Chapter 7: Internal Forces

Chapter 7: Internal Forces Chapter Objectives To show how to use the method of sections for determining the internal loadings in a member. To generalize this procedure by formulating equations that can be plotted so that they describe the internal shear and moment throughout a member. In our past discussions on trusses we have considered the internal forces in a straight two-force member. These forces produced only tension or compression in the member. The internal forces in any other type of member will usually produce shear and bending as well. This chapter is devoted to the analysis of the internal forces in beams. Beams are long, straight prismatic members designed to support loads that are applied perpendicular to the axis of the member and at various points along the member. 7.1 Internal Loadings Developed in Structural Members Consider the case of a straight two-force member. Cut the member at C. To maintain equilibrium an internal axial force must exist. Next consider the case of a multi-force member. The internal forces in beam AB are not limited to axial tension or compression as in the case of straight two-force members. The internal forces also include shear and bending. 7.1

By the method of sections we can take a cut at any point along the length of the member to find the internal resisting effects axial force (P), shear force (V), and bending moment (M). If the entire beam is in equilibrium, then any portion of the beam is in equilibrium. Equilibrium of the isolated portion of the beam is achieved by the internal resisting effects, where P axial force V shear force M bending moment Various Types of Loading and Support A beam is a structural member designed to support loads applied at various points along the member. In most cases, the loads are applied perpendicular to the axis of the beam and will cause only shear and bending. However, axial forces may be present in the beam when the applied loads are not perpendicular to the axis of the beam. Beams are usually long, straight, and symmetrical (prismatic) in cross section (such as wide flange sections, commonly called I-beams or W-sections). Designing a beam consists essentially in selecting the cross section that will provide the most effective resistance to bending, shear, and deflection produced by the applied loads. The design of a beam includes the following steps. 1. Determine the shear forces and bending moments produced by the applied loads. 2. Select the best suited cross section. 7.2

A beam may be subjected to various types of applied loads. Concentrated loads Distributed loads* Both *Distributed loads may be uniform, trapezoidal, or triangular. A beam may be supported in a number of ways. The distance between the supports, L, is referred to as the span. The following beams are statically determinate. Simply supported Overhanging Cantilever The following beams are statically indeterminate. Continuous Propped cantilever Fixed-Fixed Free-Body Diagram The following steps are used to begin the analysis or design of a beam. The entire beam is taken as a free-body diagram and the reactions are determined at the supports. Then, to determine the internal forces at any point along the length of the beam, we cut the beam and draw the free-body diagram of that portion of the beam. Using the three equations of equilibrium we may determine the axial force (P), shear force (V), and bending moment (M) at any point along the length of the beam. 7.3

Fx = 0 Fy = 0 Mcut = 0 yields the axial force (P) yields the shear force (V) yields the bending moment (M) Sign Conventions The following sign conventions are used for the internal effects. Axial force Shear force Bending moment Methods of analysis Three methods of analysis will be demonstrated. Determining values for shear and bending moment at distinct points on a beam. Writing the equations for shear and bending moment. Drawing shear and bending moment diagrams by understanding the relationships between load, shear, and bending moment. 7.4

Example Values of Shear and Moment at Specific Locations on a Beam Given: The beam loaded as shown. Find: V and M @ x = 0 +, 2, 4 -, 4 +, 6, 8 -, 8 +, 10, and 12 - x = 0 + Fy = 0 = 10 V Mcut = 0 = - 10 (0) + M V = 10 k M = 0 k-ft x = 2 Fy = 0 = 10 V V = 10 k Mcut = 0 = - 10 (2) + M M = 20.0 k-ft x = 4 - Fy = 0 = 10 V V = 10 k Mcut = 0 = - 10 (4) + M M = 40.0 k-ft x = 4 + Fy = 0 = 10 12 - V V = - 2 k Mcut = 0 = - 10 (4) + 12 (0) + M M = 40.0 k-ft x = 6 Fy = 0 = 10 12 - V V = - 2 k Mcut = 0 = - 10 (6) + 12 (2) + M M = 60 24 = 36.0 k-ft 7.5

x = 8 - Fy = 0 = 10 12 - V V = - 2 k Mcut = 0 = - 10 (8) + 12 (4) + M M = 80 48 = 32.0 k-ft x = 8 + Fy = 0 = 10 12 6 - V V = - 8 k Mcut = 0 = - 10 (8) + 12 (4) + 6 (0) + M M = 80 48 0 = 32.0 k-ft x = 10 Fy = 0 = 10 12 6 - V V = - 8 k Mcut = 0 = - 10 (10) + 12 (6) + 6 (2) + M M = 100 72 12 = 16.0 k-ft x = 12 - Fy = 0 = 10 12 6 - V V = - 8 k Mcut = 0 = - 10 (12) + 12 (8) + 6 (4) + M M = 120 96 24 = 0 k-ft To simplify the calculations, use a free body diagram from the right side. x = 10 Fy = 0 = V + 8 Mcut = 0 = - M + 8 (2) V = - 8 k M = 16.0 k-ft x = 12 - Fy = 0 = V + 8 Mcut = 0 = - M + 8 (0) V = - 8 k M = 0 k-ft Note: The moment is zero at the supports of simply supported beams. 7.6

Example Values of Shear and Moment at Specific Locations on a Beam Given: The beam loaded as shown. Find: V and M @ x = 0 +, 3, 6 -, 6 +, 7.5, and 9 - x = 0 + Fy = 0 = 19 V V = 19 k Mcut = 0 = - 19 (0) + M M = 0 k-ft x = 3 Fy = 0 = 19 ½ (6) 3 ½ (4.5) 3 V V = 19 9 6.75 = 3.25 k Mcut = 0 = - 19 (3) + ½ (6) 3 (2/3) 3 + ½ (4.5) 3 (1/3) 3 + M M = 57 18 6.75 = 32.25 k-ft x = 6 - Fy = 0 = 19 ½ (6) 6 - ½ (3) 6 V V = 19 18 9 = - 8.0 k Mcut = 0 = - 19 (6) + ½ (6) 6 (2/3) 6 + ½ (3) 6 (1/3) 6 + M M = 114 72 18 = 24.0 k-ft x = 6 + Fy = 0 = V + 8 Mcut = 0 = - M + 8 (3) V = - 8.0 k M = 24.0 k-ft 7.7

x = 7.5 Fy = 0 = V + 8 V = - 8.0 k Mcut = 0 = - M + 8 (1.5) M = 12.0 k-ft x = 9 - Fy = 0 = V + 8 V = - 8.0 k Mcut = 0 = - M + 8 (0) M = 0 k-ft 7.8

7.2 Shear and Moment Equations and Diagrams The plot of values of shear force and bending moment against a distance x are graphs called shear and bending moment diagrams. Rather than pick points along a beam and approximate these diagrams by determining values of shear and bending moment at a limited number of points, we can write equations to define functions of shear and moment at any x along the beam. The equations for shear and bending moment may be developed by the following steps. First, determine the number of sets of equations needed to define shear force and bending moment across the entire length of the beam. - The shear and bending moment equations will be discontinuous at points where a distributed load begins or ends, and where concentrated forces or concentrated couples are applied. Next, for each set of equations, draw the necessary free-body diagram. Then, for each set of equations, write the equilibrium equations and solve for shear force (V) and bending moment (M) in terms of position on the beam (i.e. x, where x is measured from the left end of the beam). - Use Fy = 0 to develop the equations for shear force. - Use Mcut = 0 to develop the equations for the bending moment. 7.9

Example Shear and Moment Equations Given: Beam loaded as shown. Find: Write the V and M equations and draw the V and M diagrams. Two sets of equations are required. First, solve for the reactions at the supports. MB = 0 = - Ay L + P (L/2) Ay = P/2 MA = 0 = By L - P (L/2) By = P/2 Note: The beam is symmetrical and symmetrically loaded; thus, the reactions are symmetrical as well. 0 < x < L/2 Fy = 0 = P/2 V V = P/2 Mcut = 0 = - (P/2) x + M M = Px/2 L/2 < x < L Fy = 0 = P/2 P - V V = - P/2 Mcut = 0 = - (P/2) x + P (x L/2) + M M = Px/2 Px + PL/2 M = P/2 (L x) 7.10

Example Shear and Moment Equations Given: Beam loaded as shown. Find: Write the V and M equations and draw the V and M diagrams. One set of equations is required. 0 < x < L Fy = 0 = wl/2 wx - V V = w (L/2 x) Mcut = 0 = - (wl/2) x + wx (x/2) + M M = - wx 2 /2 + wlx/2 M = (wx/2) (L x) 7.11

Example Shear and Moment Equations Given: Beam loaded as shown. Find: Write the V and M equations and draw the V and M diagrams. Three sets of equations are required. First, solve for the reactions at the supports. First, take moments about the roller support at the right end of the beam to find the vertical reaction at the pin support at the left end of the beam. MR = 0 = - 10 Ay + 10 (8) 46 + ½ (3) 6 [2 + (2/3) 6] + ½ (6) 6 [2 + (1/3) 6] 10 Ay = 80 46 + 9(6) + 18(4) = 160.0 Ay = 16.0 kips Note: The distributed load (a trapezoid) is treated as two triangles. Then, take moments about the pin support at the left end of the beam to find the vertical reaction at the roller support at the right end of the beam. ML = 0 = 10 By - 10 (2) 46 - ½ (3) 6 [2 + (1/3) 6] - ½ (6) 6 [2 + (2/3)6] 10 By = 20 + 46 + 9(4) + 18(6) = 210.0 By = 21.0 kips Write the equations for shear and bending moment. 0 < x < 2 Fy = 0 = 16 - V V = 16 Mcut = 0 = - 16x + M M = 16x 7.12

2 < x < 8 Since the intensity of the distributed load varies, an equation is needed to define the intensity of the distributed load (p) as a function of position x along the beam. Using the general equation of a line: p = m x + c Determine the slope m. m = (6 3)/(8 2) = 1/2 So, p = ½ x + c Next, determine the y-intercept value c. Known points on the line include: when x = 2, p = 3 and when x = 8, p = 6. Substituting the second condition into the equation p = ½ x + c: 6 = ½ (8) + c c = 6 4 = 2 Thus, p = ½ x + 2 Now write the equations for shear and moment. Fy = 0 = 16 10 ½ (3) (x 2) ½ (½ x + 2) (x 2) - V V = 16 10 (3/2) (x 2) ½ (½ x + 2) (x 2) = 6 3x/2 + 3 ½ (x 2 /2 + x 4) = 6 3x/2 + 3 - x 2 /4 - x/2 + 2 V = - x 2 /4 2x + 11 V = 0 k @ x = 3.75 Mcut = 0 = - 16x + 10 (x 2) + ½ (3) (x 2) (2/3) (x 2) + ½ (½ x + 2) (x 2) (1/3) (x 2) + M - 46 M = 16x - 10 (x 2) - (x 2) 2 (1/6) (½ x + 2) (x 2) 2 + 46 = 16x 10x + 20 - (x 2 4x + 4) (1/6) (x 3 /2 6x + 8) + 46 = 6x + 20 x 2 + 4x 4 x 3 /12 + x 1.33 + 46 M = x 3 /12 x 2 + 11x + 60.67 M = 83.5 k-ft @ x = 3.75 8 < x < 10 Fy = 0 = V + 21 V = - 21 k Mcut = 0 = - M + 21 (10 x) M = 21 (10 x) 7.13

7.3 Relations between Distributed Load, Shear, and Moment The methods outlined so far for drawing shear and bending moment diagrams become increasingly cumbersome the more complex the loading. The construction of shear and bending moment diagrams, however, can become greatly simplified by understanding the relationships that exist between the distributed load, the shear force, and the bending moment. Relation between the Distributed Load and Shear From the free body diagram: Fy = 0 = V (V + V) w x, where w = w (x) = constant for small x V = w x V/ x = - w(x), then letting x 0, by definition of a derivative, dv/dx = - w(x) If we integrate this expression between two points, then dv = - w(x) dx dv = - w(x) dx V2 V1 = - 2 1 w(x) dx Interpretation of these first two expressions: dv/dx = - w(x) The value of the slope on the shear diagram is equal to the height of the load diagram at that point times minus one. V2 V1 = - 2 1 w(x) dx The change in shear between two points is equal to the area under the load diagram times minus one. Concentrated Forces These equations are not valid under a concentrated load. The shear diagram is discontinuous at the point of a concentrated load. 7.14

Relation between the Shear and Bending Moment From the free body diagram: MR = 0 = - M V x + w ( x) 2 /2 + M + M M = V x ½ w ( x) 2 M = V x Note: When x 0, ( x) 2 0 M/ x = V, then letting x 0, by definition of a derivative, dm/dx = V If we integrate this expression between two points, then dm = V dx = V(x) dx Note: Let V = V(x) since the shear may vary. dm = V(x) dx M2 M1 = 2 1 V(x) dx Interpretation of these two expressions: dm/dx = V(x) The value of the slope on the bending moment diagram is equal to the height of the shear diagram at that point. Note: When shear is zero, the slope on the moment diagram is zero corresponding to a point of maximum bending moment. M2 M1 = 2 1 V(x) dx The change in bending moment between two points is equal to the area under the shear diagram. Concentrated Couples These equations are not valid under a concentrated couple. The bending moment diagram is discontinuous at the point of a concentrated couple. 7.15

Example Shear and Moment Diagrams Given: The beam loaded as shown. Find: Draw the shear and moment diagrams. Draw the shear diagram Between the left end and right end of the beam: The area under the load diagram = + wl V = - Area = - wl V2 = V1 + V = wl/2 wl = - wl/2 Draw the moment diagram Between the left end of the beam and mid-span: The area under the shear diagram = ½ (wl/2)(l/2) = wl 2 /8 M = Area = wl 2 /8 M2 = M1 + M = 0 + wl 2 /8 = wl 2 /8 At mid-span (i.e. x = L/2), V = 0, so dm/dx = 0. Between mid-span and the right end of the beam: The area under the shear diagram = - ½ (wl/2)(l/2) = - wl 2 /8 M = Area = - wl 2 /8 M2 = M1 + M = wl 2 /8 + (- wl 2 /8) = 0 In general, when V = 0 at a point, the slope on the moment diagram at that point is zero (i.e. a horizontal tangent). 7.16

Example Shear and Moment Diagrams Given: The beam loaded as shown. Find: Draw the shear and moment diagrams. 7.17

Example Shear and Moment Diagrams Given: The beam loaded as shown. Find: Draw the shear and moment diagrams. 7.18

Example Shear and Moment Diagrams Given: The beam loaded as shown. Find: Draw the shear and moment diagrams. Solve for the reactions at the supports. Take moments about point A to find the vertical reaction at point D. MA = 0 = - ½ (4) 12 [(2/3) 12] 20 (15) 120 + 18 Dy 18 Dy = 192 + 300 + 120 = 612.0 Dy = 34.0 k Take moments about point D to find the vertical reaction at point A. MD = 0 = ½ (4) 12 [6 + (1/3) 12] + 20 (3) 120-18 Ay 18 Ay = 240 + 60-120 = 180.0 Ay = 10.0 k In order to draw the first part of the moment diagram, the location for the point of maximum moment (i.e. V = 0 kips) needs to be determined so that the area under the shear diagram may be calculated. Write an equation for shear between x = 0 and x = 12. Fy = 0 = 10 ½ (x) (x/3) V V = 10 x 2 /6 Let V = 0, thus 0 = 10 x 2 /6 x 2 = 60.0 x = 7.75 The change in moment from x = 0 to x = 7.75 is equal to the area under the shear diagram. Δ M = (2/3) 10 (7.75 ) = 51.7 kip-ft 7.19

Determine the value of moment at x = 12 and plot this value on the moment diagram. - The area under the portion shear diagram between x = 7.75 and x = 12 cannot be calculated using standard formulas since there is no horizontal tangent at x = 7.75 or at x = 12. - Take a cut at x = 12, draw the free body diagram, and write the equilibrium equation to determine the value of moment. - Then plot this value on the moment diagram. Mcut = 0 = MB 10 (12) + ½ (12) 4 (1/3)(12) MB = 120 96 = 24.0 kip-ft The concentrated couple at point C causes the moment diagram to jump up by 120 kip-ft. Since the applied moment is acting clockwise, the internal resisting moment increases, thus the jump up on the moment diagram. 7.20

Example Shear and Moment Diagrams Given: The beam loaded as shown. Find: Draw the shear and moment diagrams. In order to draw the part of the moment diagram beyond x = 2, the location for the point of maximum moment (i.e. V = 0 kips) needs to be determined. Write an equation for shear for 2 < x < 8. Fy = 0 = 16-10 ½ (3) (x - 2) - ½ (½ x + 2) (x - 2) - V V = 6 3/2 (x 2) ½ (½ x + 2) (x 2) = 6 3x/2 + 3 ½ (x 2 /2 + x 4) = 6-3x/2 + 3 x 2 /4 - x/2 + 2 V = - x 2 /4 2x + 11 The maximum moment occurs when V = 0. Let V = 0; thus, 0 = - x 2 /4 2x + 11 Using the quadratic equation, find x when V = 0. x = - (- 2) ± [(- 2 ) 2 4 (- 1/4 ) (11)] 1/2 2 (- 1/4) x = - 11.75 (Not reasonable: not within the limits of 2 < x < 8.) x = 3.75 7.21

Find M @ x = 3.75 Mcut = 0 = - 16 (3.75) + 10 (1.75) 46 + ½ (3) 1.75 (2/3) 1.75 + ½ (3.75/2 + 2) (1.75) (1.75/3) + M M = 16 (3.75) 10 (1.75) + 46 ½ (3) 1.75 (2/3) 1.75 - ½ (3.875) 1.75 (1.75/3) M = 60 17.5 + 46 3.06 1.98 M = 83.5 k-ft 7.22