Bending Stress - Example 1Beams -Horizontal Shear StressHorizontal Shear StressSTATICS & STRENGTH OF MATERIALS - ExampleRecitation #6 October 14, 2003 TORSION OF CIRCULAR SECTIONS Assumptions 1) This analysis can only be applied to solid or hollow circular sections 2) The material must be homogeneous 3) Torque is constant and transmitted along bar by each section trying to shear over its neighbour. 4) Transverse planes remain parallel to each other. Examine the deformation of a length dx between two transverse planed of a beam with an applied torque T. For this element, assume the left end is fixed and the right end rotates by d due to the applied torque. The surface of radius 'r' rotates through angle , which is shear strain. The arc is defined as length da, which is equal to :which gives that: where : = Rate of twist Equation 27 states that: And by substituting for gamma gives that: Which relates the shear stress linearly to the distance 'r' away from the centre of the section. The shear stress distribution then looks like this. We now equate the applied torque T to the torque generated in the section by the shear stress distribution. To do this look at a small circumferential section dA.Substituting for shear stress: Since the rate of twist is constant through the section, it is not a function of r. If we assume a homogeneous material, G is also constant, so: We represent the integral term as the geometric stiffness of the cross section in a similar way to I. We call this term the Polar Second Moment of Area, J. and as for the second moment of area for beams under an applied bending moment, this term indicates the cross sectional properties to withstand the applied torque. Which when equating with the shear stress term, gives that:r1"> called Engineer's Theory of Torsion ( ETT ). Since this applies to circular bars, the standard terms for J are: For solid Shaft of radius R, diameter D : For hollow shaft , ID = Di, OD = Do Approximation for this walled tube with t < R/10:Non-Uniform TORSION OF CIRCULAR SECTIONS (Section 3.4 Gere) Can be analyzed by applying the formulas of pure torsion to finite segments of the bar and then adding the results , or by applying the formulas to differential elements of the bar and then integrating. DO Example of Non-Uniform Torque Æ 3.4-1 From GERE BEAM BENDING – Normal and Shear Stresses & Strains Bending Stress - Example 1 In Diagram 1, we have shown a simply supported 20 ft. beam with a load of 10,000 lb. acting downward at the center of the beam. The beam used is a rectangular 2" by 4" steel beam. We would like to determine the maximum bending (axial) stress which develops in the beam due to the loading.Step 1: Out first step in solving this problem is, of course, to apply static equilibrium conditions to determine the external support reactions. In this particular example, because of the symmetry of the problem, we will not go through the statics in detail, but point out that the two support forces will support the load at the center equally with forces of 5000 lb. each as shown in Diagram 2. Step 2: The second step is to draw the shear force and bending moment diagrams for the beam. We really don't need the shear force diagram at this point, except we will use it to make the bending moment diagram. We will normally be able to draw the shear force diagram by simply looking at the load forces and the support reactions. If necessary we will determine the shear force and bending moment expressions and make the shear force and bending moment diagrams from these expressions, as we did in the proceeding topic. We first draw the shear force diagram. Due to the 5000 lb. support force, the shear force value begins at +5000 lb., and since there is no additional loads for the next 10 feet, the shear force remains constant at 5000 lb. between 0 and 10 feet. At 10 feet, the 10,000 lb. downward load drives the shear force down by that amount, from + 5000 lb. to a value of - 5000 lb. Then as there are no additional loads for the next 10 feet, the shear force will remain constant over the remainder of the beam. Graphing the shear force values produces the result in Diagram 3a. Then using the fact that for non-cantilevered beams the bending moment values are equal to the area under the shear force diagram, we develop the bending moment graph shown in Diagram 3b.Step 3. We now apply the flexure formula: Bending Stress = M y / I We wish to find the maximum bending stress, which occurs at the outer edge of the beam so: M = maximum bending moment = 50,000 ft-lb. = 600,000 in-lb. (from bending moment diagram) y = distance from the neutral axis of the cross section to outer edge of beam = 2 inches I = moment of inertia of cross section; for rectangle I = (1/12) bd3 = 1/12 (2" * 4"3) = 10.67 in4. Then, Maximum Bending Stress = M y / I = (600,000 in-lb)*(2 in)/(10.67 in4) = 112, 500 lb/in2 (if you needed the strain, You can plug in Hooke’s Law to calc. for strain) This is the correct value, but it is clearly excessive for normal steel. Thus if we tried to use a rectangular 2"x4" steel beam, it would fail under the load. We will have to use a stronger beam. Notice that the maximum bending moment does not depend on the type of beam. The values of "y" and "I" in the flexure formula do depend on the beam used. Thus, if we had used a rectangular 2"x6" beam (instead of a 2" x 4" beam), the value of y would be: y = 3", and the value of I would be: I = (1/12)(2" * 6"3) = 36 in4. Then the maximum bending stress for this beam would be: Maximum Bending Stress = M y / I = (600,000 in-lb)*(3 in)/(36in4) = 50,000 lb/in2 . This value is in a more reasonable range for acceptable axial stresses in steel. Beams -Horizontal Shear Stress In addition to the bending (axial) stress which develops in a loaded beam, there is also a shear stress which develops, including both a Vertical Shear Stress, and a Horizontal (longitudinal) Shear Stress. It can be shown that at any given point in the beam. Diagram 1 shows a simply supported loaded beam.In Diagram 2a, we have cut a section dx long out of the left end of the beam, and have shown the internal horizontal forces acting on the section. In Diagram 2b, we have shown a side view of section dx. Notice that the bending moment is larger on the right hand face of the section by an amount dM. (This is clear if we make the bending moment diagram for the beam, in which we see the bending moment