1.  2005 Pearson Education South Asia Pte Ltd 6. Bending 1 CHAPTER OBJECTIVES To determine stress in members caused by bending To discuss how to establish shear and moment diagrams for a beam or shaft To determine largest shear and moment in a member, and specify where they occur To consider members that are straight, symmetric cross- section and homogeneous linear-elastic material To consider special cases of unsymmetrical bending and members made of composite materials

2.  2005 Pearson Education South Asia Pte Ltd 6. Bending 2 CHAPTER OUTLINE 1.Review Shear and Moment Diagrams 2.Graphical Method for Constructing Shear and Moment Diagrams 3.Bending Deformation of a Straight Member 4.The Flexure Formula 5.Stress Concentrations

3.  2005 Pearson Education South Asia Pte Ltd 6. Bending 3 Members that are slender and support loadings applied perpendicular to their longitudinal axis are called beams 6.1 SHEAR AND MOMENT DIAGRAMS

4.  2005 Pearson Education South Asia Pte Ltd 6. Bending 4 6.1 SHEAR AND MOMENT DIAGRAMS In order to design a beam, it is necessary to determine the maximum shear and moment in the beam Express V and M as functions of arbitrary position x along axis. These functions can be represented by graphs called shear and moment diagrams Engineers need to know the variation of shear and moment along the beam to know where to reinforce it

5.  2005 Pearson Education South Asia Pte Ltd 6. Bending 5 6.1 SHEAR AND MOMENT DIAGRAMS Shear and bending-moment functions must be determined for each region/segment of the beam between any two discontinuities of loading 3 segments: AB, BC, and CD

6.  2005 Pearson Education South Asia Pte Ltd 6. Bending 6 6.1 SHEAR AND MOMENT DIAGRAMS Beam sign convention Although choice of sign convention is arbitrary, in this course, we adopt the one often used by engineers:

7.  2005 Pearson Education South Asia Pte Ltd 6. Bending 7 6.1 SHEAR AND MOMENT DIAGRAMS IMPORTANT Beams are long straight members that carry loads perpendicular to their longitudinal axis. They are classified according to how they are supported To design a beam, we need to know the variation of the shear and moment along its axis in order to find the points where they are maximum Establishing a sign convention for positive shear and moment will allow us to draw the shear and moment diagrams

8.  2005 Pearson Education South Asia Pte Ltd 6. Bending 8 EXAMPLE 6-1 Draw the shear and moment diagrams for beam shown below.

9.  2005 Pearson Education South Asia Pte Ltd 6. Bending 9 RARA RBRB W = 25 kN EXAMPLE 6-1 Support reactions: Shown in free-body diagram. +  M A = 0; R B (10) – W(7.5) – (15)(5) – 80 = 0 R B = 34.25 kN +  F Y = 0; R A – 15 – W + R B = 0, R A = 5.75 kN

10.  2005 Pearson Education South Asia Pte Ltd 6. Bending 10 EXAMPLE 6-1 Shear and moment functions Since there is a discontinuity of distributed load and a concentrated load at beam’s center, two regions/segments of x must be considered. 0 ≤ x 1 ≤ 5 m, +↑  F y = 0;... V = 5.75 N +  M = 0;... M = (80 + 5.75x 1 ) kN·m Segment AB: A

11.  2005 Pearson Education South Asia Pte Ltd 6. Bending 11 EXAMPLE 6-1 5 m ≤ x 2 ≤ 10 m, +↑  F y = 0;...V = 5.75  15 – 5(x 2 –5) kN Segment BC: +  M = 0;...M = 80 + 5.75x 2  15(x 2  5)  5(x 2  5)(x 2  5)/2 kN·m = 80 + 5.75x 2  15(x 2  5)  2.5(x 2  5) 2 kN·m A B

12.  2005 Pearson Education South Asia Pte Ltd 6. Bending 12 EXAMPLE 6-1 Shear and moment diagrams 0 ≤ x 1 ≤ 5 m, V = 5.75 N M = (5.75x 1 + 80) kN·m V = (15.75  5x 2 ) kN 5 m ≤ x 2 ≤ 10 m, 108.75 Segment AB: Segment BC: M = 80 + 5.75x 2  15(x 2  5)  2.5(x 2  5) 2 kN·m x 1 =0, M A = 80 kN·m x 1 =5, M B = 108.75 kN·m x 2 =5, V B = – 9.25 kN x 2 =10, V C = – 34.25 kN x 2 =5, M B = 108.75 kN.m x 2 =10, M C = 0

13.  2005 Pearson Education South Asia Pte Ltd 6. Bending 13 6.2 GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT DIAGRAMS A simpler method to construct shear and moment diagram, one that is based on two differential equations that exist among distributed load, shear and moment

14.  2005 Pearson Education South Asia Pte Ltd 6. Bending 14 6.2 GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT DIAGRAMS Regions of distributed load dV dx =  w(x) Slope of shear diagram at each point =  distributed load intensity at each point dV dx =  w(x) Slope of shear diagram at each point =  distributed load intensity at each point dM dx = V Slope of moment diagram at each point = shear at each point dM dx = V Slope of moment diagram at each point = shear at each point

15.  2005 Pearson Education South Asia Pte Ltd 6. Bending 15 6.2 GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT DIAGRAMS  V =  ∫ w(x) dx Change in shear =  area under distributed loading  V =  ∫ w(x) dx Change in shear =  area under distributed loading  M = ∫ V(x) dx Change in moment = area under shear diagram  M = ∫ V(x) dx Change in moment = area under shear diagram

16.  2005 Pearson Education South Asia Pte Ltd 6. Bending 16 EXAMPLE 6-2 Draw the shear and bending moment diagrams for the shaft shown in Figure. The support at A is a thrust bearing and the support at B is a journal bearing.

17.  2005 Pearson Education South Asia Pte Ltd 6. Bending 17 EXAMPLE 6-2 Support reactions. Due to simmetry R A = R B = 3.5 kN V Shear force diagram 3.5 kN 2 kN 3 kN – 1.5 kN 1.5 kN –3.5 kN

18.  2005 Pearson Education South Asia Pte Ltd 6. Bending 18 EXAMPLE 6-2 Shear force area 1 A 1 = (3.5)(2) = 7 kN-m V 3.5 kN 1.5 kN – 3.5 kN –1.5 kN 2m A 2 = (1.5)(2) = 3 kN-m 2 3 A 3 = (– 1.5)(2) = –3 kN-m 4 A 4 = (– 3.5)(2) = –7 kN-m Bending moment diagram M M 0 = 0 M 2 = M 0 + A 1 = 7 kN-m M 4 = M 2 + A 2 = 10 kN-m M 6 = M 4 + A 3 = 7 kN-m M 8 = M 6 + A 4 = 0 7 kN-m 10 kN-m 7 kN-m

19.  2005 Pearson Education South Asia Pte Ltd 6. Bending 19 6.3 BENDING DEFORMATION OF A STRAIGHT MEMBER When a bending moment is applied to a straight prismatic beam, the longitudinal lines become curved and vertical transverse lines remain straight and yet undergo a rotation Before deformation After deformation Horizontal lines Become curved Vertical lines remains straight, yet rotate

20.  2005 Pearson Education South Asia Pte Ltd 6. Bending 20 6.3 BENDING DEFORMATION OF A STRAIGHT MEMBER The top portion is compressing. There must be a neutral surface, which is not undergoing a change in length, and lies in between top & bottom region. After deformation Horizontal lines Become curved Vertical lines remains straight, yet rotate The bottom portion of the bar is stretching.

21.  2005 Pearson Education South Asia Pte Ltd 6. Bending 21 6.3 BENDING DEFORMATION OF A STRAIGHT MEMBER A neutral surface is where longitudinal fibers of the material will not undergo a change in length.

22.  2005 Pearson Education South Asia Pte Ltd 6. Bending 22 6.3 BENDING DEFORMATION OF A STRAIGHT MEMBER Thus, we make the following assumptions: 1.Longitudinal axis x (within neutral surface) does not experience any change in length 2.All cross sections of the beam remain plane and perpendicular to longitudinal axis during the deformation 3.Any deformation of the cross-section within its own plane will be neglected In particular, the z axis, in plane of cross-section and about which the cross-section rotates, is called the neutral axis

23.  2005 Pearson Education South Asia Pte Ltd 6. Bending 23 6.3 BENDING DEFORMATION OF A STRAIGHT MEMBER For any specific cross-section, the longitudinal normal strain will vary linearly with y from the neutral axis. A contraction will occur (  ) in fibers located above the neural axis ( +y). An elongation will occur (+  ) in fibers located below the axis (  y )  =  (y/c)  max Neutral axis Eq. 6-8

24.  2005 Pearson Education South Asia Pte Ltd 6. Bending 24 6.4 THE FLEXURE FORMULA Assume that material behaves in a linear-elastic manner so that Hooke’s law applies. A linear variation of normal strain must then be the consequence of a linear variation in normal stress  =  (y/c)  max Eq. 6-9 Applying Hooke’s law to Eqn 6-8,

25.  2005 Pearson Education South Asia Pte Ltd 6. Bending 25 6.4 THE FLEXURE FORMULA By mathematical expression, equilibrium equations of moment and forces, we get Eq. 6-11 ∫ A y dA = 0 Eq. 6-10  max c M = ∫ A y 2 dA The integral represents the moment of inertia of cross-sectional area, computed about the neutral axis. We symbolize its value as I.

26.  2005 Pearson Education South Asia Pte Ltd 6. Bending 26 6.4 THE FLEXURE FORMULA Hence, Eqn 6-11 can be solved and written as Eq. 6-12 Mc I  max =  max = maximum normal stress in member, at a point on cross-sectional area farthest away from neutral axis M = resultant internal moment, computed about neutral axis of cross-section I = moment of inertia of cross-sectional area computed about neutral axis c = perpendicular distance from neutral axis to a point farthest away from neutral axis, where  max acts

27.  2005 Pearson Education South Asia Pte Ltd 6. Bending 27 6.4 THE FLEXURE FORMULA Normal stress at intermediate distance y can be written as Eq. 6-13 M yIM yI  = = The negative sign agrees with the established x,y,z axes. Equations 6-12 and 6-13 are referred to as the flexure formula. M = resultant internal moment, computed about neutral axis of cross-section  = bending stress in member, at any point on cross-sectional area away from neutral axis I = moment of inertia of cross-sectional area computed about neutral axis y = perpendicular distance from neutral axis to a point farthest away from neutral axis, where  acts

28.  2005 Pearson Education South Asia Pte Ltd 6. Bending 28 6.4 THE FLEXURE FORMULA IMPORTANT Cross-section of straight beam remains plane when beam deforms due to bending. The neutral axis is subjected to zero stress Due to deformation, longitudinal strain varies linearly from zero at neutral axis to maximum at outer fibers of beam Provided material is homogeneous and Hooke’s law applies, stress also varies linearly over the cross-section For linear-elastic material, neutral axis passes through centroid of cross-sectional area. This is based on the fact that resultant normal force acting on cross-section must be zero Flexure formula is based on requirement that resultant moment on the cross-section is equal to moment produced by linear normal stress distribution about neutral axis

29.  2005 Pearson Education South Asia Pte Ltd 6. Bending 29 6.4 THE FLEXURE FORMULA Note : The failure of metal is usually due to tensile stress, so the failure will be initiated from the point undergoing the maximum tensile stress (the most critical point). It is very important to ascertain the most critical point. Positive bending moment: Negative bending moment:

30.  2005 Pearson Education South Asia Pte Ltd 6. Bending 30 EXAMPLE 6-3 60 cm 40 cm P = 5 kN A B 1. Draw free-body diagram and find the support reaction. 2. Draw the shear force and bending moment diagram 3.Determine the bending stress maximum, either tensile or compression and show the point of each. 4.Determine the tensile stress maximum at the middle cross- section of the beam. z y 10 cm 20 cm

31.  2005 Pearson Education South Asia Pte Ltd 6. Bending 31 EXAMPLE 6-3 Free-body diagram 1. Support reactions: AYAY BYBY 60 cm 40 cm P = 5 kN +  M A = 0; B Y (100) – (5)(60) = 0 B Y = 3 kN +  F Y = 0; A Y – 5 + B Y = 0, A Y = 2 kN 2. Shear force diagram & Area x V 2 kN 5 kN –3 kN A1A1 A2A2 A 1 = (2)(0,60) = 1.2 kN.m A 2 = (3)(0,40) = –1.2 kN.m

32.  2005 Pearson Education South Asia Pte Ltd 6. Bending 32 EXAMPLE 6-3 Free-body diagram AYAY BYBY 60 cm 40 cm P = 5 kN Bending moment diagram x M M 0 = 0 M 0.6 = M 0 + A 1 = 1.2 kN-m M 1.0 = M 2 + A 2 = 1.2 – 1.2 = 0 Bending moment in the middle of the beam: 1.2 kN.m 50 cm M 0.5 M 0.5 = A Y (0.50) = 1.0 kN-m 1.2 1.0 x y M max = M 0.6 = 1.2 kN-m

33.  2005 Pearson Education South Asia Pte Ltd 6. Bending 33 EXAMPLE 6-3 Free-body diagram AYAY BYBY 60 cm 40 cm P = 5 kN 3. Maximum bending stress x M Moment of inertia: 1.2 kN.m 50 cm M 0.5 1.2 1.0 x y M max y max I  max = M max = 1.2 kN-m y max = ± 10 cm = ± 0.1 m z y 10 cm 20 cm I =I = bh 3 12 = (10)(20) 3 12 = 6666.67 cm 4 = 6.67 x 10 -5 m 4

34.  2005 Pearson Education South Asia Pte Ltd 6. Bending 34 EXAMPLE 6-3 Free-body diagram AYAY BYBY 60 cm 40 cm P = 5 kN x M 1.2 kN.m 50 cm M 0.5 1.2 1.0 x y z y 10 cm 20 cm Maximum tensile stress occur at point C. (1.2x10 3 )( –0.1) 6.67 x 10 –5  tensile = = 1.8 x 10 6 N/m 2 = 1.8 MPa C D Maximum compressive stress occur at point D. (1.2x10 3 )( 0.1) 6.67 x 10 –5  comp = = –1.8 x 10 6 N/m 2 = –1.8 MPa

35.  2005 Pearson Education South Asia Pte Ltd 6. Bending 35 EXAMPLE 6-3 Free-body diagram AYAY BYBY 60 cm 40 cm P = 5 kN 4. Maximum tensile stress at the middle cross- section is at point E. x M 1.2 kN.m 50 cm M 0.5 1.2 1.0 x y z y 10 cm 20 cm (1.0x10 3 )( –0.1) 6.67 x 10 –5  E = = 1.5 x 10 6 N/m 2 = 1.5 MPa C D M 0.5 y E I E =E = M max = 1.0 kN-m y E = –10 cm = –0.1 m E

36.  2005 Pearson Education South Asia Pte Ltd 6. Bending 36 6.4 THE FLEXURE FORMULA A B x y A B x y z 10 cm 20 cm 60 cm 40 cm P = 5 kN z y 10 cm 20 cm 60 cm 40 cm P = 5 kN WHICH ONE STRONGER? WHY ?? Maximum tensile stress occur at point C 1.  max =  C1 = 1.8 MPa C1C1 C2C2 Maximum tensile stress occur at point C 2.  max =  C2 = ??? y Case-1 Case-2

37.  2005 Pearson Education South Asia Pte Ltd 6. Bending 37 6.4 THE FLEXURE FORMULA A B x y z y 10 cm 20 cm 60 cm 40 cm P = 5 kN C2C2 Maximum tensile stress occur at point C 2. (1.2x10 3 )( –0.05) 1.67 x 10 –5  C2 = = 3.6 x 10 6 N/m 2 = 3.6 MPa M max y C2 I  max =  C2 = M C1 = M C2 = 1.2 kN-m y C2 = –5 cm = –0.05 m I =I = bh 3 12 = (0.2)(0.1) 3 12 = 1.67 x 10 -5 m 4 Case-2  C2 >  C1 Case-1 is stronger than Case-2

38.  2005 Pearson Education South Asia Pte Ltd 6. Bending 38 6.5 STRESS CONCENTRATIONS Flexure formula can only be used to determine stress distribution within regions of a member where cross- sectional area is constant or tapers slightly If the cross-section suddenly changes, normal-stress and strain distributions become nonlinear and they can only be obtained via experiment or mathematical analysis using the theory of elasticity and solved by the finite element approximation.

39.  2005 Pearson Education South Asia Pte Ltd 6. Bending 39 6.5 STRESS CONCENTRATIONS Common discontinuities include members having notches on their surfaces, holes for passage of fasteners or abrupt changes in outer dimensions of member’s cross-section The maximum normal stress at the discontinuities occur at the smallest cross-sectional area notches hole Shoulder fillet

40.  2005 Pearson Education South Asia Pte Ltd 6. Bending 40 6.5 STRESS CONCENTRATIONS For design, we only need to know the maximum normal stress developed at these sections, not the actual stress distribution Thus, the maximum normal stress due to bending can be obtained using the stress-concentration factor K  = K Mc I Eq. 6-26

41.  2005 Pearson Education South Asia Pte Ltd 6. Bending 41 6.5 STRESS CONCENTRATIONS IMPORTANT Stress concentrations in members subjected to bending occur at points of cross-sectional change, such as notches and holes, because here the stress and strain become nonlinear. The more severe the change, the larger the stress distribution For design/analysis, not necessary to know the exact stress distribution around cross-sectional change The maximum normal stress occurs at the smallest cross-sectional area The maximum normal stress can be obtained using stress concentration factor K, which is determined through experiment and is a function of the geometry of the member If material is brittle or subjected to fatigue loading, stress concentrations in the member need to be considered in design

42.  2005 Pearson Education South Asia Pte Ltd 6. Bending 42 6.5 STRESS CONCENTRATIONS

43.  2005 Pearson Education South Asia Pte Ltd 6. Bending 43 EXAMPLE 6-4 Transition in cross-sectional area of steel bar is achieved using shoulder fillets as shown. If bar is subjected to a bending moment of 5kN·m, determine the maximum normal stress developed in the steel.

44.  2005 Pearson Education South Asia Pte Ltd 6. Bending 44 EXAMPLE 6-4 Moment creates largest stress in bar at base of fillet ( c = h/2 = 0.04 m ). Referring to the graph, we get K = 1.45 I =I = t h 3 12 = (0.02)(0.08) 3 12 = 85.3 x 10 -6 m 4  = K Mc I r/h = 16/80 = 0.2 w/h =120/80 = 1.5 Geometry factor: = 1.45 (5x10 3 )(0.04) 85.3 x 10 –6 = 3.4 x 10 6 N/m 2 = 3.4 MPa