZx

z y

M zP M yA I I

yz

z y

MM Py z

I I A

(4.58)

(4.58)

' 'R R ' ' r r

r R y

4.15 Bending of Curved Members

Before bending After bending

Length of N.A. before and after bending

The elongation of JK line

' ' r R ySince

(4.59)

(4.60)

(4.61)

We have ' '( ) ( ) R y R y

If we define - = and knowing R = R , thus

y

x

y

r r

x

y

R y

x

E y

R y

(4.64)

Based on the definition of strain, we have

r R y

(4.63)

Also, x = E x

(4.62)

(4.65)

Substituting into the above equation,

x

E y

R y

r R y

x

E R r

r

Plotting

x is not a linear function of y.

Since

Substituting this eq. into Eq. (4.1)

y = R – r, therefore,

0xdA

0E R r

dAr

and 0E R r

dAr

0R r

dAr

0

dAR dA

r or cos .

Et

Therefore, R can be determined by the following equation:

AR

dAr

1r rdA

A

1 1 1dA

R A r Or in an alternative format:

(4.67)

The centroid of the section is determined by

(4.59)

(4.66)

Comparing Eqs. (4.66) and (4.67), we conclude that:

The N.A. axis does not pass through the Centroid of the cross section.

Mz = M E R r

ydA Mr

y R r

2( )

E R rdA M

r

2[ 2 ]

E dA

R RA rdA Mr

( 2 )

E

RA RA rA M

Since

Recalling Eqs. (4-66) and (4.67), we have

or

, it follows

Finally,

( )

E M

A r R

(4.68)

E M

Ae

By defining , the above equation takes the new forme r R

( )

x

My

Ae R y ( )

xM r R

Aer

(4.69)

Substituting this expression into Eqs. (4.64) and (4-65), we have

'

'

1 1

R R

'

'

1 1 1(1 ) (1 )

M

R R R EAe

and (4.70, 71)

Determination of the change in curvature:

From Eq. (4.59)

Since and from Eq. (4.69), one has

1 1'

MR R EAeR

Hence, the change of curvature is

(4.72)

End of Ch 4