Chapter Objectives To generalize the procedure by formulating equations that can be plotted so that they describe the internal shear and moment throughout a member. To use the relations between distributed load, shear, and moment to draw shear and moment diagrams. To determine the stress in elastic symmetric members subject to bending.
In-class Activities 1. Reading Quiz 2. Applications 3. Shear and moment diagrams 4. Graphical method for construction of shear and bending moment diagrams 5. Flexural formula 6. Unsymmetric bending 7. Stress Concentrations 8. Concept Quiz
READING QUIZ 1) Provided that the bending formation of a straight member is small and within elastic range, which of the following statements is incorrect? a) Plane section remains plane b) Cross section remains perpendicular c) The length of the longitudinal axis remains unchanged d) In-plane distortion of cross section to the longitudinal axis is not negligible
READING QUIZ (cont) 2) Which of the following statements is incorrect for bending of a straight member? a) Bending stress is proportional to b) Bending stress is inversely proportional c) Bending stress is inversely proportional to the moment of inertia of the section d) Bending stress is not a function to the second moment of area of the section of the location
APPLICATIONS
APPLICATIONS (cont)
SHEAR AND MOMENT DIAGRAMS Shear is obtained by summing forces perpendicular to the beam s axis up to the end of the segment. Please refer to the website for the animation: Shear and Moment Diagrams Moment is obtained by summing moments about the end of the segment. Note the sign conventions are opposite when the summing processes are carried out with opposite direction. (from left to right vs from right to left)
EXAMPLE 1 Draw the shear and moment diagrams for the beam shown in Fig. 6 4a.
EXAMPLE 1 (cont.) Solution The support reactions are shown in Fig. 6 4c. Applying the two equations of equilibrium yields F y 0; wl 2 V wx V w L x 2 0 1 M 0; wl x 2 wx x M 2 0 M w 2 2 Lx x 2
EXAMPLE 1 (cont.) Solution The point of zero shear can be found from Eq. 1: V w L x 2 L x 2 0 From the moment diagram, this value of x represents the point on the beam where the maximum moment occurs. M max w 2 L L 2 L 2 2 wl 8 2
GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT DIAGRAMS Regions of distributed load: Change in shear = area under distributed loading V w x dx Change in moment = area under shear diagram M V xdx
GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT DIAGRAMS (cont) Regions of concentrated force and moment: V F V V 0 V F M M M M Vx M 0 0 M 0
EXAMPLE 2 Draw the shear and moment diagrams for the beam shown in Fig. 6 12a.
EXAMPLE 2 (cont.) Solution The reactions are shown on the free-body diagram in Fig. 6 12b. The shear at each end is plotted first, Fig. 6 12c. Since there is no distributed load on the beam, the shear diagram has zero slope and is therefore a horizontal line.
EXAMPLE 2 (cont.) Solution The moment is zero at each end, Fig. 6 12d. The moment diagram has a constant negative slope of -M 0 /2L since this is the shear in the beam at each point. Note that the couple moment causes a jump in the moment diagram at the beam s center, but it does not affect the shear diagram at this point.
EXAMPLE 3 Draw the shear and moment diagrams for each of the beams shown in Figs. 6 13a and 6 14a.
EXAMPLE 3 (cont) Solution
BENDING DEFORMATION OF A STRAIGHT MEMBER Assumptions: 1. Plane section remains plane 2. Length of longitudinal axis remains unchanged 3. Plane section remains perpendicular to the longitudinal axis 4. In-plane distortion of section is negligible
FLEXURAL FORMULA Assumptions: Material behaves in a linear-elastic manner so that Hooke s Law Applies; i.e. σ=e.є My I M M R Z M ydf max M c Mc max I My I A y Z 2 ; A y da da A y y c max da
EXAMPLE 4 The simply supported beam in Fig. 6 26a has the crosssectional area shown in Fig. 6 26b. Determine the absolute maximum bending stress in the beam and draw the stress distribution over the cross section at this location.
EXAMPLE 4 (cont) Solution The maximum internal moment in the beam, 22.5 knm, occurs at the center. By reasons of symmetry, the neutral axis passes through the centroid C at the mid-height of the beam, Fig. 6 26b. I 2 2 I Ad 1 3 2 1 0.250.02 0.250.020.16 0.020.3 12 301.3 10 6 m 4 12 3 B My I B ; b 3 0.17 22.5 10 301.3 10 6 12.7 MPa (Ans)
EXAMPLE 4 (cont) Solution A three-dimensional view of the stress distribution is shown in Fig. 6 26d. At point B, B My I B 3 0.15 22.5 10 301.3 10 ; B 6 11.2 MPa
Example
Solution
Solution