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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition 7- 1 CE 102 Statics Chapter 5 Forces in Beams and Cables

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition 7- 2 Contents Introduction Internal Forces in Members Sample Problem 7.1 Various Types of Beam Loading and Support Shear and Bending Moment in a Beam Sample Problem 7.2 Sample Problem 7.3 Relations Among Load, Shear, and Bending Moment Sample Problem 7.4 Sample Problem 7.5 Cables With Concentrated Loads Cables With Distributed Loads Parabolic Cable Sample Problem 7.6 Catenary

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition 7- 3 Introduction Preceding chapters dealt with: a) determining external forces acting on a structure and b)determining forces which hold together the various members of a structure. The current chapter is concerned with determining the internal forces (i.e., tension/compression, shear, and bending) which hold together the various parts of a given member. Focus is on two important types of engineering structures: a)Beams - usually long, straight, prismatic members designed to support loads applied at various points along the member. b)Cables - flexible members capable of withstanding only tension, designed to support concentrated or distributed loads.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition 7- 4 Internal Forces in Members Straight two-force member AB is in equilibrium under application of F and -F. Internal forces equivalent to F and -F are required for equilibrium of free-bodies AC and CB. Multiforce member ABCD is in equil- ibrium under application of cable and member contact forces. Internal forces equivalent to a force- couple system are necessary for equil- ibrium of free-bodies JD and ABCJ. An internal force-couple system is required for equilibrium of two-force members which are not straight.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition 7- 5 Sample Problem 5.1 Determine the internal forces (a) in member ACF at point J and (b) in member BCD at K. SOLUTION: Compute reactions and forces at connections for each member. Cut member ACF at J. The internal forces at J are represented by equivalent force-couple system which is determined by considering equilibrium of either part. Cut member BCD at K. Determine force-couple system equivalent to internal forces at K by applying equilibrium conditions to either part.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition 7- 6 Sample Problem 5.1 SOLUTION: Compute reactions and connection forces. Consider entire frame as a free-body:

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition 7- 7 Sample Problem 5.1 Consider member BCD as free-body: Consider member ABE as free-body: From member BCD,

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition 7- 8 Sample Problem 5.1 Cut member ACF at J. The internal forces at J are represented by equivalent force-couple system. Consider free-body AJ:

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition 7- 9 Sample Problem 5.1 Cut member BCD at K. Determine a force-couple system equivalent to internal forces at K. Consider free-body BK:

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Various Types of Beam Loading and Support Beam - structural member designed to support loads applied at various points along its length. Beam design is two-step process: 1)determine shearing forces and bending moments produced by applied loads 2)select cross-section best suited to resist shearing forces and bending moments Beam can be subjected to concentrated loads or distributed loads or combination of both.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Various Types of Beam Loading and Support Beams are classified according to way in which they are supported. Reactions at beam supports are determinate if they involve only three unknowns. Otherwise, they are statically indeterminate.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Shear and Bending Moment in a Beam Wish to determine bending moment and shearing force at any point in a beam subjected to concentrated and distributed loads. Determine reactions at supports by treating whole beam as free-body. Cut beam at C and draw free-body diagrams for AC and CB. By definition, positive sense for internal force-couple systems are as shown. From equilibrium considerations, determine M and V or M and V.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Shear and Bending Moment Diagrams Variation of shear and bending moment along beam may be plotted. Determine reactions at supports. Cut beam at C and consider member AC, Cut beam at E and consider member EB, For a beam subjected to concentrated loads, shear is constant between loading points and moment varies linearly.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 5.2 Draw the shear and bending moment diagrams for the beam and loading shown. SOLUTION: Taking entire beam as a free-body, calculate reactions at B and D. Find equivalent internal force-couple systems for free-bodies formed by cutting beam on either side of load application points. Plot results.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 5.2 SOLUTION: Taking entire beam as a free-body, calculate reactions at B and D. Find equivalent internal force-couple systems at sections on either side of load application points. Similarly,

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 5.2 Plot results. Note that shear is of constant value between concentrated loads and bending moment varies linearly.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 5.3 Draw the shear and bending moment diagrams for the beam AB. The distributed load of 40 lb/in. extends over 12 in. of the beam, from A to C, and the 400 lb load is applied at E. SOLUTION: Taking entire beam as free-body, calculate reactions at A and B. Determine equivalent internal force- couple systems at sections cut within segments AC, CD, and DB. Plot results.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 5.3 SOLUTION: Taking entire beam as a free-body, calculate reactions at A and B. Note: The 400 lb load at E may be replaced by a 400 lb force and 1600 lb-in. couple at D.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 5.3 From C to D: Evaluate equivalent internal force-couple systems at sections cut within segments AC, CD, and DB. From A to C:

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 5.3 Evaluate equivalent internal force-couple systems at sections cut within segments AC, CD, and DB. From D to B:

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 5.3 Plot results. From A to C: From C to D: From D to B:

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Relations Among Load, Shear, and Bending Moment Relations between load and shear: Relations between shear and bending moment:

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Relations Among Load, Shear, and Bending Moment Reactions at supports, Shear curve, Moment curve,

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 5.4 Draw the shear and bending- moment diagrams for the beam and loading shown. SOLUTION: Taking entire beam as a free-body, determine reactions at supports. With uniform loading between D and E, the shear variation is linear. Between concentrated load application points, and shear is constant. Between concentrated load application points, The change in moment between load application points is equal to area under shear curve between points. With a linear shear variation between D and E, the bending moment diagram is a parabola.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 5.4 Between concentrated load application points, and shear is constant. With uniform loading between D and E, the shear variation is linear. SOLUTION: Taking entire beam as a free-body, determine reactions at supports.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 5.4 Between concentrated load application points, The change in moment between load application points is equal to area under the shear curve between points. With a linear shear variation between D and E, the bending moment diagram is a parabola.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 5.5 Sketch the shear and bending- moment diagrams for the cantilever beam and loading shown. SOLUTION: The change in shear between A and B is equal to the negative of area under load curve between points. The linear load curve results in a parabolic shear curve. With zero load, change in shear between B and C is zero. The change in moment between A and B is equal to area under shear curve between points. The parabolic shear curve results in a cubic moment curve. The change in moment between B and C is equal to area under shear curve between points. The constant shear curve results in a linear moment curve.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 5.5 With zero load, change in shear between B and C is zero. SOLUTION: The change in shear between A and B is equal to negative of area under load curve between points. The linear load curve results in a parabolic shear curve.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 5.5 The change in moment between A and B is equal to area under shear curve between the points. The parabolic shear curve results in a cubic moment curve. The change in moment between B and C is equal to area under shear curve between points. The constant shear curve results in a linear moment curve.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Cables With Concentrated Loads Cables are applied as structural elements in suspension bridges, transmission lines, aerial tramways, guy wires for high towers, etc. For analysis, assume: a)concentrated vertical loads on given vertical lines, b)weight of cable is negligible, c)cable is flexible, i.e., resistance to bending is small, d)portions of cable between successive loads may be treated as two force members Wish to determine shape of cable, i.e., vertical distance from support A to each load point.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Cables With Concentrated Loads Consider entire cable as free-body. Slopes of cable at A and B are not known - two reaction components required at each support. Four unknowns are involved and three equations of equilibrium are not sufficient to determine the reactions. For other points on cable, Additional equation is obtained by considering equilibrium of portion of cable AD and assuming that coordinates of point D on the cable are known. The additional equation is

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Cables With Distributed Loads For cable carrying a distributed load: a)cable hangs in shape of a curve b)internal force is a tension force directed along tangent to curve. Consider free-body for portion of cable extending from lowest point C to given point D. Forces are horizontal force T 0 at C and tangential force T at D. From force triangle: Horizontal component of T is uniform over cable. Vertical component of T is equal to magnitude of W measured from lowest point. Tension is minimum at lowest point and maximum at A and B.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Parabolic Cable Consider a cable supporting a uniform, horizontally distributed load, e.g., support cables for a suspension bridge. With loading on cable from lowest point C to a point D given byinternal tension force magnitude and direction are Summing moments about D, or The cable forms a parabolic curve.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 5.6 The cable AE supports three vertical loads from the points indicated. If point C is 5 ft below the left support, determine (a) the elevation of points B and D, and (b) the maximum slope and maximum tension in the cable. SOLUTION: Determine reaction force components at A from solution of two equations formed from taking entire cable as free-body and summing moments about E, and from taking cable portion ABC as a free- body and summing moments about C. Calculate elevation of B by considering AB as a free-body and summing moments B. Similarly, calculate elevation of D using ABCD as a free- body. Evaluate maximum slope and maximum tension which occur in DE.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 5.6 SOLUTION: Determine two reaction force components at A from solution of two equations formed from taking entire cable as a free-body and summing moments about E, and from taking cable portion ABC as a free-body and summing moments about C. Solving simultaneously,

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 5.6 Calculate elevation of B by considering AB as a free-body and summing moments B. Similarly, calculate elevation of D using ABCD as a free-body.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 5.6 Evaluate maximum slope and maximum tension which occur in DE.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Catenary Consider a cable uniformly loaded along the cable itself, e.g., cables hanging under their own weight. With loading on the cable from lowest point C to a point D given bythe internal tension force magnitude is To relate horizontal distance x to cable length s,

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Catenary To relate x and y cable coordinates, which is the equation of a catenary.

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40 Problem 5.7 A B C D E 2 ft 3 ft 5 ft 7.5 ft 200 lb F H G A 200-lb load is applied at point G of beam EFGH, which is attached to cable ABCD by vertical hangers BF and CH. Determine (a) the tension in each hanger, (b) the maximum tension in the cable, (c) the bending moment at F and G.

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41 A B C D E 2 ft 3 ft 5 ft 7.5 ft 200 lb F H G A 200-lb load is applied at point G of beam EFGH, which is attached to cable ABCD by vertical hangers BF and CH. Determine (a) the tension in each hanger, (b) Solving Problems on Your Own 1. Identify points of the cable where useful information (position, slope,etc.) exists. Cut the cable at these points and draw a free-body diagram of one of the two portions of the cable. 2. Use M = 0 if you know the position, or F x = 0 and F y = 0 if you know the slope, to generate needed equations of equilibrium. Problem the maximum tension in the cable, (c) the bending moment at F and G.

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42 4. For a cable supporting vertical loads only, the horizontal component of the tension force is the same at any point. For such a cable, the maximum tension occurs in the steepest portion of the cable. 3. The tension in each section can be determined from the equations of equilibrium. A B C D E 2 ft 3 ft 5 ft 7.5 ft 200 lb F H G A 200-lb load is applied at point G of beam EFGH, which is attached to cable ABCD by vertical hangers BF and CH. Determine (a) the tension in each hanger, (b) Solving Problems on Your Own Problem the maximum tension in the cable, (c) the bending moment at F and G.

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43 + A B C D E 2 ft 3 ft 5 ft 7.5 ft 200 lb F H G Problem Solution Identify points of the cable where useful information exists. Draw a free-body diagram of one of the two portions of the cable. Use M = 0 to generate the equations of equilibrium. A B 2 ft 5 ft F BF T BC T0T0 AyAy Free Body: Portion AB M B = 0: T 0 (2 ft) - A y (5 ft) = 0 A y = 0.4 T 0

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44 Problem Solution + A B C D E 2 ft 3 ft 5 ft 7.5 ft 200 lb F H G Identify points of the cable where useful information exists. Draw a free-body diagram of one of the two portions of the cable. Use M = 0 to generate the equations of equilibrium. A B 3 ft 5 ft F BF T CD T0T0 Free Body: Portion ABC M C = 0: T 0 (3 ft) T 0 (10 ft) + F BF (5 ft) = 0 0.4T 0 C 5 ft F CH F BF = 0.2 T 0

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45 + A B C D E 2 ft 3 ft 5 ft 7.5 ft 200 lb F H G Identify points of the cable where useful information exists. Draw a free-body diagram of one of the two portions of the cable. Use M = 0 to generate the equations of equilibrium. 3 ft T BC T0T0 Free Body: Portion CD M C = 0: D y (5 ft) - T 0 (3 ft) = 0 C 5 ft F CH F BF = 0.2 T 0 DyDy D y = 0.6 T 0 D Problem Solution

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46 + A B C D E 2 ft 3 ft 5 ft 7.5 ft 200 lb F H G Identify points of the cable where useful information exists. Draw a free-body diagram of one of the two portions of the cable. Use M = 0 to generate the equations of equilibrium. 2 ft T0T0 Free Body: Portion BCD M B = 0: 0.6T 0 (10 ft) - T 0 (2 ft) - F CH (5ft) = 0 C 5 ft F CH F BF = 0.2 T T 0 5 ft T AB 0.2 T 0 B D F CH = 0.8 T 0 Problem Solution

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47 A B C D E 3 ft 5 ft 7.5 ft 200 lb F H G F CH = 0.8 T 0 2 ft The tension in each section can be determined from the equations of equilibrium. Free Body: Beam EFH E 7.5 ft 200 lb F H G EyEy ExEx 0.8 T T 0 5 ft F BF = 0.2T T 0 (5 ft) T 0 (10 ft) - (200 lb)(7.5 ft) = 0 M E = 0: T 0 = lb F CH = 0.8(166.67) F CH = lb T F BF = 0.2(166.67) F BF = lb T Problem Solution

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48 A B C D E 3 ft 5 ft 7.5 ft 200 lb F H G 2 ft The maximum tension occurs in the steepest portion of the cable. T0T0 TmTm D y = 0.6 T 0 T 0 = lb T m = T (0.6T 0 ) 2 = T 0 = (166.67) T m = lb D Problem Solution

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49 + E 7.5 ft 200 lb F H G EyEy ExEx 5 ft F CH = lb F BF = lb Beam EFH lb 2.5 ft V MGMG G M G = 0: ( lb)(2.5 ft) - M G = 0 M G = lb-ft lb 2.5 ft V MFMF G 200 lb + M F = 0: ( lb)(5 ft) - (200 lb)(2.5 ft) - M F = 0 M F = lb-ft The moments at points G and F are determined by using free-body diagrams for two sections of the beam. Problem Solution

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50 Problem 5.8 For the beam and loading shown, (a) write the equations of the shear and bending- moment curves, (b) determine the magnitude and location of the maximum bending moment. x y L A B w = w 0 cos x 2L

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51 Solving Problems on Your Own For beams supporting a distributed load expressed as a function w(x), the shear V can be obtained by integrating the function -w(x), and the moment M can be obtained by integrating V (x). For the beam and loading shown, (a) write the equations of the shear and bending- moment curves, (b) determine the magnitude and location of the maximum bending moment. x y L A B w = w 0 cos x 2L Problem 7.159

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52 Problem Solution x y L A B w = w 0 cos x 2L The shear V can be obtained by integrating the function -w(x) and the moment M can be obtained by integrating V (x). x = -w = -w 0 cos dV dx x 2L V = - wdx = - w 0 sin + C 1 ( ) 2L x 2L dM dx = V = - w 0 sin + C 1 ( ) 2L x 2L M = Vdx = w 0 cos + C 1 x + C 2 ( ) 2L 2 x 2L

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53 x y L A B w = w 0 cos x 2L The shear V can be obtained by integrating the function -w(x) and the moment M can be obtained by integrating V (x). x V = - w 0 sin + C 1 ( ) 2L x 2L M = w 0 cos + C 1 x + C 2 ( ) 2L 2 x 2L Boundary conditions At x = 0 : V = C 1 = 0 C 1 = 0 At x = 0 : M = w 0 (2L/ ) 2 cos (0) + C 2 = 0 C 2 = - w 0 (2L/ ) 2 Problem Solution

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54 x y L A B w = w 0 cos x 2L The shear V can be obtained by integrating the function -w(x) and the moment M can be obtained by integrating V (x). x V = - w 0 sin ( ) 2L x 2L M = w 0 cos - w 0 ( ) 2L 2 x 2L ( ) 2L 2 M = w 0 ( -1 + cos ) ( ) 2L 2 x 2L M max at x = L: M max = w 0 [-1 + 0] ( ) 2L Problem Solution M max = w 0 L 2

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55 Problem 5.9 It has been experimentally determined that the bending moment at point K of the frame shown is 300 N-m. Determine (a) the tension in rods AE and FD, (b) the corresponding internal forces at point J. A B D E J k F 100 mm 120 mm 280 mm C

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56 Solving Problems on Your Own 1. Cut the member at a point, and draw the free-body diagram of each of the two portions. 2. Select one of the two free-body diagrams and use it to write the equations of equilibrium. It has been experimentally determined that the bending moment at point K of the frame shown is 300 N-m. Determine (a) the tension in rods AE and FD, (b) the corresponding internal forces at point J. A B D E J k F 100 mm 120 mm 280 mm C Problem 7.161

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57 Problem Solution A B D TxTx J T 100 mm 120 mm Cut the member at a point, and draw the free-body diagram of each of the two portions. D 100 mm 280 mm C TyTy k V F M K = 300 N-m k V F TxTx T TyTy = = 340 A D A D T y = T T x = T 8 17

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58 + A B D J 200 mm 120 mm k V F M K = 300 N-m T y = T T x = T 8 17 Select one of the two free-body diagrams and use it to write the equations of equilibrium. M k = 0: 300 N-m - T(0.2 m) - T(0.12 m) = T = 1500 N Free Body: ABK Problem Solution

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59 A B D J 100 mm 120 mm T y = (1500) = N T x = (1500) = N 8 17 Free Body: ABJ V F MJMJ + M J = 0: M J - ( N)(0.1 m) - ( N)(0.12 m) = 0 M J = 229 N-m F x = 0: N - V = 0 V = 706 N + F y = 0: -F N = 0 F = 1324 N + Problem Solution

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60 Problem 5.10 Cable ACB supports a load uniformly distributed along the horizontal as shown. The lowest point C is located 9 m to the right of A. Determine (a) the vertical B 2.25 m C A a 9 m 6 m distance a, (b) the length of the cable, (c) the components of the reaction at A. 60 kg/m

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61 Solving Problems on Your Own Cable ACB supports a load uniformly distributed along the horizontal as shown. The lowest point C is located 9 m to the right of A. B 2.25 m C A a 9 m 6 m Determine (a) the vertical distance a, (b) the length of the cable, (c) the components of the reaction at A. 1. Identify points of the cable where useful information (position, slope,etc.) exists. Cut the cable at these points and draw a free-body diagram of the two portions of the cable. 2. Use M = 0 if you know the position, or F x = 0 and F y = 0 if you know the slope, to generate needed equations of equilibrium. 3. The length of the cable can be determined from (7.10). 60 kg/m Problem 7.162

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62 + Problem Solution C A a 9 m AyAy T0T0 T0T0 Identify points of the cable where useful information exists. Cut the cable at these points and draw a free-body diagram. Use M = 0 if you know the position, or F x = 0 and F y = 0 if you know the slope to generate needed equations of equilibrium. Free Body : Portion AC F y = 0: A y - 9w =0, A y = 9w 4.5 m 9w9w M A = 0: T o a - (9w)(4.5 m) = 0, T o a = 40.5w (1) +

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63 B C 6 m Identify points of the cable where useful information exists. Cut the cable at these points and draw a free-body diagram. Use M = 0 if you know the position, or F x = 0 and F y = 0 if you know the slope to generate needed equations of equilibrium. Free Body : Portion CB T0T0 T0T0 ByBy yByB 3 m 6w6w + F y = 0: B y - 6w = 0, B y = 6w M B = 0: (6w)(3 m) - T o y B = 0, T o y B = 18w + Problem Solution

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64 B 2.25 m C A y B = a m 6 m Identify points of the cable where useful information exists. Cut the cable at these points and draw a free-body diagram. Use M = 0 if you know the position, or F x = 0 and F y = 0 if you know the slope, to generate needed equations of equilibrium. T0T0 6w6w T0T0 9w9w 15w 7.5 m + M A = 0: 6w (15 m) - 15w(7.5 m) + T o (2.25 m) = 0 T o a = 40.5w (1) Free Body : Entire Cable T o = 10w Using (1) (10w) a = 40.5w a = 4.05 m Problem Solution

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65 B 2.25 m C A m 6 m 6w6w 9w9w 15w 7.5 m (b) Length of AC & CB The length of the cable can be determined from (7.10). 10w Portion ACx A = 9 m, y A = a = 4.05 m; = = 0.45 yAxAyAxA S AC = x A yAxAyAxA 2323 ( ) 4 [ ] yAxAyAxA ( ) S AC = 9 m [ 1 + (2/3)(0.45) 2 - (2/5)(0.45) ]= m Portion CBx B = 6 m, y B = = 1.8 m; = 0.3 yBxByBxB S CB = 6 m [ 1 + (2/3)(0.3) 2 - (2/5)(0.3) ]= m S ABC = S AC + S CB = S ABC = m Problem Solution

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66 B 2.25 m C A m 6 m 6w6w 9w9w 15w 7.5 m (c) Components of Reaction at A 10w A y = 9w = 9(60 kg/m)(9.81 m/s 2 ) = N A x = 10w = 10(60 kg/m)(9.81 m/s 2 ) = 5886 N A y = 5300 N A x = 5890 N Problem Solution

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67 Problem 5.13 For the beam and loading shown, (a) draw the shear and bending- moment diagrams, (b) determine the magnitude and location of the maximum absolute value of the bending moment. 4 ft 6 ft 6000 lb 1500 lb/ft A B C

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68 Solving Problems on Your Own 1. Draw a free-body diagram for the entire beam, and use it to determine the reactions at the beam supports. 2. Draw the shear diagram. 3. Draw the bending-moment diagram by computing the area under each portion of the shear curve. For the beam and loading shown, (a) draw the shear and bending- moment diagrams, (b) determine the magnitude and location of the maximum absolute value of the bending moment. 4 ft 6 ft 6000 lb 1500 lb/ft A B C Problem 7.163

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69 + Problem Solution 4 ft 6 ft 6 kips 9 kips A B C Draw a free-body diagram for the entire beam, and use it to determine the reactions at the beam supports. 3 ft M A = 0 : (6 kips)(10 ft) - (9 kips)(7 ft) + B (4 ft) = 0 B = 0.75 kips F y = 0 : A kips + 6 kips - 9 kips = 0 A = 2.25 kips +

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70 6 kips 1.5 kips/ft A B C 2.25 kips 0.75 kips 4 ft 6 ft Draw the shear diagram. x kip + 3 kip - 6 kip A B C D 2 ft 4 ft V (kips) BD = = VBwVBw 3 kips 1.5 kips/ft Problem Solution

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71 6 kips 1.5 kips/ft A B C 2.25 kips 0.75 kips 4 ft 6 ft Draw the bending-moment diagram. x kip + 3 kip - 6 kip A B C D 2 ft 4 ft V (kips) BD = = VBwVBw 3 kips 1.5 kips/ft x A B C D M (kip-ft) (+9 kip-ft) (3 kips)(2 ft) = (+3 kip-ft) 1212 (- 12 kip-ft) (+9 kip-ft) (+12 kip-ft) M max = 12 kip-ft Problem Solution 6 ft from A

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