# BEAMS: SHEAR AND MOMENT DIAGRAMS

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BEAMS: SHEAR AND MOMENT DIAGRAMS

LECTURE Third Edition BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) • A. J. Clark School of Engineering •Department of Civil and Environmental Engineering 13 Chapter 5.3 by Dr. Ibrahim A. Assakkaf SPRING 2003 ENES 220 – Mechanics of Materials Department of Civil and Environmental Engineering University of Maryland, College Park Slide No. 1 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) ENES 220 ©Assakkaf Shear Forces and Bending Moments in Beams Example 8 The beam is loaded and supported as shown in the figure. Write equations for the shear V and bending moment M for any section of the beam in the interval AB. 6 kN/m y A x B 9m 1 Slide No. 2 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) ENES 220 ©Assakkaf Shear Forces and Bending Moments in Beams Example 8 (cont’d) A free-body diagram for the beam is shown Fig. 17. The reactions shown on the diagram are determined from equilibrium equations as follows: 6 × 9 1 = 0; RA (9) − 9 × = 0 2 3 ∴ RA = 9 kN + ∑M B + ↑ ∑ Fy − 0; RB + 9 − 6×9 =0 2 ∴ RB = 18 kN Slide No. 3 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) ENES 220 ©Assakkaf Shear Forces and Bending Moments in Beams Example 8 (cont’d) 6 kN/m Figure 17 y w 6 2 = ⇒w= x x 9 3 6 kN/m x A x B RA = 9 kN x 3 y A 9 kN x x S Rb = 18 kN 2 x kN/m 3 M V 9m 2 1 + ↑ ∑ Fy = 0; − V + 9 − x (x ) = 0 3 2 2 x ∴V = 9 − for 0 < x < 9 3 2 x x + ∑ M S = 0; − M + 9 x − x = 0 3 2 3 3 x ∴ M = 9x − for 0 < x < 9 9 2 Slide No. 4 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) ENES 220 ©Assakkaf Shear Forces and Bending Moments in Beams Example 9 A timber beam is loaded as shown in Fig. 18a. The beam has the cross section shown in Fig.18.b. On a transverse cross section 1 ft from the left end, determine a) The flexural stress at point A of the cross section b) The flexural stress at point B of the cross section. Slide No. 5 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) ENES 220 ©Assakkaf Shear Forces and Bending Moments in Beams Example 9 (cont’d) y 8′′ 600 lb · B 300 lb/ft 5600 ft-lb A 3500 ft-lb 2′′ x 6 ft 6′′ A · 10 ft 900 lb Figure 18a 5 ft 2′′ 8′′ 1500 lb Figure 18b 3 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear Forces and Bending Moments in Beams Example 9 (cont’d) First, we have to determine the moment of inertia Ix. From symmetry, the neutral axis is located at a distance y = 5 in. either from the bottom or the upper edge. Therefore, 8′′ · 5′′ Slide No. 6 ENES 220 ©Assakkaf 2′′ 8(5)3 6(3)3 4 − I x = 2 = 558.7 in 3 3 6′′ 2′′ 8′′ LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear Forces and Bending Moments in Beams Slide No. 7 ENES 220 ©Assakkaf Example 9 (cont’d) 600 lb y 300 lb/ft 5600 ft-lb A 3500 ft-lb x 1500 lb 1 ft 5600 ft-lb A 900 lb M + ↑ ∑ Fy = 0; − V + 1( 200) = 0 ∴V = 200 lb + V ∑M S for 0 < x < 6 = 0; − M − 5600 + (1)(200)(0.5) = 0 ∴ M = −5600 + 100 = −5500 ft - lb at x = 1 ft 4 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear Forces and Bending Moments in Beams Example 9 (cont’d) a) Stress at A 8′′ 5′′ 2′′ · Slide No. 8 ENES 220 ©Assakkaf σA = − 6′′ 2′′ 8′′ (− 5500 ×12)(− 3) = −354.4 psi Mr y =− Ix 558.7 = 354.4 psi Compression (C) b) Stress at B σB = − (− 5500 ×12)(5) = +591 psi Mr y =− 558.7 Ix = 591 psi Tension (T) LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear Forces and Bending Moments in Beams Slide No. 9 ENES 220 ©Assakkaf Load, Shear Force, and Bending Moment Relationships – In cases where a beam is subjected to several concentrated forces, couples, and distributed loads, the equilibrium approach discussed previously can be tedious because it would then require several cuts and several free-body diagrams. – In this section, a simpler method for constructing shear and moment diagrams are discussed. 5 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear Forces and Bending Moments in Beams Slide No. 10 ENES 220 ©Assakkaf Load, Shear Force, and Bending Moment Relationships L b a w2 P w1 O x1 x2 x3 Figure 19 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear Forces and Bending Moments in Beams Slide No. 11 ENES 220 ©Assakkaf Load, Shear Force, and Bending Moment Relationships – The beam shown in the Figure 20 is subjected to an arbitrary distributed loading w = w(x) and a series of concentrated forces and couple moments. – We will consider the distributed load w to positive when the loading acts upward as shown in Figure 20. 6 Slide No. 12 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) ENES 220 ©Assakkaf Shear Forces and Bending Moments in Beams Load, Shear Force, and Bending Moment Relationships P P ML ∆x w w M R = M L + ∆M VL C C x x Figure 20a ∆x 2 ∆x 2 Figure 20b LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear Forces and Bending Moments in Beams VR = VL + ∆V Slide No. 13 ENES 220 ©Assakkaf Load, Shear Force, and Bending Moment Relationships – In reference to Fig. 20a at some location x, the beam is acted upon by a distributed load w(x), a concentrated load force P, and a concentrated couple C. – A free-body diagram segment of the beam centered at the location x is shown in Figure 20b. 7 Slide No. 14 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) ENES 220 ©Assakkaf Shear Forces and Bending Moments in Beams Load, Shear Force, and Bending Moment Relationships The element must be in equilibrium, therefore, + ↑ ∑ Fy = 0; VL + wavg ∆x + P − (VL + ∆V ) = 0 From which ∆V = P + wavg ∆x (28) LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear Forces and Bending Moments in Beams Slide No. 15 ENES 220 ©Assakkaf Load, Shear Force, and Bending Moment Relationships In Eq. 28, if the concentrated force P and distributed force w are both zero in some region of the beam, then ∆V = 0 or VL = VR (29) This implies that the shear force is constant in any segment of the beam where there are no loads. 8 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear Forces and Bending Moments in Beams Slide No. 16 ENES 220 ©Assakkaf Load, Shear Force, and Bending Moment Relationships If the concentrated load P is not zero, then in the limit as ∆x → 0, (30) ∆V = P or VR = VL + P That is, across any concentrated load P, the shear force graph (shear force versus x) jumps by the amount of the concentrated load. LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear Forces and Bending Moments in Beams Slide No. 17 ENES 220 ©Assakkaf Load, Shear Force, and Bending Moment Relationships Furthermore, moving from left to right along the beam, the shear force graph jumps in the direction of the concentrated load. If the concentrated load is zero, then in the limit as ∆x → 0, we have (31) ∆V = w ∆x → 0 avg 9 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear Forces and Bending Moments in Beams Slide No. 18 ENES 220 ©Assakkaf Load, Shear Force, and Bending Moment Relationships And the shear force is a continuous function at x. Dividing through by ∆x in Eq. 31, gives ∆V dV = =w lim (32) dx ∆x →0 ∆X That is, the slope of the shear force graph at any section x in the beam is equal to the intensity of loading at that section of the beam. LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear Forces and Bending Moments in Beams Slide No. 19 ENES 220 ©Assakkaf Load, Shear Force, and Bending Moment Relationships Moving from left to right along the beam, if the distributed force is upward, the slope of the shear force graph (dV/dx = w) is positive and the shear force graph is increasing (moving upward). If the distributed force is zero, then the slope of the shear graph (dV/dx = 0) and the shear force is constant. 10 Slide No. 20 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear Forces and Bending Moments in Beams ENES 220 ©Assakkaf Load, Shear Force, and Bending Moment Relationships In any region of the beam in which Eq. 32 is valid (any region in which there are no concentrated loads), the equation can be integrated between definite limits to obtain V2 x2 V1 x1 ∆V = V2 − V1 = ∫ dV = ∫ w dx Slide No. 21 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear Forces and Bending Moments in Beams ENES 220 ©Assakkaf Load and Shear Force Relationships dV = w( x) (33) dx Slope of Distributed = Shear Diagram Load Intensity 11 Slide No. 22 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) ENES 220 ©Assakkaf Shear Forces and Bending Moments in Beams Load and Shear Force Relationships V2 x2 V1 x1 ∆V2−1 = V2 − V1 = ∫ dV = ∫ w( x) dx Change in Shear = (34) Area under Loading Curve between x1 and x2 Slide No. 23 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) ENES 220 ©Assakkaf Shear Forces and Bending Moments in Beams Load, Shear Force, and Bending Moment Relationships Similarly, applying moment equilibrium to the free-body diagram of Fig. 20b we obtain + ∑M c = 0; - (M L + ∆M ) − M L − VL ∆x ∆x − (VL + ∆V ) + a (wavg ∆x ) = 0 2 2 From which ∆x ∆M = C + VL ∆x + ∆V − a wavg ∆x 2 ( ) (35) 12 Slide No. 24 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) ENES 220 ©Assakkaf Shear Forces and Bending Moments in Beams Load, Shear Force, and Bending Moment Relationships P P ML ∆x w w M R = M L + ∆M VL C C x x Figure 20a ∆x 2 ∆x 2 Figure 20b LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear Forces and Bending Moments in Beams VR = VL + ∆V Slide No. 25 ENES 220 ©Assakkaf Load, Shear Force, and Bending Moment Relationships In which -∆x/2 < a < ∆x/2 and in the limit as ∆x → 0, a → 0 and wavg → w. Three important relationships are clear from Eq. 35. First, if the concentrated couple is not zero, then in the limit as ∆x → 0, ∆M = C or M R = M L + C (36) 13 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear Forces and Bending Moments in Beams Slide No. 26 ENES 220 ©Assakkaf Load, Shear Force, and Bending Moment Relationships That is, across any concentrated couple C, the bending moment graph (bending moment versus x) jumps by the amount of the concentrated couple. Furthermore, moving from left to right along the beam, the bending moment graph jumps upward for a clockwise concentrated couple and jumps downward for a concentrated counterclockwise C. LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear Forces and Bending Moments in Beams Slide No. 27 ENES 220 ©Assakkaf Load, Shear Force, and Bending Moment Relationships Second, if the concentrated couple C and concentrated force P are both zero, then in the limit as ∆x → 0, we have ∆M = VL ∆x + ∆V ∆x − a (wavg ∆x ) → 0 2 and dividing by ∆x, gives lim ∆x →0 ∆M dM = =V ∆x dx (37) (38) 14 Slide No. 28 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) ENES 220 ©Assakkaf Shear Forces and Bending Moments in Beams Load, Shear Force, and Bending Moment Relationships That is, the slope of the bending moment graph at any location x in the beam is equal to the value of the shear force at that section of the beam. Moving from left to right along the beam, if V is positive, then dM/dx = V is positive, and the bending moment graph is increasing Slide No. 29 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) ENES 220 ©Assakkaf Shear Forces and Bending Moments in Beams Load, Shear Force, and Bending Moment Relationships – In the region of the beam I which Eq. 38 is valid (any region in which there are no concentrated loads or couples), the equation can be integrated between definite limits to get ∆M = M 2 − M 1 = M2 x2 ∫ dM = ∫ V dx M1 (39) x1 15 Slide No. 30 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) ENES 220 ©Assakkaf Shear Forces and Bending Moments in Beams Shear Force and Bending Moment Relationships dM =V dx Slope of Moment Diagram (40) = Shear Slide No. 31 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) ENES 220 ©Assakkaf Shear Forces and Bending Moments in Beams Shear Force and Bending Moment Relationships ∆M 2−1 = M 2 − M 1 = Change in Moment = M2 x2 M1 x1 ∫ dM = ∫ V dx Area under Shear (41) diagram between x1 and x2 16 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear Forces and Bending Moments in Beams Slide No. 32 ENES 220 ©Assakkaf Shear and Moment Diagrams – Procedure for Analysis • Support Reactions – Determine the support reactions and resolve the forces acting on the beam into components which are perpendicular and parallel to the beam’s axis • Shear Diagram – Establish the V and x axes and plot the values of the shear at two ends of the beam. Since dV/dx = w, the slope of the shear diagram at any point is equal to the intensity of the distributed loading at the point. LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear Forces and Bending Moments in Beams Slide No. 33 ENES 220 ©Assakkaf Shear and Moment Diagrams – If a numerical value of the shear is to be determined at the point, one can find this value either by using the methods of establishing equations (formulas)for each section under study or by using Eq. 34, which states that the change in the shear force is equal to the area under the distributed loading diagram. Since w(x) is integrated to obtain V, if w(x) is a curve of degree n, then V(x) will be a curve of degree n + 1. For example, if w(x) is uniform, V(x) will be linear. • Moment Diagram – Establish the M and x axes and plot the values of the moment at the ends of the beam. 17 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear Forces and Bending Moments in Beams Slide No. 34 ENES 220 ©Assakkaf Shear and Moment Diagrams – Since dM/dx = V, the slope of the moment diagram at any point is equal to the intensity of the shear at the point. In particular, note that at the point where the shear is zero, that is dM/dx = 0, and therefore this may be a point of maximum or minimum moment. If the numerical value of the moment is to be determined at a point, one can find this value either by using the method of establishing equations (formulas) for each section under study or by using Eq. 41, which states that the change in the moment is equal to the area under the shear diagram. Since w(x) is integrated to obtain V, if w(x) is a curve of degree n, then V(x) will be a curve of degree n + 1. For example, if w(x) is uniform, V(x) will be linear. LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear Forces and Bending Moments in Beams Slide No. 35 ENES 220 ©Assakkaf Shear and Moment Diagrams Loading Shear Diagram, dV =w dx dM Moment Diagram, dx =V 18 Slide No. 36 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear Forces and Bending Moments in Beams ENES 220 ©Assakkaf Shear and Moment Diagrams Loading Shear Diagram, dV =w dx dM Moment Diagram, dx Slide No. 37 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear and Moment Diagrams =V ENES 220 ©Assakkaf Example 10 Draw the shear and bending moment diagrams for the beam shown in Figure 21a. 600 lb 40 lb/ft A 12 ft 20 ft 1000 lb · in Figure 21a 19 Slide No. 38 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear and Moment Diagrams ENES 220 ©Assakkaf Example 10 (cont’d) – Support Reactions • The reactions at the fixed support can be calculated as follows: + ↑ ∑ Fy = 0; RA − 40(12 ) − 600 = 0 → RA = 1080 lb + ∑ M A = 0; − M + 40(12)(6 ) + 600(20) + 1000 = 0 Figure 21b ∴ M = 15,880 lb ⋅ in 600 lb 40 lb/ft A 12 ft M = 15,880 lb · ft 1000 lb · ft 20 ft RA = 1080 lb Slide No. 39 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear and Moment Diagrams ENES 220 ©Assakkaf Example 10 (cont’d) – Shear Diagram • Using the established sign convention, the shear at the ends of the beam is plotted first. For example, when x = 0, V = 1080; and when x = 20, V = 600 40 lb/ft V ( x ) = 1080 − 40 x for 0 < x < 12 M A 15,880 lb · ft x 1080 lb 2 V M = 1080 x − 40 x − 15,880 for 0 < x < 12 2 20 Slide No. 40 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear and Moment Diagrams ENES 220 ©Assakkaf Example 10 (cont’d) 40 lb/ft A 15,880 lb · in M 12 ft x V 1080 lb V ( x ) = 600 for 12 < x < 20 M = −15,880 − 40(12 )(x - 6 ) + 1080 x for 12 < x < 20 Slide No. 41 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear and Moment Diagrams Example 10 (cont’d) 40 lb/ft Shear Digram 12 ft ENES 220 ©Assakkaf 600 lb 20 ft 1000 lb · in 1000 V (lb) 800 600 400 V ( x ) = 1080 − 40 x for 0 < x < 12 10 15 200 0 0 5 20 x (ft) 21 Slide No. 42 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) ENES 220 ©Assakkaf Shear and Moment Diagrams Example 10 (cont’d) 40 lb/ft 600 lb Bending Moment Diagram 12 ft 0 -2000 0 5 10 15 20 ft 1000 lb · in 20 M (ft . lb) -4000 -6000 -8000 -10000 -12000 -14000 -16000 x (ft) Slide No. 43 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) ENES 220 ©Assakkaf Shear and Moment Diagrams Example 10 (cont’d) 600 lb 40 lb/ft 12 ft M = 15,880 lb · ft RA = 1080 lb 20 ft 1000 lb · ft 1080 V (lb) (+) 600 (-) -1000 M (ft·lb) x x 22 Slide No. 44 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear and Moment Diagrams ENES 220 ©Assakkaf Example 11 Draw complete shear and bending moment diagrams for the beam shown in Fig. 22 8000 lb y 2000 lb/ft A B C x 12 ft 4 ft D Figure 22a 8 ft Slide No. 45 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear and Moment Diagrams ENES 220 ©Assakkaf Example 11 (cont’d) – The support reactions were computed from equilibrium as shown in Fig. 22.b. y 8000 lb 2000 lb/ft A 12 ft RA = 11,000 lb B C x 4 ft D Figure 22a 8 ft RC = 21,000 lb 23 Slide No. 46 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear and Moment Diagrams Example 11 (cont’d) 2000 lb/ft A 8000 lb B C x 12 ft 11,000 lb ENES 220 ©Assakkaf D 4 ft 8 ft 21,000 lb 11,000 V (lb) (+) 8,000 lb (+) 5.5 ft (-) 13,000 lb 30,250 M (ft -lb) (-) (-) 12,000 64,000 LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Shear and Moment Diagrams Slide No. 47 ENES 220 ©Assakkaf Example 11 (cont’d) 12 11,000 x x 12 − x 12 = ⇒x= = 5.5 11,000 13,000 2.18 13,000 24