Lecture 2

Bending stresses

Structure II (AR-106 G)

BY- AR. KHURRAM ALI Asst. Professor Gateway College of architecture Sonipat, Haryana

Introduction

• As we’d discussed earlier, the stress produced in a body to resist the bending moment are known as bending stress.

Pure bending or Simple bending

• The bending of beam which is not accompanied by any shear force is termed as pure bending or simple bending.

• Figure shows a beam ABCD with equal overhangs and supported at B and C .

• Let a point load W be applied at each end of the beam. SF and BM diagrams are drawn for the given loading.

Pure bending or Simple bending

• From these diagram, it is clear that there is no shear force between B and C but the BM between B and C is constant.

• This means that between B and C, the beam is subjected to a constant bending moment only.

• This condition of the beam between B and C is known as pure or Simple bending.

Assumptions in the theory of simple bending

• The material of beam is homogenous and isotropic.

• The section of a beam, which is plane before bending will remain plane after bending.

• The value of Young’s modulus of elasticity is the same in tension and compression.

• The loads are applied in the plane of bending.

• The beam material is stressed within its elastic limit and thus obeys Hooke’s law.

Neutral Axis (Neutral layer)

• Neutral axis of a beam is the axis at which the stress is zero. It is written as N.A.

• Neutral axis is always passes through the centroid of the cross-section of beam and is parallel to the base of the cross-section.

• The fibres of the beam in bending are in a state of tension on one side of the N.A. and compression on the other side.

Bending Equation

The equation of bending is:

M/I = σb/y = E/R

where,

M = B.M. or Moment of Resistance of the section in Nmm.

I = MOI of the section about N.A. in mm4

σb= Bending stress at distance y from N.A. in N/mm2

y = distance of fibre from N.A. in mm

E = Young’s modulus of elasticity in N/mm2

R = Radius of curvature of N.A. in mm

Moment of Resistance (MR)

• When a beam is subjected to BM, the section above and below N.A. develops internal forces or stresses of opposite nature.

• These internal stresses or forces have certain moments about N.A.

• The algebraic sum of these moments is equal and opposite to the bending moment acting on the section thus keeping the section of beam in equilibrium.

• The sum of moments of the internal forces about N.A. is known as the Moment of Resistance.

• MR is equivalent to Max BM in the beam.

• From bending equation, we have

M/I = σb/y

or M = (σb/y)xI

or M = σb x Z

where Z is called Section Modulus.

Section Modulus (Modulus of Section)

• Section Modulus is the ratio of MOI of a section about the N.A. to the distance of the outermost layer from the N.A. It is denoted by Z.

• Hence mathematically, Z = I/y

• Note: Section Modulus is of practically greater use than the MOI. The strength of the beam section depends mainly on section modulus.

Section Modulus of various shapes

• Rectangular section

• Z = I/y

• MOI of rectangle about its N.A.

• I = bd3/12

• y = distance of outermost layer from N.A. = d/2

• hence, Z = (bd3/12)/(d/2)

• Z = bd2/6

Similarly, find the section modulus of Hollow rectangular section, circular section, hollow circular section etc.

Illustrative examples

A steel plate of width 60mm and of thickness 10mm is bent into a

circular arc of radius 10m. Determine the maximum stress induced and the bending moment which will produce the maximum stress. Take E = 2x105N/mm2.

Solution- width of plate, b = 60mm

thickness of plate, t = 10mm

MOI, I = bt3/12 = 60x103/12

= 5000mm4

Radius of curvature, R = 10m = 10000mm

Using the relation,

σb/y = E/R

σb = E.y/R

Contd.

Stress will be maximum, when y is maximum. But y will be maximum at the top layer or bottom layer.

y = t/2 = 10/2 = 5mm

hence, σb = E.y/R

= 2x105x5/10x10000 = 100N/mm2

again using relation,

M/I = E/R

M = ExI/R

= (2x105x5000)/(10x103)

= 100x103Nmm = 100Nm.

Contd.

A rectangular beam 100mm deep and 75mm wide is simply supported over a span of 6m. It carries a UDL of 4000N/m over the whole span. Calculate the max bending stress developed in the section at:

i) supports

ii) 1.5m from the supports

iii) mid span of the beam

Solution- depth of beam, d = 100mm

width of beam, b = 75mm

max bending stress will be at the extreme fibre i.e. d/2 from NA in all the cases.

now, section modulus for rectangular section,

Z = bd2/6

= 75x1002/6 = 125000mm3

reaction, RA = RB = total load/2

= 4000x6/2 = 12000N

i) Bending stress at supports:

BM at supports (A or B), M = 0

Bending stress, σb = M/Z = 0

Contd.

ii) Bending stress at 1.5m from supports:

BM at C (1.5m from A), M = 12000x1.5-4000x1.5x (1.5/2)

= 13500Nm = 13.5x106 Nmm

Bending stress, σb = M/Z = 13.5x106/125000

= 108 N/mm2

iii) Bending stress at mid span of beam:

BM at D (mid span), M = wl2/8

= 4000x62/8

= 18000Nm = 18x106 Nmm

bending stress, σb = M/Z = 18x106/125000

= 144 N/mm2

Contd.

Test

1. A rectangular beam 300mm deep 150mm wide is simply supported over a span of 8m. Find the max UDL that the beam can carry, if the bending stress is not to exceed 120 N/mm2

2. A rectangular beam 350mm deep 175mm wide is simply supported over a span of 7m. Find the max UDL that the beam can carry, if the bending stress is not to exceed 120 N/mm2