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.


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1 BENDING STRESS The effect of a bending moment applied to a crosssection 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 crosssection. 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. 10 kn 10 kn A B C D L L Shear force 10 kn Bending moment B C 10 kn B PL C Page 61
2 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 crosssection of the beam is termed the NEUTRAL AXIS. Neutral Axis: It is the axis of the crosssection of a beam at which both bending strain and bending stress are zero. The neutral axis passes through the centroid of the crosssection. Page 62
3 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. Page 63
4 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. Stressstrain 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. Page 64
5 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. Page 65
6 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. Page 66
7 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. Page 67
8 Result 6: The elastic Flexure formula for beam bending is as follows: The bending stress σ x at a distance y from the Neutral Axis is given by: σ x = My/I where M = Bending Moment I = Moment of inertia of the section (about its centroidal axis) and the Maximum normal stress σ max is given by: σ max = 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, σ max = Mc/I Page 68
9 Example 1 A simply supported beam is subject a point load of 1500 N at the midspan 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 N Solution 1.7 m 1.7 m Step 1 The maximum bending moment occurs at the midspan 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, Step 4 c = 50 mm I = 25(100) 3 /12 = 2.08 x 10 6 mm 4 Step 5 The maximum tensile and compressive stresses due to bending are: σ max, = Mc/I = 1275 x 10 3 (50)/ 2.08 x 10 6 = 30.6 N/mm 2 or 30.6 MPa Page 69
10 Example 2 A simply supported Ibeam 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 midspan of the beam. 115 mm 115 mm 20 mm B 20 mm 300 mm D 20 mm Solution Step 1 The supported reaction = 5 * 6 / 2 = 15 kn Consider the following free body diagram. 5 kn/m H X M X 15 kn 3 m V X The bending moment occurs at the midspan of the beam. M = 15 * 3 5 * 3 * 1.5 = 22.5 knm Page 610
11 Step 2 The centroid of the Isection is at the intersection of its two axes of symmetry. Step 3 For the Isection, Moment of inertia, I = 250(340) 3 /12 2*115(300) 3 /12 = 3.01 x 10 8 mm 4 Step 4 At point B, y b = 150 mm At point D, y d = 170 mm Step 5 The compressive stress due to bending at point B is: σ b, = M y b /I = 22.5 x 10 6 * 150 / 3.01 x 10 8 = 11.2 N/mm 2 or 11.2 MPa The tensile stress due to bending at point D is: σ d, = M y d /I = 22.5 x 10 6 * 170 / 3.01 x 10 8 = 12.7 N/mm 2 or 12.7 MPa The two dimensional view of the stress distribution is shown in the following figure. B D Page 611
12 NONSYMMETRICAL SECTIONS UNDER BENDING The critical difference between the nonsymmetrical section and symmetrical section under bending is that in a nonsymmetrical 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 midheight 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. Page 612
13 Page 613
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 c 1 and c 2 the distances from the neutral axis to the extreme elements in the positive and negative y directions, respectively. Then the maximum normal stresses: σ 1 = M c 1 /I = M / Z 1 σ 2 = M c 2 /I = M / Z 2 in which, Z 1 = I / c 1 Z 2 = I / c 2 The quantities Z 1 are Z 2 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 nonsymmetrical section Page 614
15 If the cross section is symmetric with respect to the z axis, then c 1 = c 2 = c, and the maximum tensile and compressive stresses are equal numerically: σ 1 = σ 2 = 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 = bh 3 /12 Z = bh 2 /6 Stress distribution on a symmetrical section Page 615
16 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. 8 kn A 2 kn/m B C H A V A V C 5m 3m 700 mm 220 mm Cross section of the beam Page 616
17 Solution The shear force and bending moment diagrams for the beam are shown in the following figures. 8 kn A 2 kn/m C B 11 kn 13 kn 5m 3m +ve 11 A B Shear Force (kn) C A +ve B C Bending Moment (knm) Page 617
18 Method 1 : use section modulus for calculations Section modulus of the crosssectional area, Z = bh 2 /6 = 220*700 2 /6 =1.8 x 10 7 mm 3 Maximum bending tensile and compressive stresses, σ = M max / Z = 30 x 10 6 /1.8 x 10 7 = 1.67 N/mm 2 Method 2 : use moment of inertia for calculations Moment inertia of the crosssectional area, I = bh 3 /12 = 220*700 3 /12 =6.29 x 10 9 mm 4 Maximum bending tensile and compressive stresses, σ = M max c/ I = 30 x 10 6 * 350/6.29 x 10 9 = 1.67 N/mm 2 Page 618
19 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. 50 mm 50 mm 50 mm 50 mm 300 mm 225 mm Solution The maximum bending moment, M max = wl 2 /8 = w * /8 = 2 x 10 6 w Nmm Moment inertia of the beam section, I = 225*300 3 /12125*200 3 /12 = x 10 6 mm 4 The section modulus of the beam section, Z = x 10 6 /150 = x 10 6 mm 3 Page 619
20 From the bending stress formula, M max = σ Z 2 x 10 6 w = 70 * x 10 6 w = N/mm w = kn/m Example 5 A simply supported Tbeam 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 Centroidal Axis 40 mm 100 mm 120 mm Page 620
21 Solution The maximum bending moment, M max = 1*6 2 /8 = 4.5 knm The moment inertia of beam, I = 120(40) 3 / *40*( ) (120) 3 / *120*(10060) 2 = 2.18 x 10 7 mm 4 The maximum tensile stress in the bottom fibre, σ = M c 2 /I = 4.5 x 10 6 *100/2.18 x 10 7 = N/mm 2 The maximum compressive stress in the top fibre, σ = M c 1 /I = 4.5 x 10 6 *60/2.18 x 10 7 = 12.4 N/mm 2 Page 621
22 Example 6 A Tbeam is loaded as shown in the following figure. Calculate the maximum bending stresses induced in the beam. 10 kn 20 kn 4 kn/m A B C D E H B VB V D 2m 2m 4m 2m 120 mm Centroidal Axis 40 mm 100 mm 120 mm Page 622
23 Solution The shear force and bending moment diagrams for the beam are shown in the following figures. 10 kn 20 kn 4 kn/m A B C D E A B C D Shear Force (kn) E 0 10 A 120 B C 8 D 0 E Bending Moment (knm) Page 623
24 The moment inertia of beam, I = 120(40) 3 / *40*( ) (120) 3 / *120*(10060) 2 = 2.18 x 10 7 mm 4 When M = 20kNm at point B, The maximum tensile stress in the top fibre, σ T = M c 1 /I = 20 x 10 6 *60/2.18 x 10 7 = 55 N/mm 2 The maximum compressive stress in the bottom fibre, σ C = M c 2 /I = 20 x 10 6 *100/2.18 x 10 7 = 91.7 N/mm 2 When M = 26.7 Nm at point C, The maximum tensile stress in the bottom fibre, σ T = M c 2 /I = 26.7 x 10 6 *100/2.18 x 10 7 = N/mm 2 The maximum compressive stress in the top fibre, σ C = M c 1 /I = 26.7 x 10 6 *60/2.18 x 10 7 = 73.5 N/mm 2 Therefore, the maximum tensile stress and compressive stress are N/mm 2 and 91.7 N/mm 2 respectively. Page 624
25 PROBLEMS for bending stress Page 625
26 Page 626
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