Shear Forces and Bending Moments

1 4. Shear Forces and Bending Moments Shear Forces and Bending Moments 800 lb 1600 lb Problem Calculate the Shear force V and Bending moment M. A B. at a cross section just to the left of the 1600-lb load acting on the simple beam AB shown in the figure. 30 in. 60 in. 30 in. 120 in. Solution Simple beam Free-body diagram of segment DB. 800 lb 1600 lb 1600 lb D V. A B D. B. M. 30 in. 30 in. 60 in. 30 in. RB. RA RB FVERT 0: V 1600 lb 1400 lb 200 lb MA 0: RB 1400 lb MD 0: M (1400 lb)(30 in.). MB 0: RA 1000 lb 42,000 lb-in. kN kN/m Problem Determine the Shear force V and Bending moment M. at the midpoint C of the simple beam AB shown in the figure. C. A B. m m m m Solution Simple beam kN kN/m Free-body diagram of segment AC. C kN. A B M. A C V. m m m m m RA RB RA. FVERT 0: V kN. MA 0: RB kN. MC 0: M kN m MB 0: RA kN. 259. 260 CHAPTER 4 Shear Forces and Bending Moments Problem Determine the Shear force V and Bending moment M at P P. the midpoint of the beam with overhangs (see figure). Note that one load acts downward and the other upward.

2 B L b Solution beam with overhangs 2b P P MA 0: RB P 1 (downward). L. A B P. M. A C. b L b b L/2. RA RB RA V. Free-body diagram (C is the midpoint). MB 0. FVERT 0: 1 2b 2bP. RA [P(L b b) ] V RA P P 1 P . L L L. MC 0: 2b P 1 (upward) 2b L L. L M P 1 P b . L 2 2. PL PL. M Pb Pb 0. 2 2. Problem Calculate the Shear force V and Bending moment M at a kN kN/m cross section located m from the fixed support of the cantilever beam A. AB shown in the figure. B. m m m Solution Cantilever beam kN kN/m A FVERT 0: B V kN ( kN m)( m). kN kN kN. m m m MD 0: M ( kN)( m). ( kN m)( m)( m). Free-body diagram of segment DB kN m kN m kN m Point D is m from support A. kN kN/m V. D. M B. m m m SECTION Shear Forces and Bending Moments 261. Problem Determine the Shear force V and Bending moment M 400 lb/ft 200 lb/ft at a cross section located 16 ft from the left-hand end A of the beam B. with an overhang shown in the figure. A C. 10 ft 10 ft 6 ft 6 ft Solution beam with an overhang 400 lb/ft 200 lb/ft Free-body diagram of segment AD.

3 B. A C 400 lb/ft D. A M. 10 ft 10 ft 6 ft 6 ft 10 ft 6 ft RA RB RA V. MB 0: RA 2460 lb MA 0: RB 2740 lb Point D is 16 ft from support A. FVERT 0: V 2460 lb (400 lb ft)(10 ft). 1540 lb MD 0: M (2460 lb)(16 ft). (400 lb ft)(10 ft)(11 ft). 4640 lb-ft Problem The beam ABC shown in the figure is simply P = kN. 1. supported at A and B and has an overhang from B to C. The P2 = kN. loads consist of a horizontal force P1 kN acting at the end of a vertical arm and a vertical force P2 kN acting at m the end of the overhang. A B C. Determine the Shear force V and Bending moment M at a cross section located m from the left-hand support. (Note: Disregard the widths of the beam and vertical arm and m m use centerline dimensions when making calculations.). Solution beam with vertical arm P1 = kN Free-body diagram of segment AD. P2 = kN. Point D is m from support A. m A B M. A D. kN m m RA V. m m RA RB. FVERT 0: V RA kN. MB 0: RA kN (downward) MD 0: M RA ( m) kN m MA 0: RB kN (upward) kN m 262 CHAPTER 4 Shear Forces and Bending Moments Problem The beam ABCD shown in the figure has overhangs q at each end and carries a uniform load of intensity q.

4 For what ratio b/L will the Bending moment at the midpoint of the A D. beam be zero? B C. b L b Solution beam with overhangs q Free-body diagram of left-hand half of beam : Point E is at the midpoint of the beam . A D. B C q b L b M = 0 (Given). RB RC A. E. b L/2 V. From symmetry and equilibrium of vertical Forces : RB. L. RB RC q b . 2 ME 0 . L 1 L 2. RB q b 0. 2 2 2. L L 1 L 2. q b q b 0. 2 2 2 2. Solve for b/L : b 1.. L 2. Problem At full draw, an archer applies a pull of 130 N to the bowstring of the bow shown in the figure. Determine the Bending moment at the midpoint of the bow. 70 . 1400 mm 350 mm SECTION Shear Forces and Bending Moments 263. Solution Archer's bow B Free-body diagram of segment BC. B.. H. H T. 2. P C. A. b C. M. MC 0 . H. b T(sin b) (b) M 0. T(cos b) . 2. P 130 N H. M T cosb b sin b . 2. 70 P H. b tan b . H 1400 mm 2 2. m Substitute numerical values: b 350 mm 130 N m M B ( m)(tan 70 )R. m 2 2. M 108 N m Free-body diagram of point A. T.. P. A. T. T tensile force in the bowstring FHORIZ 0: 2T cos P 0.

5 P. T . 2 cos b 264 CHAPTER 4 Shear Forces and Bending Moments Problem A curved bar ABC is subjected to loads in the form M N. of two equal and opposite Forces P, as shown in the figure. The axis of B. the bar forms a semicircle of radius r. V. Determine the axial force N, Shear force V, and Bending moment M r P P P . acting at a cross section defined by the angle . A O C A. Solution Curved bar M N FN 0 Q b N P sin u 0. B. B. P cos N P sin u V. r . P P FV 0 R a V P cos u 0. P O. A O C A V P cos u P sin . MO 0 M Nr 0. M Nr Pr sin u Problem Under cruising conditions the distributed load acting on the wing of a small airplane has the idealized variation shown in the figure. 1600 N/m 900 N/m Calculate the Shear force V and Bending moment M at the inboard end of the wing. m m m Solution Airplane wing 1600 N/m Loading (in three parts). 900 N/m M V 700 N/m 1. 900 N/m 2 3. A B A B. m m m Shear Force Bending Moment . FVERT 0 c T MA 0 . 1 1 m V (700 N m)( m) (900 N m)( m) M (700 N m)( m) . 2 2 3.

6 1. (900 N m)( m) 0 (900 N m)( m)( m). 2. 1 m V 6040 N kN (900 N m)( m) m 0. 2 3. (Minus means the Shear force acts opposite to the M N m 12,168 N m 2490 N m direction shown in the figure.). 15,450 N m kN m SECTION Shear Forces and Bending Moments 265. Problem A beam ABCD with a vertical arm CE is supported as E P. a simple beam at A and D (see figure). A cable passes over a small pulley that is attached to the arm at E. One end of the cable is attached to the beam at point B. Cable 8 ft What is the force P in the cable if the Bending moment in the A B C D. beam just to the left of point C is equal numerically to 640 lb-ft? (Note: Disregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.). 6 ft 6 ft 6 ft Solution beam with a cable E P Free-body diagram of section AC. P. Cable 8 ft 4P. __ 3P. __ M. P A 5 5 C. P A B C D. N. 6 ft B 6 ft 6 ft 6 ft 6 ft 4P. __. 9 V. 4P. __ 4P. __ MC 0 . 9 9. 4P 4P. M (6 ft) (12 ft) 0. UNITS: 5 9. P in lb 8P. M lb-ft 15.

7 M in lb-ft Numerical value of M equals 640 lb-ft. 8P. 640 lb-ft lb-ft 15. and P 1200 lb Problem A simply supported beam AB supports a trapezoidally 50 kN/m distributed load (see figure). The intensity of the load varies linearly 30 kN/m from 50 kN/m at support A to 30 kN/m at support B. Calculate the Shear force V and Bending moment M at the midpoint A B. of the beam . 3m 266 CHAPTER 4 Shear Forces and Bending Moments Solution beam with trapezoidal load 50 kN/m Free-body diagram of section CB. 30 kN/m Point C is at the midpoint of the beam . 40 kN/m A B. 30 kN/m V. 3m M C B. RA RB m 55 kN. Reactions FVERT 0 c T . V (30 kN m)( m) 12 (10 kN m)( m). MB 0 RA (3 m) (30 kN m)(3 m)( m). (20 kN m)(3 m)( 1 2 )(2 m) 0 55 kN 0. RA 65 kN V kN. MC 0 . FVERT 0 c M (30 kN/m)( m)( m). RA RB 1 2 (50 kN m 30 kN m)(3 m) 0. 1 2 (10 kN m)( m)( m). RB 55 kN. (55 kN)( m) 0. M kN m Problem beam ABCD represents a reinforced-concrete q1 = 3500 lb/ft foundation beam that supports a uniform load of intensity q1 3500 lb/ft B C.

8 (see figure). Assume that the soil pressure on the underside of the beam is A D. uniformly distributed with intensity q2. (a) Find the Shear force VB and Bending moment MB at point B. q2. (b) Find the Shear force Vm and Bending moment Mm at the midpoint ft ft ft of the beam . Solution Foundation beam q1 = 3500 lb/ft (b) V and M at midpoint E. A B C D. 3500 lb/ft A B E Mm q2. ft ft ft Vm 2000 lb/ft FVERT 0: q2(14 ft) q1(8 ft) 3 ft 4 ft 8. q2 q 2000 lb ft 14 1 FVERT 0: Vm (2000 lb/ft)(7 ft) (3500 lb/ft)(4 ft). (a) V and M at point B. Vm 0. A B. MB ME 0: FVERT 0: Mm (2000 lb/ft)(7 ft)( ft). 2000 lb/ft VB. VB 6000 lb 3 ft (3500 lb/ft)(4 ft)(2 ft). MB 0: MB 9000 lb-ft Mm 21,000 lb-ft SECTION Shear Forces and Bending Moments 267. Problem The simply-supported beam ABCD is loaded by E. a weight W 27 kN through the arrangement shown in the figure. Cable The cable passes over a small frictionless pulley at B and is attached m at E to the end of the vertical arm. A B C D. Calculate the axial force N, Shear force V, and Bending moment M at section C, which is just to the left of the vertical arm.

9 (Note: Disregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.) m m m W = 27 kN. Solution beam with cable and weight E. Free-body diagram of pulley at B. Cable m 27 kN. A B C D. kN. m m m kN. 27 kN. 27 kN. RA RD. RA 18 kN RD 9 kN. Free-body diagram of segment ABC of beam kN. kN. A B C M. N. m m V. 18 kN. FHORIZ 0: N kN (compression). FVERT 0: V kN. MC 0: M kN m 268 CHAPTER 4 Shear Forces and Bending Moments Problem The centrifuge shown in the figure rotates in a horizontal plane (the xy plane) on a smooth surface about the z axis (which is vertical) y c with an angular acceleration . Each of the two arms has weight w per unit L. length and supports a weight W wL at its end. Derive formulas for the maximum Shear force and maximum Bending b W. moment in the arms, assuming b L/9 and c L/10. x . W. Solution Rotating centrifuge c L. b W (L + b + c) . __. g w x __. x g Tangential acceleration r Substitute numerical data: W L L. Inertial force Mr g r W wL b c.

10 9 10. Maximum V and M occur at x b. 91wL2 . Vmax . L b 30g . W w . Vmax (L b c) x dx 229wL3 . g b g Mmax . 75g W . (L b c). g wL . (L 2b). 2g W . Mmax (L b c)(L c). g L b . w . x(x b)dx b g W . (L b c)(L c). g w L2 . (2L 3b). 6g SECTION Shear -Force and Bending -Moment Diagrams 269. Shear -Force and Bending -Moment Diagrams When solving the problems for Section , draw the Shear -force and Bending -moment diagrams approximately to scale and label all critical ordinates, including the maximum and minimum values. Probs. through are symbolic problems and Probs. through are numerical problems. The remaining problems ( through ) involve specialized topics, such as optimization, beams with hinges, and moving loads. Problem Draw the Shear -force and Bending -moment diagrams for a P P a a simple beam AB supporting two equal concentrated loads P (see figure). A B. L. Solution Simple beam a P P a A B. RA = P L RB = P. P. V. 0. P. Pa M 0. 270 CHAPTER 4 Shear Forces and Bending Moments Problem A simple beam AB is subjected to a counterclockwise M0.