4 Shear Forces and Bending Moments

Transcription

1 4 Shear Forces and ending oments Shear Forces and ending oments 8 lb 16 lb roblem alculate the shear force and bending moment at a cross section just to the left of the 16-lb load acting on the simple beam shown in the figure. 3 in. 6 in. 3 in. 1 in. Solution Simple beam 8 lb 16 lb 3 in. 6 in. 3 in. R : R 14 lb : R 1 lb R Free-body diagram of segment 16 lb 3 in. R F ERT : 16 lb 14 lb lb : (14 lb)(3 in.) 4, lb-in. roblem 4.3- etermine the shear force and bending moment at the midpoint of the simple beam shown in the figure. 6. kn. kn/m 1. m 1. m 4. m. m Solution 4.3- Simple beam 6. kn. kn/m Free-body diagram of segment 6. kn 1. m 1. m. m R R R 1. m 1. m : : R 4.5 kn R 5.5 kn F ERT :.5 kn : 5. kn m 59

2 6 HTER 4 Shear Forces and ending oments roblem etermine the shear force and bending moment at the midpoint of the beam with overhangs (see figure). Note that one load acts downward and the other upward. b b Solution eam with overhangs :R 1 b (downward) b R R 1 [( b b)] 1 b (upward) R b b / R Free-body diagram ( is the midpoint) F ERT : R 1 b : 1 b b b b b roblem alculate the shear force and bending moment at a cross section located.5 m from the fixed support of the cantilever beam shown in the figure. 4. kn 1.5 kn/m 1. m 1. m. m Solution antilever beam 4. kn 1.5 kn/m 1. m 1. m. m Free-body diagram of segment oint is.5 m from support. F ERT : 4. kn (1.5 knm)(. m) 4. kn 3. kn 7. kn : (4. kn)(.5 m) (1.5 knm)(. m)(.5 m). kn m 7.5 kn m 9.5 kn m 4. kn 1.5 kn/m.5 m 1. m. m

3 SETION 4.3 Shear Forces and ending oments 61 roblem etermine the shear force and bending moment at a cross section located 16 ft from the left-hand end of the beam with an overhang shown in the figure. 4 lb/ft lb/ft 1 ft 1 ft 6 ft 6 ft Solution eam with an overhang 4 lb/ft lb/ft R 1 ft 1 ft 6 ft R 6 ft Free-body diagram of segment 4 lb/ft 1 ft 6 ft R : : R 46 lb R 74 lb oint is 16 ft from support. F ERT : 46 lb (4 lbft)(1 ft) 154 lb : (46 lb)(16 ft) (4 lbft)(1 ft)(11 ft) 464 lb-ft roblem The beam shown in the figure is simply supported at and and has an overhang from to. The loads consist of a horizontal force 1 4. kn acting at the end of a vertical arm and a vertical force 8. kn acting at the end of the overhang. etermine the shear force and bending moment at a cross section located 3. m from the left-hand support. (Note: isregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.) 1 = 4. kn 1. m = 8. kn 4. m 1. m Solution eam with vertical arm 1 = 4. kn = 8. kn 1. m 4. m 1. m R R : R 1. kn (downward) : R 9. kn (upward) Free-body diagram of segment oint is 3. m from support. 4. kn m 3. m R F ERT : R 1. kn : R (3. m) 4. kn m 7. kn m

4 6 HTER 4 Shear Forces and ending oments roblem The beam shown in the figure has overhangs at each end and carries a uniform load of intensity q. For what ratio b/ will the bending moment at the midpoint of the beam be zero? q b b Solution eam with overhangs q Free-body diagram of left-hand half of beam: oint E is at the midpoint of the beam. q b b R R From symmetry and equilibrium of vertical forces: b / R E = (Given) R R q b E R q 1 b q b q 1 b Solve for b/ : b 1 roblem t full draw, an archer applies a pull of 13 N to the bowstring of the bow shown in the figure. etermine the bending moment at the midpoint of the bow mm 35 mm

5 SETION 4.3 Shear Forces and ending oments 63 Solution rcher s bow Free-body diagram of segment H T H b b 13 N 7 H 14 mm 1.4 m b 35 mm.35 m Free-body diagram of point T(cos H b) T(sin b)(b) T H cosb b sin b H b tan b Substitute numerical values: 13 N 1.4 m (.35 m)(tan 7)R 18 N m T T T tensile force in the bowstring F HORIZ : T cos T cos b

6 64 HTER 4 Shear Forces and ending oments roblem curved bar is subjected to loads in the form of two equal and opposite forces, as shown in the figure. The axis of the bar forms a semicircle of radius r. etermine the axial force N, shear force, and bending moment acting at a cross section defined by the angle. O r N Solution urved bar O r cos sin N O F N Q b N sin u N sin u F R a cos u cos u O Nr Nr r sin u roblem Under cruising conditions the distributed load acting on the wing of a small airplane has the idealized variation shown in the figure. alculate the shear force and bending moment at the inboard end of the wing. 16 N/m 9 N/m.6 m.6 m 1. m Solution irplane wing 16 N/m 9 N/m.6 m.6 m 1. m oading (in three parts) 7 N/m 1 9 N/m 3 Shear Force F ERT c T 1 (7 Nm)(.6 m) (9 Nm)(5. m) 1 (9 Nm)(1. m) 64 N 6.4 kn (inus means the shear force acts opposite to the direction shown in the figure.) ending oment 1 (9 Nm)(5. m)(.6 m) 1 m (7 Nm)(.6 m) m (9 Nm)(1. m) 5. m N m 1,168 N m 49 N m 15,45 N m kn m

7 SETION 4.3 Shear Forces and ending oments 65 roblem beam with a vertical arm E is supported as a simple beam at and (see figure). 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. What is the force in the cable if the bending moment in the beam just to the left of point is equal numerically to 64 lb-ft? (Note: isregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.) able E 8 ft 6 ft 6 ft 6 ft Solution eam with a cable E Free-body diagram of section 4 9 able 6 ft 6 ft 6 ft 8 ft ft 6 ft N UNITS: in lb in lb-ft 4 5 (6 ft) 4 9 (1 ft) 8 15 lb-ft Numerical value of equals 64 lb-ft. 64 lb-ft 8 15 lb-ft and 1 lb roblem simply supported beam supports a trapezoidally distributed load (see figure). The intensity of the load varies linearly from 5 kn/m at support to 3 kn/m at support. alculate the shear force and bending moment at the midpoint of the beam. 5 kn/m 3 kn/m 3 m

8 66 HTER 4 Shear Forces and ending oments Solution eam with trapezoidal load 5 kn/m Free-body diagram of section 3 kn/m oint is at the midpoint of the beam. 4 kn/m 3 kn/m 3 m R R 1.5 m 55 kn Reactions R (3 m) (3 knm)(3 m)(1.5 m) ( knm)(3 m)( 1 )( m) R 65 kn F ERT c R R 1 (5 knm 3 knm)(3 m) R 55 kn F ERT c T (3 knm)(1.5 m) 1 (1 knm)(1.5 m) 55 kn.5 kn (3 kn/m)(1.5 m)(.75 m) 1 (1 knm)(1.5 m)(.5 m) (55 kn)(1.5 m) 45. kn m roblem eam represents a reinforced-concrete foundation beam that supports a uniform load of intensity q 1 35 lb/ft (see figure). ssume that the soil pressure on the underside of the beam is uniformly distributed with intensity q. (a) Find the shear force and bending moment at point. (b) Find the shear force m and bending moment m at the midpoint of the beam. q 1 = 35 lb/ft q 3. ft 8. ft 3. ft Solution Foundation beam q 1 = 35 lb/ft q 3. ft 8. ft 3. ft F ERT : q (14 ft) q 1 (8 ft) q 8 14 q 1 lbft (a) and at point lb/ft 3 ft F ERT : : 9 lb-ft 6 lb (b) and at midpoint E 35 lb/ft E m lb/ft m 3 ft 4 ft F ERT : m ( lb/ft)(7 ft) (35 lb/ft)(4 ft) m E : m ( lb/ft)(7 ft)(3.5 ft) (35 lb/ft)(4 ft)( ft) m 1, lb-ft

9 SETION 4.3 Shear Forces and ending oments 67 roblem The simply-supported beam is loaded by a weight W 7 kn through the arrangement shown in the figure. The cable passes over a small frictionless pulley at and is attached at E to the end of the vertical arm. alculate the axial force N, shear force, and bending moment at section, which is just to the left of the vertical arm. (Note: isregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.) E able 1.5 m. m. m. m W = 7 kn Solution eam with cable and weight E Free-body diagram of pulley at able 1.5 m 7 kn 1.6 kn. m. m. m 7 kn 7 kn 1.8 kn R R R 18 kn R 9kN Free-body diagram of segment of beam 1.8 kn 1.6 kn. m. m 18 kn F HORIZ :N 1.6 kn (compression) F ERT : 7. kn : 5.4 kn m N

10 68 HTER 4 Shear Forces and ending oments roblem 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) with an angular acceleration. Each of the two arms has weight w per unit length and supports a weight W. w at its end. erive formulas for the maximum shear force and maximum bending moment in the arms, assuming b /9 and c /1. y b c W x W Solution Rotating centrifuge b c W g ( + b + c) x wx g Tangential acceleration r Substitute numerical data: W Inertial force r g r aximum and occur at x b. max W g ( b c) b w g x dx W g ( b c) w ( b) g max W ( b c)( c) g b w x(x b)dx g b W ( b c)( c) g w ( 3b) 6g b W. wb 9 91w max 3g max 9w 3 75g c 1

11 SETION 4.5 Shear-Force and ending-oment iagrams 69 Shear-Force and ending-oment iagrams When solving the problems for Section 4.5, draw the shear-force and bending-moment diagrams approximately to scale and label all critical ordinates, including the maximum and minimum values. robs through are symbolic problems and robs through are numerical problems. The remaining problems (4.5-5 through 4.5-3) involve specialized topics, such as optimization, beams with hinges, and moving loads. roblem raw the shear-force and bending-moment diagrams for a simple beam supporting two equal concentrated loads (see figure). a a Solution Simple beam a a R = R = a

12 7 HTER 4 Shear Forces and ending oments roblem 4.5- simple beam is subjected to a counterclockwise couple of moment acting at distance a from the left-hand support (see figure). raw the shear-force and bending-moment diagrams for this beam. a Solution 4.5- Simple beam R = a R = (1 a ) a roblem raw the shear-force and bending-moment diagrams for a cantilever beam carrying a uniform load of intensity q over one-half of its length (see figure). q Solution antilever beam = 3q 8 q R = q q 3q 8 q 8

13 SETION 4.5 Shear-Force and ending-oment iagrams 71 roblem The cantilever beam shown in the figure is subjected to a concentrated load at the midpoint and a counterclockwise couple of moment 1 /4 at the free end. raw the shear-force and bending-moment diagrams for this beam. 1 = 4 Solution antilever beam 1 4 R R / / roblem The simple beam shown in the figure is subjected to a concentrated load and a clockwise couple 1 /4 acting at the third points. raw the shear-force and bending-moment diagrams for this beam. 1 = Solution Simple beam 1 = 4 R = R = 7 1 5/1 7/1 5/36 7/36 /18

14 7 HTER 4 Shear Forces and ending oments roblem simple beam subjected to clockwise couples 1 and 1 acting at the third points is shown in the figure. raw the shear-force and bending-moment diagrams for this beam Solution Simple beam 1 1 R = R 3 = roblem simply supported beam is loaded by a vertical load acting at the end of a bracket E (see figure). raw the shear-force and bending-moment diagrams for beam. E 4 4 Solution eam with bracket 4 R = R = 8 3 8

15 SETION 4.5 Shear-Force and ending-oment iagrams 73 roblem beam is simply supported at and and has an overhang (see figure). The beam is loaded by two forces and a clockwise couple of moment a that act through the arrangement shown. raw the shear-force and bending-moment diagrams for beam. a a a a a Solution eam with overhang upper beam: a a a a lower beam: a a a a roblem eam is simply supported at and and has overhangs at each end (see figure). The span length is and each overhang has length /3. uniform load of intensity q acts along the entire length of the beam. raw the shear-force and bending-moment diagrams for this beam. 3 q 3 Solution eam with overhangs q /3 5q R = 6 /3 5q R = 6 q q/3 q q 3 5q 7 q /18 X 1 q /18 x

16 74 HTER 4 Shear Forces and ending oments roblem raw the shear-force and bending-moment diagrams for a cantilever beam supporting a linearly varying load of maximum intensity q (see figure). q Solution antilever beam x q=q q q = 6 x q R = q x = q q x = 3 6 q 6 roblem The simple beam supports a uniform load of intensity q 1 lb/in. acting over one-half of the span and a concentrated load 8 lb acting at midspan (see figure). raw the shear-force and bending-moment diagrams for this beam. = 8 lb q = 1 lb/in. = 4 in. = 4 in. Solution Simple beam = 8 lb 1 lb/in. R =14 lb 4 in. 4 in. R = 34 lb 14 (lb) 6 6 in. 34 (lb/in.) 56 max = in.

17 SETION 4.5 Shear-Force and ending-oment iagrams 75 roblem The beam shown in the figure supports a uniform load of intensity 3 N/m acting over half the length of the beam. The beam rests on a foundation that produces a uniformly distributed load over the entire length. raw the shear-force and bending-moment diagrams for this beam. 3 N/m.8 m 1.6 m.8 m Solution eam with distributed loads 3 N/m 15 N/m.8 m m.8 m (N) 96 1 (N. m) roblem cantilever beam supports a couple and a concentrated load, as shown in the figure. raw the shear-force and bending-moment diagrams for this beam. 4 lb-ft lb 5 ft 5 ft Solution antilever beam = 16 lb-ft 4 lb-ft lb R = lb 5 ft 5 ft + (lb) (lb-ft)

18 76 HTER 4 Shear Forces and ending oments roblem The cantilever beam shown in the figure is subjected to a uniform load acting throughout one-half of its length and a concentrated load acting at the free end. raw the shear-force and bending-moment diagrams for this beam.. kn/m m m.5 kn Solution antilever beam. kn/m.5 kn = 14 kn. m R = 6.5 kn (kn) (kn. m) m m roblem The uniformly loaded beam has simple supports at and and an overhang (see figure). raw the shear-force and bending-moment diagrams for this beam. 5 lb/in. 7 in. 48 in. Solution eam with an overhang 5 lb/in. 7 in. R = 5 lb 5 (lb) in. 5 (lb-in.) in. 4 in in. R = 5 lb 13 8,8

19 SETION 4.5 Shear-Force and ending-oment iagrams 77 roblem beam with an overhang at one end supports a uniform load of intensity 1 kn/m and a concentrated load of magnitude.4 kn (see figure). raw the shear-force and bending-moment diagrams for this beam. 1 kn/m.4 kn 1.6 m 1.6 m 1.6 m Solution eam with an overhang 1 kn/m.4 kn (kn) (kn. m) 1.6 m 1.6 m 1.6 m R = 13. kn R = 8.4 kn m max m max = m 3.84 roblem The beam shown in the figure is simply supported at and and has an overhang from to. The loads consist of a horizontal force 1 4 lb acting at the end of the vertical arm and a vertical force 9 lb acting at the end of the overhang. raw the shear-force and bending-moment diagrams for this beam. (Note: isregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.) 1 = 4 lb 1. ft = 9 lb 4. ft 1. ft Solution eam with vertical arm 1. ft 1 = 4 lb R = 15 lb 4 lb-ft 15 lb = 9 lb 4. ft 1. ft R = 15 lb 9 lb 15 lb (lb) (lb)

20 78 HTER 4 Shear Forces and ending oments roblem simple beam is loaded by two segments of uniform load and two horizontal forces acting at the ends of a vertical arm (see figure). raw the shear-force and bending-moment diagrams for this beam. 4 kn/m 1 m 8 kn 1 m 4 kn/m 8 kn m m m m Solution Simple beam 4 kn/m 16 kn. m m m m R = 6 kn 4 kn/m m R = 1 kn (kn) m (kn. m) 1.5 m roblem beam with a vertical arm E is supported as a simple beam at and (see figure). 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. The tensile force in the cable is 18 lb. raw the shear-force and bending-moment diagrams for beam. (Note: isregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.) able E 18 lb 8 ft 6 ft 6 ft 6 ft Solution eam with a cable E 18 lb Free-body diagram of beam 18 lb able 8 ft lb-ft ft 6 ft 6 ft R = 8 lb R = 8 lb 64 Note: ll forces have units of pounds. 48 (lb) (lb-ft)

21 SETION 4.5 Shear-Force and ending-oment iagrams 79 roblem 4.5- The beam shown in the figure has overhangs that extend in both directions for a distance of 4. m from the supports at and, which are 1. m apart. raw the shear-force and bending-moment diagrams for this overhanging beam. 5.1 kn/m 1.6 kn/m 5.1 kn/m 4. m 4. m 1. m Solution 4.5- eam with overhangs kn/m 1.6 kn/m 5.1 kn/m (kn) m 4. m R 1. m = kn R = kn (kn. m) roblem The simple beam shown in the figure supports a concentrated load and a segment of uniform load. raw the shear-force and bending-moment diagrams for this beam. 4. k. k/ft 5 ft ft 1 ft Solution Simple beam 4. k. k/ft R = 8 k 5 ft 5 ft 1 ft R = 16 k (k) ft ft 16 max = 64 k-ft (k-ft) 1 ft 8 ft

22 8 HTER 4 Shear Forces and ending oments roblem 4.5- The cantilever beam shown in the figure supports a concentrated load and a segment of uniform load. raw the shear-force and bending-moment diagrams for this cantilever beam. 3 kn.8 m.8 m 1. kn/m 1.6 m Solution 4.5- antilever beam 4.6 = 6.4 kn. m 3 kn 1. kn/m (kn) m.8 m R = 4.6 kn 1.6 m (kn. m) roblem The simple beam shown in the figure is subjected to a triangular load of maximum intensity 18 lb/ft. raw the shear-force and bending-moment diagrams for this beam. 18 lb/ft 6. ft 7. ft Solution Simple beam 18 lb/ft (lb) 4 x 1 = 4. ft 3 39 max = 64 R = 4 lb 6. ft 1. ft R = 39 lb (lb-ft) 36 roblem beam with simple supports is subjected to a trapezoidally distributed load (see figure). The intensity of the load varies from 1. kn/m at support to 3. kn/m at support. raw the shear-force and bending-moment diagrams for this beam. 1. kn/m 3. kn/m.4 m

23 SETION 4.5 Shear-Force and ending-oment iagrams 81 Solution kn/m Simple beam 3. kn/m (kn). x 1 = 1.98 m x R =. kn.4 m R =.8 kn Set :. x x.4 (x meters; kn).8 max = 1.45 x m (kn. m) roblem beam of length is being designed to support a uniform load of intensity q (see figure). If the supports of the beam are placed at the ends, creating a simple beam, the maximum bending moment in the beam is q /8. However, if the supports of the beam are moved symmetrically toward the middle of the beam (as pictured), the maximum bending moment is reduced. etermine the distance a between the supports so that the maximum bending moment in the beam has the smallest possible numerical value. raw the shear-force and bending-moment diagrams for this condition. q a Solution eam with overhangs q Solve for a: a ( ).5858 ( a)/ ( a)/ a R = q/ R = q/ 1 1 The maximum bending moment is smallest when 1 (numerically). q( a) 1 8 R a q 8 q (a ) 8 1 ( a) (a ) 1 q ( a) 8 q 8 (3 ).145q q q.145 q x 1 x 1.71 q.99 q.145 q.145 q x 1 =.3536 a =.71

24 8 HTER 4 Shear Forces and ending oments roblem The compound beam E shown in the figure consists of two beams ( and E) joined by a hinged connection at. The hinge can transmit a shear force but not a bending moment. The loads on the beam consist of a 4-kN force at the end of a bracket attached at point and a -kn force at the midpoint of beam E. raw the shear-force and bending-moment diagrams for this compound beam. 4 kn 1 m kn 1 m E m m m m Solution ompound beam 4 kn 4 kn. m Hinge kn E m m m 1 m 1 m R =.5 kn R =.5 kn R E = 1 kn (kn) (kn. m).67 m roblem The compound beam E shown in the figure consists of two beams ( and E) joined by a hinged connection at. The hinge can transmit a shear force but not a bending moment. force acts upward at and a uniform load of intensity q acts downward on beam E. raw the shear-force and bending-moment diagrams for this compound beam. q E Solution ompound beam q E Hinge R = + q R = + q q R E = q q q q q

25 SETION 4.5 Shear-Force and ending-oment iagrams 83 roblem The shear-force diagram for a simple beam is shown in the figure. etermine the loading on the beam and draw the bendingmoment diagram, assuming that no couples act as loads on the beam. 1 kn 1 kn. m 1. m 1. m Solution Simple beam ( is given) 6. kn/m 1 kn R = 1kN m 1 m 1 m R = 1kN (kn) (kn. m) roblem The shear-force diagram for a beam is shown in the figure. ssuming that no couples act as loads on the beam, determine the forces acting on the beam and draw the bendingmoment diagram. 65 lb 57 lb 58 lb 5 lb 18 lb 448 lb 4 ft 16 ft 4 ft Solution Force diagram Forces on a beam ( is given) lb/ft (lb) ft 16 ft 4 ft 65 lb 7 lb 18 lb 5 lb (lb-ft) ft 16

26 84 HTER 4 Shear Forces and ending oments roblem simple beam supports two connected wheel loads and that are distance d apart (see figure). The wheels may be placed at any distance x from the left-hand support of the beam. (a) etermine the distance x that will produce the maximum shear force in the beam, and also determine the maximum shear force max. (b) etermine the distance x that will produce the maximum bending moment in the beam, and also draw the corresponding bendingmoment diagram. (ssume 1 kn, d.4 m, and 1 m.) x d Solution x oving loads on a beam d 1 kn d.4 m 1 m Reaction at support : R x (x d) (d 3x) ending moment at : R ( x d) (d 3x)( x d) (a) aximum shear force y inspection, the maximum shear force occurs at support when the larger load is placed close to, but not directly over, that support. x d 9.6 m max R 3 d 8 kn (b) aximum bending moment y inspection, the maximum bending moment occurs at point, under the larger load. R = d x x = d d R = (3 d ) d R d dx [3x (3 5d)x d( d)] Solve for x: x 6 3 5d 4. m Substitute x into Eq (1): max d (3 5d) (kn. m) (6x 3 5d) 6 3 5d d( d)r 1 3 d 78.4 kn m 64 Note:R 3 d 16 kn R 3 d 14 kn max = m.4 m 5.6 m Eq.(1)