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**Forces in Beams and Cables**

CE 102 Statics Chapter 5 Forces in Beams and Cables

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**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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**Introduction Preceding chapters dealt with:**

determining external forces acting on a structure and 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: Beams - usually long, straight, prismatic members designed to support loads applied at various points along the member. Cables - flexible members capable of withstanding only tension, designed to support concentrated or distributed loads.

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**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. Internal forces equivalent to a force-couple system are necessary for equil-ibrium of free-bodies JD and ABCJ. Multiforce member ABCD is in equil-ibrium under application of cable and member contact forces. An internal force-couple system is required for equilibrium of two-force members which are not straight.

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**Sample Problem 5.1 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. Determine the internal forces (a) in member ACF at point J and (b) in member BCD at K.

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**Sample Problem 5.1 SOLUTION: Compute reactions and connection forces.**

Consider entire frame as a free-body:

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**Sample Problem 5.1 Consider member BCD as free-body:**

Consider member ABE as free-body: From member BCD,

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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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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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**Various Types of Beam Loading and Support**

Beam - structural member designed to support loads applied at various points along its length. Beam can be subjected to concentrated loads or distributed loads or combination of both. Beam design is two-step process: determine shearing forces and bending moments produced by applied loads select cross-section best suited to resist shearing forces and bending moments

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**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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**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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**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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**Sample Problem 5.2 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. Draw the shear and bending moment diagrams for the beam and loading shown. Plot results.

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**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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**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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**Sample Problem 5.3 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. 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. Plot results.

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**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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Sample Problem 5.3 Evaluate equivalent internal force-couple systems at sections cut within segments AC, CD, and DB. From A to C: From C to D:

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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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**Sample Problem 5.3 Plot results. From A to C: From C to D:**

From D to B:

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**Relations Among Load, Shear, and Bending Moment**

Relations between load and shear: Relations between shear and bending moment:

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**Relations Among Load, Shear, and Bending Moment**

Reactions at supports, Shear curve, Moment curve,

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**Sample Problem 5.4 SOLUTION:**

Taking entire beam as a free-body, determine reactions at supports. Between concentrated load application points, and shear is constant. With uniform loading between D and E, the shear variation is linear. Draw the shear and bending-moment diagrams for the beam and loading shown. 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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**Sample Problem 5.4 SOLUTION:**

Taking entire beam as a free-body, determine reactions at supports. Between concentrated load application points, and shear is constant. With uniform loading between D and E, the shear variation is linear.

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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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**Sample Problem 5.5 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. Sketch the shear and bending-moment diagrams for the cantilever beam and loading shown. 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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**Sample Problem 5.5 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. With zero load, change in shear between B and C is zero.

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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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**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: concentrated vertical loads on given vertical lines, weight of cable is negligible, cable is flexible, i.e., resistance to bending is small, 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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**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. 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 For other points on cable,

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**Cables With Distributed Loads**

For cable carrying a distributed load: cable hangs in shape of a curve 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 T0 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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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 by internal tension force magnitude and direction are Summing moments about D, or The cable forms a parabolic curve.

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**Sample Problem 5.6 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. 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. Evaluate maximum slope and maximum tension which occur in DE.

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**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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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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Sample Problem 5.6 Evaluate maximum slope and maximum tension which occur in DE.

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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 by the internal tension force magnitude is To relate horizontal distance x to cable length s,

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**Catenary To relate x and y cable coordinates,**

which is the equation of a catenary.

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Problem 5.7 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. 5 ft 5 ft 5 ft A D 2 ft B 3 ft C F H G E 7.5 ft 200 lb

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**Solving Problems on Your Own**

5 ft 5 ft 5 ft Solving Problems on Your Own A D 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) 2 ft B 3 ft C F H G E 7.5 ft 200 lb the maximum tension in the cable, (c) the bending moment at F and G. 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 S M = 0 if you know the position, or S Fx = 0 and S Fy = 0 if you know the slope, to generate needed equations of equilibrium.

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**Solving Problems on Your Own**

5 ft 5 ft 5 ft Solving Problems on Your Own A D 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) 2 ft B 3 ft C F H G E 7.5 ft 200 lb the maximum tension in the cable, (c) the bending moment at F and G. 3. The tension in each section can be determined from the equations of equilibrium. 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.

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**S MB = 0: T0(2 ft) - Ay(5 ft) = 0 Ay = 0.4 T0**

Problem Solution A D Identify points of the cable where useful information exists. Draw a free-body diagram of one of the two portions of the cable. Use S M = 0 to generate the equations of equilibrium. 2 ft B 3 ft C F H G E 7.5 ft 200 lb Free Body: Portion AB Ay 5 ft + S MB = 0: T0(2 ft) - Ay(5 ft) = 0 A Ay = 0.4 T0 2 ft T0 B TBC FBF

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**S MC = 0: T0(3 ft) - 0.4 T0 (10 ft) + FBF (5 ft) = 0 FBF = 0.2 T0**

Problem Solution A D Identify points of the cable where useful information exists. Draw a free-body diagram of one of the two portions of the cable. Use S M = 0 to generate the equations of equilibrium. 2 ft B 3 ft C F H G E 7.5 ft 200 lb Free Body: Portion ABC 0.4T0 + S MC = 0: T0(3 ft) T0 (10 ft) + FBF (5 ft) = 0 5 ft 5 ft A TCD T0 3 ft FBF = 0.2 T0 B C FBF FCH

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**S MC = 0: Dy(5 ft) - T0 (3 ft) = 0 Dy= 0.6 T0**

Problem Solution A D Identify points of the cable where useful information exists. Draw a free-body diagram of one of the two portions of the cable. Use S M = 0 to generate the equations of equilibrium. 2 ft B 3 ft C F H G E 7.5 ft 200 lb FBF = 0.2 T0 Free Body: Portion CD Dy + S MC = 0: Dy(5 ft) - T0 (3 ft) = 0 5 ft Dy= 0.6 T0 TBC D T0 3 ft C FCH

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**S MB = 0: 0.6T0 (10 ft) - T0 (2 ft) - FCH (5ft) = 0 FCH = 0.8 T0**

Problem Solution A D Identify points of the cable where useful information exists. Draw a free-body diagram of one of the two portions of the cable. Use S M = 0 to generate the equations of equilibrium. 2 ft B 3 ft C F H G E 7.5 ft 200 lb FBF = 0.2 T0 Free Body: Portion BCD 0.6 T0 + S MB = 0: 0.6T0 (10 ft) - T0 (2 ft) - FCH (5ft) = 0 5 ft 5 ft TAB 2 ft D T0 B FCH = 0.8 T0 C 0.2 T0 FCH

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**S ME = 0: 0.2T0 (5 ft) + 0.8 T0 (10 ft) - (200 lb)(7.5 ft) = 0**

Problem Solution A D The tension in each section can be determined from the equations of equilibrium. 2 ft B 3 ft C F H FCH = 0.8 T0 FBF = 0.2T0 G E 7.5 ft 200 lb Free Body: Beam EFH + S ME = 0: 0.2 T0 0.8 T0 Ey 0.2T0 (5 ft) T0 (10 ft) - (200 lb)(7.5 ft) = 0 5 ft 5 ft H F Ex G E T0 = lb 7.5 ft 200 lb FCH = 0.8(166.67) FCH = lb T FBF = 0.2(166.67) FBF = lb T

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5 ft 5 ft 5 ft Problem Solution A D The maximum tension occurs in the steepest portion of the cable. 2 ft B 3 ft C F H T0 = lb G E 7.5 ft 200 lb Tm Dy= 0.6 T0 Tm = T02 + (0.6T0)2 = T0 = (166.67) D T0 Tm = lb

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**S MG = 0: (133.33 lb)(2.5 ft) - MG = 0 MG = + 333 lb-ft**

Beam EFH Problem Solution FCH = lb FBF = lb The moments at points G and F are determined by using free-body diagrams for two sections of the beam. Ey 5 ft 5 ft F H Ex G E 7.5 ft 200 lb lb V + MG S MG = 0: ( lb)(2.5 ft) - MG = 0 MG = lb-ft G 2.5 ft + S MF = 0: ( lb)(5 ft) - (200 lb)(2.5 ft) - MF = 0 MF = lb-ft lb V G MF 2.5 ft 2.5 ft 200 lb

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Problem 5.8 w = w0 cos px 2L y 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. A x B L

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**Solving Problems on Your Own**

w = w0 cos px 2L y 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. A B x L 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).

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**( ) ( ) ( ) ò ò dV dx = -w = -w0 cos V = - wdx = - w0 sin + C1 dM**

y w = w0 cos px 2L Problem Solution The shear V can be obtained by integrating the function -w(x) and the moment M can be obtained by integrating V (x). A x B x L dV dx px 2L = -w = -w0 cos ( ) 2L p px 2L V = - wdx = - w sin C1 ò dM dx ( ) 2L p px 2L = V = - w sin C1 ( ) 2L p 2 px 2L M = Vdx = w cos C1x + C2 ò

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**( ) ( ) V = - w0 sin + C1 2 M = w0 cos + C1x + C2 w = w0 cos**

y w = w0 cos px 2L Problem Solution The shear V can be obtained by integrating the function -w(x) and the moment M can be obtained by integrating V (x). A x B x L V = - w sin C1 ( ) 2L p px M = w cos C1x + C2 ( ) 2L p 2 px Boundary conditions At x = 0 : V = C1 = C1 = 0 At x = 0 : M = w0 (2L/p)2 cos (0) + C2 = 0 C2 = -w0 (2L/p)2

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**( ) ( ) ( ) ( ) ( ) V = - w0 sin 2 2 M = w0 cos - w0 2**

y w = w0 cos px 2L Problem Solution The shear V can be obtained by integrating the function -w(x) and the moment M can be obtained by integrating V (x). A B x x L V = - w sin ( ) 2L p px ( ) 2L p 2 px 2L ( ) 2L p 2 M = w cos w0 ( ) 2L p 2 px 2L M = w ( -1 + cos ) ( ) 2L p 2 Mmax = w [-1 + 0] Mmax at x = L: 4 p2 Mmax = w0 L2

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Problem 5.9 120 mm 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. B A 100 mm E J F 100 mm k 100 mm D C 280 mm

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**Solving Problems on Your Own**

120 mm Problem 7.161 B A Solving Problems on Your Own 100 mm E 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. J F 100 mm k 100 mm D C 280 mm 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.

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120 mm Problem Solution B A Cut the member at a point, and draw the free-body diagram of each of the two portions. Tx 100 mm J T 100 mm Ty k A V MK = 300 N-m F D = 340 300 D = 160 A F 8 MK = 300 N-m T 15 17 V Ty k D 100 mm Tx 15 17 D Ty = T C 8 17 280 mm Tx = T

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**SMk = 0: 300 N-m - T(0.2 m) - T(0.12 m) = 0 T = 1500 N**

120 mm Problem Solution 8 17 Tx = T B A Select one of the two free-body diagrams and use it to write the equations of equilibrium. 200 mm J 15 17 Ty = T k V MK = 300 N-m Free Body: ABK F D + 8 17 15 17 SMk = 0: N-m T(0.2 m) T(0.12 m) = 0 T = 1500 N

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**SMJ = 0: MJ - (705.88 N)(0.1 m) - (1323.53 N)(0.12 m) = 0**

120 mm Problem Solution B A 8 17 Tx = (1500) = N 100 mm J 15 17 Ty = (1500) = N V MJ F Free Body: ABJ D + SMJ = 0: MJ - ( N)(0.1 m) - ( N)(0.12 m) = 0 MJ = 229 N-m + SFx = 0: N - V = V = 706 N + SFy = 0: -F N = F = 1324 N

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Problem 5.10 9 m 6 m 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 A 2.25 m a B C 60 kg/m distance a, (b) the length of the cable, (c) the components of the reaction at A.

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**Solving Problems on Your Own**

A 2.25 m 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. a B C 60 kg/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 S M = 0 if you know the position, or S Fx = 0 and S Fy = 0 if you know the slope, to generate needed equations of equilibrium. 3. The length of the cable can be determined from (7.10).

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**S MA = 0: To a - (9w)(4.5 m) = 0, To a = 40.5w (1)**

Free Body : Portion AC Problem Solution Identify points of the cable where useful information exists. Cut the cable at these points and draw a free-body diagram. Use S M = 0 if you know the position, or S Fx = 0 and S Fy = 0 if you know the slope to generate needed equations of equilibrium. 9 m Ay T0 A a T0 C 9w 4.5 m S Fy = 0: Ay - 9w =0, Ay = 9w + + S MA = 0: To a - (9w)(4.5 m) = 0, To a = 40.5w (1)

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**S MB = 0: (6w)(3 m) - To yB = 0, To yB = 18w**

Free Body : Portion CB Problem Solution By Identify points of the cable where useful information exists. Cut the cable at these points and draw a free-body diagram. Use S M = 0 if you know the position, or S Fx = 0 and S Fy = 0 if you know the slope to generate needed equations of equilibrium. 6 m yB B T0 T0 C 6w 3 m S Fy = 0: By - 6w = 0, By = 6w + + S MB = 0: (6w)(3 m) - To yB = 0, To yB = 18w

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**S MA = 0: 6w (15 m) - 15w(7.5 m) + To (2.25 m) = 0 To = 10w**

Free Body : Entire Cable Problem Solution 9 m 6 m 9w 6w A To a = 40.5w (1) T0 2.25 m B C T0 yB = a Identify points of the cable where useful information exists. Cut the cable at these points and draw a free-body diagram. Use S M = 0 if you know the position, or S Fx = 0 and S Fy = 0 if you know the slope, to generate needed equations of equilibrium. 7.5 m 15w + S MA = 0: 6w (15 m) - 15w(7.5 m) + To (2.25 m) = 0 To = 10w Using (1) (10w) a = 40.5w a = 4.05 m

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9 m 6 m Problem Solution 9w 10w 6w (b) Length of AC & CB A 2.25 m The length of the cable can be determined from (7.10). 4.05 B C 10w 7.5 m 15w = = 0.45 yA xA 4.05 9 Portion AC xA = 9 m, yA = a = 4.05 m; 2 3 yA xA 2 [ ] ( ) yA xA 4 2 5 ( ) SAC = xA SAC = 9 m [ 1 + (2/3)(0.45)2 - (2/5)(0.45) ]= m yB xB Portion CB xB = 6 m, yB = = 1.8 m; = 0.3 SCB = 6 m [ 1 + (2/3)(0.3)2 - (2/5)(0.3) ]= m SABC = SAC + SCB = SABC = m

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**(c) Components of Reaction at A**

Problem Solution 9 m 6 m 9w 10w 6w A 2.25 m 4.05 B C 10w 7.5 m 15w Ay = 9w = 9(60 kg/m)(9.81 m/s2) = N Ax = 10w = 10(60 kg/m)(9.81 m/s2) = 5886 N Ay = 5300 N Ax = 5890 N

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Problem 5.13 1500 lb/ft 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. A B C 6000 lb 4 ft 6 ft

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**Solving Problems on Your Own**

1500 lb/ft Solving Problems on Your Own A B 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. C 6000 lb 4 ft 6 ft 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.

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**S MA = 0 : (6 kips)(10 ft) - (9 kips)(7 ft) + B (4 ft) = 0 **

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

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**Draw the shear diagram. V (kips) x VB 3 kips BD = = w 1.5 kips/ft**

Problem Solution 1.5 kips/ft A B C 2.25 kips 0.75 kips 6 kips 4 ft 6 ft V (kips) + 3 kip D kip x A B C VB w 3 kips 1.5 kips/ft BD = = - 6 kip 2 ft 4 ft

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**Draw the bending-moment diagram.**

Problem Solution 1.5 kips/ft A B C 2.25 kips 0.75 kips 6 kips 4 ft 6 ft V (kips) + 3 kip 1 2 (3 kips)(2 ft) = (+3 kip-ft) kip (+9 kip-ft) D x A B C (- 12 kip-ft) VB w 3 kips 1.5 kips/ft BD = = - 6 kip 2 ft 4 ft (+12 kip-ft) Mmax = 12 kip-ft (+9 kip-ft) M (kip-ft) 6 ft from A x A B D C

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