6 - Bending Stress - VTCtycnw01.vtc.edu.hk/cbe5029/6 - Bending Stress.pdf · Chapter 6 – BENDING STRESS Page 6 -2 The above beam segment BC is subject to a pure sagging moment

Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 6 – BENDING STRESS

Page 6 -1

BENDING STRESS

The effect of a bending moment applied to a cross-section of a beam is to induce a state of stress across that section. These stresses are known as bending stresses and they act normally to the plane of the cross-section.

PURE BENDING

If a simply supported beam carries two point loads of 10 kN as shown in the following figure, pure bending occurs at segment BC.

AB C D

10 kN 10 kN

L L

10 kN

10 kN

Shear force

PL

Bending moment

BC

B C


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The above beam segment BC is subject to a pure sagging moment. The beam segment will bend into the shape shown in the following figures in which the upper surface is concave and the lower convex. It can be seen that the upper longitudinal fibres of the beam are compressed while the lower fibres are stretched. It follows that between these two extremes there is a fibre that remains unchanged in length. Thus the direct stress varies through the depth of the beam from compression in the upper fibres to tension in the lower. Clearly the direct stress is zero at the fibre which does not change in length. The surface, which contains this fibre and runs through the length of the beam, is known as the NEUTRAL SURFACE or NEUTRAL PLANE; the line of intersection of the neutral surface and any cross-section of the beam is termed the NEUTRAL AXIS.

Neutral Axis: It is the axis of the cross-section of a beam at which both bending strain and bending stress are zero. The neutral axis passes through the centroid of the cross-section.


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Deformation Geometry for a symmetrical beam in pure bending

Cross sections of a beam that are plane and normal to the axis of the beam before bending remain plane and normal to the axis after bending.


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SIMPLE THEORY OF BENDING

Assumptions:

1. Beams are initially straight. 2. The material is homogenous and isotropic (i.e. its mechanical properties

are the same in all directions.) 3. Stress-strain relationship is linear and elastic. 4. Young’s modulus is the same in tension as in compression. 5. Sections are symmetrical about the plane of bending. 6. Sections which are plane before bending remain plane after bending. Implications of the last assumption: 1. Each section rotates during bending about a neutral axis. 2. The distribution of strain across the section is linear, with zero strain at

the neutral axis. 3. The section is divided into tensile and compressive zones separated by a

neutral surface. The theory gives very accurate results for stresses and deformations for most practical beams provided that deformations are small.


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Bending Stresses in Beams Basic Assumption: “Plane sections through a beam taken normal to its axis remain plane after the beam is subject to bending.”

Result 1: The normal strain varies along the beam depth linearly with the

distance y. Result 2: (Assuming Linear Elasticity, by using Hooke’s Law) The normal stress varies along the beam depth linearly with the

distance y.


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Result 3 : For the top and bottom edges of the section, one would be in tension while the other would be in compression. A plane of zero deformation and zero bending stresses (Neutral Axis plane) exists between the two.


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Result 4 : (From considerations of equilibrium, it can be shown that :) The Neutral Axis passes through the Centroid of the section.

Result 5 : The stress and strain varies linearly from zero at the Neutral Axis to their absolute maximum values at the largest value of y.


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Result 6: The elastic Flexure formula for beam bending is as follows: The bending stress σx at a distance y from the Neutral Axis is given by:

σx = -My/I

where M = Bending Moment I = Moment of inertia of the section (about its centroidal axis)

and the Maximum normal stress σmax is given by:

σmax = -Mc/I

where c = |y|max

Use of the flexure formula The method of solving any beam stress problem involves the following steps: 1. Determine the maximum bending moment on the beam by drawing the

shear and bending moment diagrams. 2. Locate the centroid of the cross section of the beam. 3. Compute the moment of inertia of the cross section with respect to its

centroidal axis. 4. Compute the distance c from the centroid axis to the top or bottom of the

beam, whichever is greater. 5. Compute the stress from the flexural formula, σmax = -Mc/I


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Example 1 A simply supported beam is subject a point load of 1500 N at the mid-span of the beam as shown in the following figure. The cross section of the beam is a rectangular 100 mm high and 25 mm wide. Calculate the maximum stresses due to bending.

Solution Step 1 The maximum bending moment occurs at the mid-span of the

beam. M = 750 * 1.7 = 1275 Nm Step 2 The centroid of the rectangular section is at the intersection of its

two axes of symmetry. Step 3 For the rectangular section, I = 25(100)3/12 = 2.08 x 106 mm4 Step 4 c = 50 mm Step 5 The maximum tensile and compressive stresses due to bending are: σmax, = -Mc/I = 1275 x 103 (50)/ 2.08 x 106 = 30.6 N/mm2 or 30.6 MPa

1500 N

1.7 m 1.7 m


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Example 2 A simply supported I-beam carries a uniformly distributed load of 5 kN/m over the entire span of 6 m. The cross section of the beam is shown in the following figure. Determine the bending stresses that acts at points B and D, located at the mid-span of the beam.

300 mm

20 mm

20 mm

20 mm

115 mm 115 mm

B

D Solution Step 1 The supported reaction = 5 * 6 / 2 = 15 kN Consider the following free body diagram.

5 kN/m

15 kN3 m

HX

V

MX

X

The bending moment occurs at the mid-span of the beam.

M = 15 * 3 – 5 * 3 * 1.5 = 22.5 kNm


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Step 2 The centroid of the I-section is at the intersection of its two axes of symmetry.

Step 3 For the I-section, Moment of inertia, I

= 250(340)3/12 – 2*115(300)3/12 = 3.01 x 108 mm4 Step 4 At point B, yb = 150 mm

At point D, yd = 170 mm

Step 5 The compressive stress due to bending at point B is: σb, = -M yb /I = 22.5 x 106 * 150 / 3.01 x 108 = 11.2 N/mm2 or 11.2 MPa

The tensile stress due to bending at point D is: σd, = -M yd /I = 22.5 x 106 * 170 / 3.01 x 108 = 12.7 N/mm2 or 12.7 MPa The two dimensional view of the stress distribution is shown in the following figure.

B

D


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NON-SYMMETRICAL SECTIONS UNDER BENDING The critical difference between the non-symmetrical section and symmetrical section under bending is that in a non-symmetrical section, such as a T beam or triangular shape beam, the location of the centroid is no longer obvious and is usually never at the mid-height of the section. In the T beam, the centroid is located near the top flange of the member. Deformations and bending stresses in the member still vary linearly in the member and are proportional to the distance from the neutral axis of the member. This implies that the stress levels at the top and bottom of the beam are no longer equal as they typically are in symmetrical sections. The stresses are greater at the bottom of the beam than they are at the top because of the larger y distance. The two and three dimensional views of bending stress distribution in a T beam are shown in the following figures.


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Page 6 -14

SECTION MODULUS The maximum tensile and compressive stresses in the beam occur at points located farthest from the neutral axis. Let us denote by c1 and c2 the distances from the neutral axis to the extreme elements in the positive and negative y directions, respectively. Then the maximum normal stresses:

σ1 = M c1 /I = M / Z1

σ2 = -M c2 /I = M / Z2

in which,

Z1 = I / c1

Z2 = I / c2

The quantities Z1 are Z2 are known as the section moduli of the cross sectional area. We see that a section modulus has dimensions of length to the third power.

Stress distribution on a non-symmetrical section


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If the cross section is symmetric with respect to the z axis, then c1 = c2 = c, and the maximum tensile and compressive stresses are equal numerically:

σ1 = -σ2 = M c /I = M / Z

in which Z = I / c

is the section modulus. For a beam of rectangular cross section with width b and height h, the moment of inertia and section modulus are I = bh3/12 Z = bh2/6

Stress distribution on a symmetrical section


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Example 3 A simply supported beam of span length of 8 m is subject to a uniformly distributed load of 2 kN/m over the entire span and a point load of 8 kN at 5m from the left support of the beam as shown in the following figure. Determine the maximum tensile and compressive stresses in the beam due to bending.

HA VA VC

B

5m 3m

8 kN

2 kN/mA C

700 mm

220 mm

Cross section of the beam


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Solution The shear force and bending moment diagrams for the beam are shown in the following figures.

B

5m 3m

8 kN

2 kN/m

11 kN 13 kN

Shear Force (kN)11 9 7 5

3 1

-7-9 -11

-13

0 010

1824 28 30

12

22

Bending Moment (kNm)

A C

AB

C

CBA

+ve

+ve


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Method 1 : use section modulus for calculations Section modulus of the cross-sectional area, Z = bh2/6 = 220*7002/6 =1.8 x 107 mm3 Maximum bending tensile and compressive stresses, σ = Mmax / Z = 30 x 106/1.8 x 107

= 1.67 N/mm2 Method 2 : use moment of inertia for calculations Moment inertia of the cross-sectional area, I = bh3/12 = 220*7003/12 =6.29 x 109 mm4 Maximum bending tensile and compressive stresses, σ = Mmax c/ I = 30 x 106* 350/6.29 x 109

= 1.67 N/mm2


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Example 4 A simple supported beam with a span of 4 m supports a uniformly distribution load w. The cross section of the beam is shown in the following figure. If the allowable bending stress is 70 MPa, find the allowable uniform load w.

Solution The maximum bending moment, Mmax = wL2/8 = w *40002/8 = 2 x 106w Nmm Moment inertia of the beam section, I = 225*3003/12-125*2003/12 = 422.9 x 106 mm4 The section modulus of the beam section, Z = 422.9 x 106/150 = 2.819 x 106 mm3

300 mm

225 mm

50 mm

50 mm

50 mm

50 mm


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From the bending stress formula, Mmax = σ Z 2 x 106 w = 70 * 2.819 x 106 w = 98.665 N/mm w = 98.665 kN/m Example 5 A simply supported T-beam shown in the following figure carries a uniformly distributed load of 1 kN/m over the entire span of 6 m. Calculate the maximum bending stresses induced in the beam.

120 mm

40 mm

120 mm100 mm

Centroidal Axis


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Solution The maximum bending moment,

Mmax = 1*62/8 = 4.5 kNm The moment inertia of beam, I = 120(40)3/12 + 120*40*(140-100)2 + 40(120)3/12 + 40*120*(100-60)2 = 2.18 x 107 mm4 The maximum tensile stress in the bottom fibre, σ = M c2 /I = 4.5 x 106*100/2.18 x 107

= 20.64 N/mm2 The maximum compressive stress in the top fibre, σ = M c1 /I

= 4.5 x 106*60/2.18 x 107

= 12.4 N/mm2


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Example 6 A T-beam is loaded as shown in the following figure. Calculate the maximum bending stresses induced in the beam.

HB VB VD

A B C D E

10 kN 20 kN4 kN/m

2m 2m 4m 2m

120 mm

40 mm

120 mm100 mm

Centroidal Axis


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Solution The shear force and bending moment diagrams for the beam are shown in the following figures.

37.3

A B C D E

10 kN 20 kN4 kN/m

24.7

-10 -10

27.319.3

-0.7

-16.7

8

0

Shear Force (kN)

-20

+46.7

-34.7

+8

14

14

14

-20

+26.7

-80

101

Bending Moment (kNm)

A B C DE

E DC

BA


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The moment inertia of beam, I = 120(40)3/12 + 120*40*(140-100)2 + 40(120)3/12 + 40*120*(100-60)2 = 2.18 x 107 mm4 When M = -20kNm at point B, The maximum tensile stress in the top fibre, σT = M c1 /I = 20 x 106*60/2.18 x 107

= 55 N/mm2 The maximum compressive stress in the bottom fibre, σC = M c2 /I

= 20 x 106*100/2.18 x 107

= 91.7 N/mm2 When M = 26.7 Nm at point C, The maximum tensile stress in the bottom fibre, σT = M c2 /I = 26.7 x 106*100/2.18 x 107

= 122.5 N/mm2 The maximum compressive stress in the top fibre, σC = M c1 /I

= 26.7 x 106*60/2.18 x 107

= 73.5 N/mm2

Therefore, the maximum tensile stress and compressive stress are 122.5 N/mm2 and 91.7 N/mm2 respectively.


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PROBLEMS for bending stress


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Documents

6 - Bending Stress - VTCtycnw01.vtc.edu.hk/cbe5029/6 - Bending Stress.pdf · Chapter 6 – BENDING STRESS Page 6 -2 The above beam segment BC is subject to a pure sagging moment