CHAPTER 4: PURE BENDING Lecture 1:

1. Introduction

2. Shear and moment diagrams

3. Relation of load, shear and bending moment

Lecture 2:

4. Deformations of symmetric member in pure bending

5. Stress and strain due to pure bending

6. Design beam for bending

Lecture 3:

7. Composite sections

8. Eccentric bending

9. Unsymmetrical bending

1. Introduction

- To date we have studied:

+ Axial loading (bars and cables)

+ Torsion loading (shafts)

- Today’s objective:

Member subjected to bending.

- Pure bending:

+ Member subjected to equal and opposite couples M and M’ acting in the same longitudinal plane.

+ In this topic: Assume that each member possesses a symmetric plane.

Example 1:

An athlete holds the barbell.

Distances from the weights to hands are equal (a).

Pure bending occurs on

center portion of bar.

PA = PB

M = M’ = PA.a = PB.a

a a b

A B C D

PA PB

RC RD

C D M M’

Center portion in pure bending will form a circular arc

Beam and shaft:

Important mechanical elements in engineering.

Shaft:

Beam: slender and support loadings that are applied perpendicular to their longitudinal axis.

General beam: long, straight, constant cross-section area

Classification of beams:

Statically Determinate beams

Simply support beam

Overhanging beam

Cantilever beam

Statically Indeterminate beams

Continuous beam

Beam fixed at one end and simply supported at the other end

Fixed beam

Beam connected by hinges

2. Shear and bending-moment diagrams

Sign convection:

- Analysis of a simply supported beam:

Example 2: Draw shear and moment-bending diagrams

Example 3: Draw shear and moment-bending diagrams

Example 4: Draw shear and moment-bending diagrams

Example 5: Draw shear and moment-bending diagrams

3. Relation of load, shear and bending moment

A simply supported beam AB with distribution load w per unit length.

𝐹𝑦 = 0 V – (V + V) – w x = 0

V = - w x 𝑑𝑉

𝑑𝑥= −𝑤

the shear curve is negative

VD – VC = - 𝑤𝑑𝑥𝑥𝐷𝑥𝐶

= - (area under load curve between C&D)

Free body diagram of the portion of beam CC’

Relations between shear and bending moment

𝑀𝐶′ = 0 (M + M) – M - Vx + w x x2

= 0

M = V x - 1

2 𝑤 𝑥 2

Dividing by x and letting x approach 0

dM

dx= V

MD – MC = 𝑉𝑑𝑥𝑥𝐷𝑥𝐶

= area under shear curve between C & D

Example 7: Draw shear and moment-bending diagrams

1/ Draw shear and bending moment diagrams

2/ Example 4

Symmetric member in pure bending

x: Normal stress xy, xz: Shear stress

The system of the elementary internal forces exerted on the section is equivalent to the couple M.

Tension on the bottom

Compression on the top

𝐹𝑥 = 𝑥𝑑𝐴 = 0

𝑀

𝑦

= 𝑧𝑥𝑑𝐴 = 0;

𝑀

𝑧

= −𝑦𝑥𝑑𝐴 = 𝑀

4. Deformations in a symmetric member in pure bending

A prismatic member processing a

symmetric plane and subjected to

couples M and M’.

+ M: same in any cross section

bend uniformly

+ Top portion symmetric plane

= AB (a portion of a circle of center C).

+ AB and A’B’ change in length.

Consider a rubber (highly deformable material) straight prismatic member subjected to a bending moment.

3 assumptions:

- The longitudinal axis x (within the neutral surface) does not experience any change in length.

- All cross sections of the beam remain plane and perpendicular to the longitudinal axis during deformation.

- Any deformation of the cross section within its own plane will be neglected.

5. Stress and Strain due to bending

- Neutral surface: x = 0; x = 0

- Neutral axis: Neutral surface

Intersects a transverse section.

Determine longitudinal strain

of JK (above neutral surface)?

Length of original JK (DE):

L =

Length of JK: L’ = (-y)

Elongated deformation of JK:

= L’ – L = -y

x = 𝐿=;𝑦

= ;𝑦

(Geometry of deformation) Transverse section

- Longitudinal normal strain x varies linearly with distance y from the neutral surface.

x = −𝑦

The largest distance from the neutral surface: c

Maximum absolute value: 𝑚 =𝑐

𝑥 = −

𝑦

𝑐𝑚

Transverse section

Transverse Strain

Poisson’s ratio effect: 𝑦 = 𝑧 = −𝑥

Consider transverse strain 𝑧:

Above neutral axis, beam is compressed

𝑥 < 0 𝑧 > 0

Below neutral axis, beam is elongated

𝑥 > 0 𝑧 < 0

𝑥 = −𝑦

; 𝑧 =

𝑦

The neutral axis of transverse section

will be bent into a circle of radius ′ =

Stress distribution:

How stress distribution in a beam relates to the internal resultant bending moment acting on the beam’s cross section?

Assumptions:

- Stress under bending remains below yield strength y

(proportional limit and elastic limit)

- Homogeneous material 𝑥 = 𝐸𝑥

with 𝑥 = −𝑦

𝑐𝑚

𝑥 = −𝑦

𝑐𝐸𝑚 = −

𝑦

𝑐𝑚

(Stress distribution over cross-

sectional area)

The location of neutral surface:

𝐹𝑥 = 𝑥𝑑𝐴 = 0 −𝑚𝑐 𝑦𝑑𝐴 = 0 𝑦𝑑𝐴 = 0

In elastic range, neutral axis pass through the centroid of the section.

Elastic flexure formula

𝑀 𝑧 = −𝑦𝑥𝑑𝐴 = 𝑀;

−𝑦 −𝑦𝑚𝑐𝑑𝐴 =

𝑚𝑐 𝑦2𝑑𝐴 = 𝑀

Iz = 𝑦2𝑑𝐴𝐴: Area moment of inertia

𝑚 =𝑀.𝑐

𝐼𝑧 𝑥 = −

𝑀𝑦

𝐼𝑧

𝑚 =𝑀. 𝑐

𝐼𝑧

where

𝑚: The maximum normal stress in the member, which occurs at a point on the cross-sectional area farthest away from the neutral axis.

M: The resultant internal moment, determined from the equations of equilibrium.

c: The perpendicular distance from the neutral axis to a point farthest away from the neutral axis.

Iz: The moment of inertia of the cross-sectional area about the neutral axis.

Area moment of inertia of a rectangular area:

Iz = 𝑦2𝑑𝐴𝐴= 𝑦2(𝑏𝑑𝑦)𝑕/2

;𝑕/2=𝑏𝑕3

12

Maximum absolute value: 𝑚 =𝑐

Definition: Curvature of the neutral surface

1

=𝑚

𝑐=

𝑚/𝐸

𝑐=

𝑀𝑐𝐼𝑧𝐸

𝑐= 𝑀

𝐼𝑧𝐸

Example 1:

If the beam is subjected to a bending moment of M = 50 kN.m, determine the maximum bending stress in the beam?

Solution:

The centroid moment of inertia:

Example 2:

Bending in xz plane

determine Iz in yz plane.

A1 = A3 = 2400 mm2 A2 = 3060 mm2

Q1 = Q3 = 2400 x 50 = 120000 mm3

Q2 = 3060 x 15 = 45900 mm3

𝑧 =2𝑄1:𝑄2

2𝐴1:𝐴2= 36.37 mm

Iz,1 = Iz,3 = 24 ∗ 1003

12 + 2400*13.632

Iz,2 = 102∗303

12 + 3060*21.372

Iz = 2Iz,1 + Iz,2 = 6.519 x 106 [mm4]

Example 3: Determine centroid moment of inertia?

A1 = 1800 mm2; A2 = 1200 mm2

Q1 = 50 x 1800 = 90000 mm3

Q2 = 20 x 1200 = 24000 mm3

𝑌 = 90000:24000

1800:1200= 38 mm

Ix,1 = I1 + A1d12 Ix,2 = I2 + A2d2

2

I = Ix,1 + Ix,2

d1

d2

6. Design of prismatic beam for bending

The design of beam is controlled by the maximum absolute value 𝑀 𝑚𝑎𝑥 of bending moment.

𝑚 =𝑀 𝑚𝑎𝑥.𝑐

𝐼 𝑚 =

𝑀 𝑚𝑎𝑥

𝑆

For a safe design: 𝑚 ≤ 𝑎𝑙𝑙 < 𝑌 (elastic range)

where Y is yield strength,

𝑎𝑙𝑙 =𝑈

𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦 , U is ultimate stress

1/ Determine the centroidal area moment of inertia (I,z)?

2/ Knowing that the allowable normal stress for the steel used is 24 ksi. Determine the minimum flange width b that can be used.

7. Composite sections

- Member’s material is made of homogeneous, with modulus E:

- The member subjected to pure bending is made of 2 or

more materials with different moduli of elasticity: x = −𝑦

bonded

E1

E2

8. Eccentric bending

Centric loading: Line of action of loads P and P’ passes through the centroid of the cross section

Eccentric loading: Line of action of loads P and P’ does not pass through the centroid of the cross section.

P P

Vertical forces exerted on the press cause an eccentric loading

Member with eccentric loading: F = P’ and M = P’d

Internal forces in member of eccentric loading

Stress distribution in eccentric loading: 𝑥 𝑐𝑒𝑛𝑡𝑟𝑖𝑐 + 𝑥 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 = 𝑥

𝑥 = 𝑥 𝑐𝑒𝑛𝑡𝑟𝑖𝑐 + 𝑥 𝑏𝑒𝑛𝑑𝑖𝑛𝑔

𝑥 =𝑃

𝐴−𝑀.𝑦

𝐼𝑧

- The distribution of stresses across the section is linear but not uniform.

Another example of stress distribution in eccentric loading:

- The neutral axis does not coincide with the centroid axis of the section.

At centroid axis: y = 0 but 𝑥 0

Neutral axis corresponds to points 𝑥 = 0

Neutral axis

Example 3: Determine max. value of P?

all = 30 Mpa (in tension), all = 120 Mpa (in compressive)

P uniform stress distribution 0 =𝑃

𝐴

M= Pd linear stress distribution: 1 =𝑀.𝑐𝐴

𝐼; 2 =

𝑀.𝑐𝐵

𝐼

Superposition: 𝐴 = - 𝑃

𝐴 + 𝑀.𝑐𝐴

𝐼 = 377P < 30

𝐵 = - 𝑃

𝐴 - 𝑀.𝑐𝐵

𝐼 = - 1559P < -120

Example 4: P = 8 kN

Determine the stress at point A.

Solution:

9. Unsymmetrical bending

Symmetrical bending:

+ Member possesses at least one plane of symmetry

+ Member is subjected to couples acting in that plane.

+ The members remained symmetric and bend in the plane of symmetry

+ The neutral axis of the cross section coincided with the axis of the couple.

Un-symmetric bending:

+ Bending couples do not act in the plane of symmetry of the member.

Bending couples act in different plane

Member does not possess any plane of symmetry

An unsymmetrical case: - Resolve M into Mz and My

Mz = Mcos; My = Msin

𝑥 = −𝑀𝑧.𝑦

𝐼𝑧 𝑥 = +

𝑀𝑦.𝑧

𝐼𝑦

𝑥 = −𝑀𝑧.𝑦

𝐼𝑧+𝑀𝑦.𝑧

𝐼𝑦

Example 5:

Couple M is applied to a beam of the cross section shown in a plane forming an angle with the vertical.

Determine stress at:

(a) Point A

(b) Point B

(c) Point D

1/ Determine the stress at points A, B, D and E.