[8] Bending and Shear Loading of Beams


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1 [8] Bending and Shear Loading of Beams Page 1 of 28 [8] Bending and Shear Loading of Beams [8.1] Bending of Beams (will not be covered in class) [8.2] Bending Strain and Stress [8.3] Shear in Straight Members
2 [8] Bending and Shear Loading of Beams Page 2 of 28 [8.1] Bending of Beams STRAIGHT BEAM MEMBERS Q: What are the beams? A: Members that are slender and support loadings that are applied perpendicular to their longitudinal axis are called beams 1) There are several types of beams, and they are often classified based upon how they are supported: 2) The internal shear and moment of the beams can be represented as shear and moment diagrams
3 [8] Bending and Shear Loading of Beams Page 3 of 28 SHEAR AND MOMENT DIAGRAMS 1) Beams are structural members which are designed to support loadings applied perpendicular to their axes: a) The design of beams requires a detailed knowledge of the variation of the internal shear force V and bending moment M. b) The variations of V and M as functions of the position x along the beam s axis can be obtained by the method of sections; shear and moment diagrams. a) The normal forces are ignored in shear and moment diagrams. 2) Plotting the shear and moment diagrams requires to establish a sign convention: 3) Shear and moment as functions of the position x can be determined by cutting the beam at the location x and applying the equations of equilibrium on that cut freebody diagram. 4) Considering a beam under the arbitrary loadings, the relations between distributed load, shear, and moment can be mathematically derived. a) A small element x at location x is cut out as a freebody diagram. b) Applying equations of equilibrium on this freebody diagram yields: + F y = 0 : V w( x) x ( V + V ) = 0 V = w( x) x
4 [8] Bending and Shear Loading of Beams Page 4 of 28 + M O = 0 : V x M + w( x) x[ k( x)] + ( M + M ) = 0 2 M = Vx w( x) k( x) c) Dividing by x and taking the limit as x 0, these two equations become: dv = wx ( ) dx (Slope of shear diagram = Distributed load magnitude) dm = V ( x ) dx (Slope of moment diagram = Shear magnitude) d) These equations can be integrated as: V = w( x) dx (Change in shear = Area under the curve of distributed load) M = V ( x) dx (Change in moment = Area under the curve of shear diagram) 5) A freebody diagram of a small element of a beam, taken at one of the forces provides: + F y = 0 : V = F Therefore, the shear will jump downward when force F acts downward on the beam. 6) A freebody diagram of a small element, taken at the couple moment provides:
5 [8] Bending and Shear Loading of Beams Page 5 of 28 + M O = 0 : M = M O Therefore, the moment will jump upward when clockwise moment MO acts on the beam. 7) The summary of shear and moment diagram:
6 [8] Bending and Shear Loading of Beams Page 6 of 28 CLASS EXAMPLE Draw the shear and moment diagrams for the beam. It is supported by a smooth plate at B which slides within the groove and so it cannot support a vertical force, although it can support a moment and axial load.
7 [8] Bending and Shear Loading of Beams Page 7 of 28 Name: Student ID: HOMEWORK R41 Draw the shear and moment diagrams for the beam shown in the figure.
8 [8] Bending and Shear Loading of Beams Page 8 of 28 CLASS EXAMPLE A standard Tbeam is subjected to the loading shown. Draw the shear and moment diagrams.
9 [8] Bending and Shear Loading of Beams Page 9 of 28 Name: Student ID: HOMEWORK Draw the shear and moment diagrams for the beam shown in the figure.
10 [8] Bending and Shear Loading of Beams Page 10 of 28 [8.2] Bending Strain and Stress BENDING DEFORMATION OF A BEAM 1) The deformations of a straight and prismatic beam (made of a homogeneous material) due to bending can be analyzed under the following assumptions: a) The beam has at least one axis of symmetry, and the bending moment is applied perpendicular to the axis of symmetry
11 [8] Bending and Shear Loading of Beams Page 11 of 28 b) The bottom portion of the bar is stretched and the top portion of the bar is compressed, when the bending moment is applied c) The boundary between bottom and top portion undergoes no change in length, just bends: this boundary is called neutral plane d) The longitudinal axis of the beam is defined such that it goes through the centroid of all cross sections of the beam and lies on the neutral plane Q: What is the centroid? A: The geometric center of the crosssectional area: y da = 0 A d) At a cross section, the neutral surface becomes neutral axis e) It is assumed that all cross sections of the beam remain plane and perpendicular to the longitudinal axis during the deformation NORMAL STRAIN DUE TO BENDING 1) The normal strain is developed within a beam due to the bending 2) The segment x lies on the longitudinal axis (neutral plane) and undergoes no change in its length, but the segment s (originally equal length with x ) is contracted after the bending and becomes s'
12 [8] Bending and Shear Loading of Beams Page 12 of 28 3) The normal strain is determined as: = s' s s ( = y) y = 4) The result indicates that the longitudinal normal strain of an element within a beam depends on its location y (measured from neutral axis) and the radius of curvature at the cross section: linear variation of strain 5) Assuming that the linear variation of strain, the maximum normal strain occurs at the outermost surface away from the neutral axis: y = c max y = c y = c max [Normal Strain due to Bending]
13 [8] Bending and Shear Loading of Beams Page 13 of 28 THE FLEXURE FORMULA 1) If a bending moment M is applied to a beam, it develops a corresponding internal bending moment within the beam: normal stress 2) The assumption of linear variation in normal strain leads to a corresponding linear variation in normal stress as well 3) Using Hooke s law: = E, substituted into the normal strain equation yields the normal stress equation as: y = max c
14 [8] Bending and Shear Loading of Beams Page 14 of 28 4) The resultant moment due to the stress distribution over the entire cross section must be equal to the internal moment at the cross section: M = A y y ( da) = y da max M = c y c A max A 2 da Define moment of inertia, I = A y 2 da Mc = max [Flexure Formula] I where, max M : I : c : : Maximum normal stress developed in the shaft (occurs at the outermost surface away from the neutral axis) Resultant internal bending moment at the cross section Moment of inertia of the crosssectional area Distance to the outermost surface away from the neutral axis 5) The normal stress at any intermediate distance y can be determined by the similar procedure as: My = [Generalized Flexure Formula] I
15 [8] Bending and Shear Loading of Beams Page 15 of 28 Q: Why there is a negative sign in the flexure formula? A: In order to make it consistent to the coordinate system given 6) The upper portion of the beam (+ y coordinate) is under compressive ( ) and bottom portion of the beam ( y coordinate) is under tensile (+) stress Note: Usually the calculation of stress magnitude is performed first, and the sign (+ or ) is determined by the observation either tensile (+) compressive ( )
16 [8] Bending and Shear Loading of Beams Page 16 of 28 CLASS EXAMPLE The member has a cross section with the dimensions shown. Determine the largest bending stress developed in the member if it is subjected to an internal bending moment of M = 85 knm.
17 [8] Bending and Shear Loading of Beams Page 17 of 28 Name: Student ID: HOMEWORK A cantilever supported beam, in the figure (a), has a crosssectional area in the shape of a channel shown in the figure (b). Determine the maximum bending stress that occurs in the beam at section aa.
18 [8] Bending and Shear Loading of Beams Page 18 of 28 CLASS EXAMPLE The rod is supported by smooth journal bearings at A and B that only exert vertical reactions on the shaft. Determine its smallest diameter d, if the allowable bending stress is allow = 180 MPa.
19 [8] Bending and Shear Loading of Beams Page 19 of 28 CLASS EXAMPLE Determine the moment of inertia for symmetric cross sections.
20 [8] Bending and Shear Loading of Beams Page 20 of 28 Name: Student ID: HOMEWORK 8.2.2/8.2.3 R44 A simply supported beam, in the figure (a), has the crosssectional area shown in the figure (b). Determine the absolute maximum (the largest value of ) bending stress in the beam.
21 [8] Bending and Shear Loading of Beams Page 21 of 28 [8.3] Shear in Straight Members TRANSVERSE SHEAR 1) Recall that beams can support both shear and moment loadings 2) The internal shear force V at the cross section of a beam is the resultant of a transverse shear stress distribution acts over the cross section 3) The transverse shear stress of a beam can be analyzed by utilizing two previously developed analytical tools: My a) The flexure formula: = I b) The relationship between moment and shear: V = dm dx
22 [8] Bending and Shear Loading of Beams Page 22 of 28 THE SHEAR FORMULA 1) Considering a differential element taken from a beam, the horizontal force balance requires: F x = 0 2) The element will indeed satisfy the force balance: Now considering the top segment of the element that has been sectioned at a distance y from the neutral axis of the cross section, what about the requirement: F = 0? x
23 [8] Bending and Shear Loading of Beams Page 23 of ) Applying the equation of equilibrium: F = 0 yields, ' da da ( t dx) = 0 A' A' My Substituting the flexure formula: = I M + dm I M I y da y da ( t dx) = A' dm I A' A' y da = ( t dx) Solving for the shear stress: 0 x = 1 It dm dx y da A' 4) Recall that: dm = V, and define first moment of the area: Q = y da = y ' A ' dx VQ = [The Shear Formula] It where, : Shear stress at the point located a distance y from the neutral axis of the cross section V : Internal resultant shear force at the cross section I : Moment of inertia of the entire cross section t : Width of the member s crosssectional area, measured at the point where is to be determined Q : First moment of the area: Q = y da = y' A' A' A'
24 [8] Bending and Shear Loading of Beams Page 24 of 28 BEAM WITH A RECTANGULAR CROSS SECTION 1) Considering a beam with a rectangular cross section of width b and height h 2) The distribution of the shear stress throughout the cross section can be determined by computing the shear stress at an arbitrary height y from the neutral axis as: Q = 2 1 h h 1 h y' A' = y + y yb = y 2 b = VQ It 2 2 V (1 2)[( h 4) y ] b = 3 [(1 12) bh ] b 6V = 3 bh 2 h 4 y 2 3) The result indicates that the shear stress distribution over the cross section is parabolic, the maximum value at the neutral axis (y = 0): 3 V = 2 bh max = = 1. 5 y= 0 V A = 1.5 [Maximum Shear Stress for Rectangular Cross Section] max avg
25 [8] Bending and Shear Loading of Beams Page 25 of 28 4) The maximum shear stress of the beam with a rectangular cross section is 1.5 times larger than the average shear stress over the cross section 5) The shear stress at the outermost surface of the cross section (y = h 2 ) is zero: There is no transverse shear stress on the top and bottom surfaces of the beam under shear loading (zero surface shear stress condition)
26 [8] Bending and Shear Loading of Beams Page 26 of 28 CLASS EXAMPLE If the wideflange beam is subjected to a shear of V = 25 kip, determine the maximum shear stress in the beam.
27 [8] Bending and Shear Loading of Beams Page 27 of 28 CLASS EXAMPLE Railroad ties must be designed to resist large shear loadings. If the tie is subjected to the 30kip rail loadings and the gravel bed exerts a distributed reaction as shown, determine the intensity w for equilibrium, and find the maximum shear stress in the tie.
28 [8] Bending and Shear Loading of Beams Page 28 of 28 Name: Student ID: HOMEWORK 8.3.1/ The beam, shown in the figure, is made of wood and is subjected to a resultant internal vertical shear force of 3 kip. (a) Determine the shear stress in the beam at point P, and (b) compute the maximum shear stress in the beam.
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