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1、Shearing Stress in Beams06IntroductionFor most beams, both the bending moment and shear force will present.When shearing stresses are exerted on the vertical faces of an element, equal stresses must be exerted on the horizontal faces.The possible stress components at any point on the cross section w
2、ill be sx, txy and txz.xyzsxtxztxyIf V = 0 (pure bending), txy = txz =0Now V 0, both txy and txz may not be zero.Distribution of shearing stresses is unknown but satisfies( txy 0)txz may not be zero, but the average is zero2Calculation of Shearing Stress: PreliminaryThe distribution of txy can not b
3、e determined from statics alone. Analysis of deformation may not help too much.However, we know that shearing stress comes in pair, arrow to arrow and tail to tail, as shown. If we can determine tyx, then from txy = tyx, we can obtain txy.xyzsxtxztxyWe know that txz may not be zero, but the average
4、is zero. However, for a narrow beam (width much smaller than height), we can assume that txz = 0 zero, since tzx 0 on the left and right surfaces. Therefore, this topic will be focusing on txy only.txztzx = 03Shear on the Horizontal Face of a Beam ElementMVV+dVM+dMtxytyxss + dsdH (Resultant of tyx)F
5、1F2DCDdxA*yFor the beam shown, consider an element CDDC to calculate shear stress at y1.CCDA*4Shear on the Horizontal Face: First Moment of AreaShear FlowFirst Moment of AreaI is the second moment of area of the entire section about the centroidal axis of the entire section.A* is area of the isolate
6、d area, from where the shear flow is calculated to a free surface.Horizontal shear flow, the horizontal shear force per unit length, in a beam is given byqss + ds= the 1st moment of area of the isolated area A*.yA*C*y*y1For built up sections y* is the distance from the centroidal axis (neutral axis,
7、 N.A.) to the centroid (C*) of the isolated area.5Example 1ProblemA beam is made of three planks, nailed together. Knowing that the spacing between nails is 25 mm and that the vertical shear in the beam is V = 500 N, determine the shear force in each nail.6Example 1: Solution Strategy If the beam we
8、re one piece, there would be shear stress (shear flow) between the cap and the web.Solution StrategyThe nails will carry the force resulted from the shear flow.7First Moment of Area. For the isolated area (the flange), Q can be calculated as*Example 1: Solution Shear Flow. The horizontal shear flow
9、q between the flange and web can be calculated asMoment of Inertia. The moment of inertia of the entire area can be calculated as the moment of outer rectangle subtracting that of the 2 small rectangles.Shear Force in Nail. From the solution strategy, we haveF = 92.6 N8Shearing Stress in a BeamOn th
10、e upper and lower surfaces of the beam, tyx= 0. It follows that txy= 0 on the upper and lower edges of the transverse sections.If the width of the beam is comparable or large relative to its depth, the shearing stresses at D1 and D2 are significantly higher than at D.Lets consider the beam element C
11、DCD. On the surface CD, we have the same horizontal force dH.dxCDdHdxdAThe average shearing stress on the horizontal face of the element is obtained by dividing the shearing force on the element by the area of the face.For a narrow beam (width height), we can assume that txy is uniform through the t
12、hickness, and9Shear Stress in a Narrow Rectangular BeamFree Body DiagramShear StressparabolicA*: Area of the shaded area.A: Area of the entire cross section*txysx10*I Beam: Shear Stress txyShear StressFor Wide Beams11Example 2ProblemA timber beam is to support the three concentrated loads shown. Kno
13、wing that for the grade of timber used, the allowable stresses are sall = 1800 psi and tall = 120 psi, determine the minimum required depth d of the beam based on the maximum normal stress and shear stress.Solution StrategyDevelop shear and bending moment diagrams. Identify the maximums.Determine th
14、e beam depth based on allowable normal stress.Determine the beam depth based on allowable shear stress.Required beam depth is equal to the larger of the two depths found.12Example 2: Solution Shear and Bending Moment. Develop shear and bending moment diagrams. The maximums areDesign Based on Normal
15、Stress. For Mmax = 90 kipin and sall = 1800 psi, we haved = 9.26 in. Design Based on shear Stress. For Vmax = 3 kips and tall = 1200 psi, we haved = 10.71 in. Summary. Required beam depth is the larger of the two obtained above, i.e., d = 10.71 in. Use d = 11.25 in.13Longitudinal Shear on a Beam Ele
16、ment of Arbitrary ShapedxA*yWe have examined the shear flow on a horizontal surface parallel to the z-axis in a beam. We now wish to consider the shear on an arbitrary longitudinal surface, as shown.dHss + dsdHEquilibrium analysis in the horizontal direction shows that except for the differences in
17、integration areas, the same result can be obtained.14Example 3ProblemA square box beam is constructed from four planks as shown. Knowing that the spacing between nails is s =1.5 in. and the beam is subjected to a vertical shear of magnitude V = 600 lb, determine the shearing force in each nail.Solut
18、ion Strategy This problem is similar to Example 1. The main difference is that the nails in this problem will carry the shear flow on a vertical longitudinal surface at the connection.(Ex. 1)Obtain the shear flow at the connection.The shearing force in each nail equals qs, where q is the shear flow
19、carried by one row of nails.15Moment of Inertia. The moment of inertia of the section can be easily calculated asExample 3: Solution Shear Force in Nail. From the solution strategy, we haveF = 80.8 lb*First Moment of Area. Use the upper plank as the isolated area,Shear Flow. The total shear flow q b
20、etween the upper and the two vertical planks flange can be calculated asNote that q is carried by two rows of nails. The shear flow carried by one row of nails equals q/2.16Shearing Stresses in Thin-Walled MembersConsider a segment of a wide-flange beam subjected to the vertical shear V.The longitudinal shear force on the element isThe corresponding shear stress isNOTE:in the flangesin the webPreviously found a similar expression for the shearing stress in the web17Shear
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