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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition CE 102 Statics Chapter 6 Analysis of Structures

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Contents Introduction Definition of a Truss Simple Trusses Analysis of Trusses by the Method of Joints Joints Under Special Loading Conditions Space Trusses Sample Problem 6.1 Analysis of Trusses by the Method of Sections Trusses Made of Several Simple Trusses Sample Problem 6.2 Analysis of Frames Frames Which Cease to be Rigid When Detached From Their Supports Sample Problem 6.3 Machines

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Introduction For the equilibrium of structures made of several connected parts, the internal forces as well the external forces are considered. In the interaction between connected parts, Newton’s 3 rd Law states that the forces of action and reaction between bodies in contact have the same magnitude, same line of action, and opposite sense. Three categories of engineering structures are considered: a)Frames: contain at least one one multi-force member, i.e., member acted upon by 3 or more forces. b)Trusses: formed from two-force members, i.e., straight members with end point connections c)Machines: structures containing moving parts designed to transmit and modify forces.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Definition of a Truss A truss consists of straight members connected at joints. No member is continuous through a joint. Bolted or welded connections are assumed to be pinned together. Forces acting at the member ends reduce to a single force and no couple. Only two- force members are considered. Most structures are made of several trusses joined together to form a space framework. Each truss carries those loads which act in its plane and may be treated as a two-dimensional structure. When forces tend to pull the member apart, it is in tension. When the forces tend to compress the member, it is in compression.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Definition of a Truss Members of a truss are slender and not capable of supporting large lateral loads. Loads must be applied at the joints.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Definition of a Truss

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Simple Trusses A rigid truss will not collapse under the application of a load. A simple truss is constructed by successively adding two members and one connection to the basic triangular truss. In a simple truss, m = 2n - 3 where m is the total number of members and n is the number of joints.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Analysis of Trusses by the Method of Joints Dismember the truss and create a freebody diagram for each member and pin. The two forces exerted on each member are equal, have the same line of action, and opposite sense. Forces exerted by a member on the pins or joints at its ends are directed along the member and equal and opposite. Conditions of equilibrium on the pins provide 2n equations for 2n unknowns. For a simple truss, 2n = m + 3. May solve for m member forces and 3 reaction forces at the supports. Conditions for equilibrium for the entire truss provide 3 additional equations which are not independent of the pin equations.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Joints Under Special Loading Conditions Forces in opposite members intersecting in two straight lines at a joint are equal. The forces in two opposite members are equal when a load is aligned with a third member. The third member force is equal to the load (including zero load). The forces in two members connected at a joint are equal if the members are aligned and zero otherwise. Recognition of joints under special loading conditions simplifies a truss analysis.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Space Trusses An elementary space truss consists of 6 members connected at 4 joints to form a tetrahedron. A simple space truss is formed and can be extended when 3 new members and 1 joint are added at the same time. Equilibrium for the entire truss provides 6 additional equations which are not independent of the joint equations. In a simple space truss, m = 3n - 6 where m is the number of members and n is the number of joints. Conditions of equilibrium for the joints provide 3n equations. For a simple truss, 3n = m + 6 and the equations can be solved for m member forces and 6 support reactions.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 6.1 Using the method of joints, determine the force in each member of the truss. SOLUTION: Based on a free-body diagram of the entire truss, solve the 3 equilibrium equations for the reactions at E and C. Joint A is subjected to only two unknown member forces. Determine these from the joint equilibrium requirements. In succession, determine unknown member forces at joints D, B, and E from joint equilibrium requirements. All member forces and support reactions are known at joint C. However, the joint equilibrium requirements may be applied to check the results.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 6.1 SOLUTION: Based on a free-body diagram of the entire truss, solve the 3 equilibrium equations for the reactions at E and C.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 6.1 Joint A is subjected to only two unknown member forces. Determine these from the joint equilibrium requirements. There are now only two unknown member forces at joint D.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 6.1 There are now only two unknown member forces at joint B. Assume both are in tension. There is one unknown member force at joint E. Assume the member is in tension.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 6.1 All member forces and support reactions are known at joint C. However, the joint equilibrium requirements may be applied to check the results.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Analysis of Trusses by the Method of Sections When the force in only one member or the forces in a very few members are desired, the method of sections works well. To determine the force in member BD, pass a section through the truss as shown and create a free body diagram for the left side. With only three members cut by the section, the equations for static equilibrium may be applied to determine the unknown member forces, including F BD.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Trusses Made of Several Simple Trusses Compound trusses are statically determinant, rigid, and completely constrained. Truss contains a redundant member and is statically indeterminate. Necessary but insufficient condition for a compound truss to be statically determinant, rigid, and completely constrained, non-rigid rigid Additional reaction forces may be necessary for a rigid truss.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 6.2 Determine the force in members FH, GH, and GI. SOLUTION: Take the entire truss as a free body. Apply the conditions for static equilib- rium to solve for the reactions at A and L. Pass a section through members FH, GH, and GI and take the right-hand section as a free body. Apply the conditions for static equilibrium to determine the desired member forces.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 6.2 SOLUTION: Take the entire truss as a free body. Apply the conditions for static equilib- rium to solve for the reactions at A and L.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 6.2 Pass a section through members FH, GH, and GI and take the right-hand section as a free body. Apply the conditions for static equilibrium to determine the desired member forces.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 6.2

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Analysis of Frames Frames and machines are structures with at least one multiforce member. Frames are designed to support loads and are usually stationary. Machines contain moving parts and are designed to transmit and modify forces. A free body diagram of the complete frame is used to determine the external forces acting on the frame. Internal forces are determined by dismembering the frame and creating free-body diagrams for each component. Forces between connected components are equal, have the same line of action, and opposite sense. Forces on two force members have known lines of action but unknown magnitude and sense. Forces on multiforce members have unknown magnitude and line of action. They must be represented with two unknown components.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Frames Which Cease To Be Rigid When Detached From Their Supports Some frames may collapse if removed from their supports. Such frames can not be treated as rigid bodies. A free-body diagram of the complete frame indicates four unknown force components which can not be determined from the three equilibrium conditions. The frame must be considered as two distinct, but related, rigid bodies. With equal and opposite reactions at the contact point between members, the two free-body diagrams indicate 6 unknown force components. Equilibrium requirements for the two rigid bodies yield 6 independent equations.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 6.3 Members ACE and BCD are connected by a pin at C and by the link DE. For the loading shown, determine the force in link DE and the components of the force exerted at C on member BCD. SOLUTION: Create a free-body diagram for the complete frame and solve for the support reactions. Define a free-body diagram for member BCD. The force exerted by the link DE has a known line of action but unknown magnitude. It is determined by summing moments about C. With the force on the link DE known, the sum of forces in the x and y directions may be used to find the force components at C. With member ACE as a free-body, check the solution by summing moments about A.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 6.3 SOLUTION: Create a free-body diagram for the complete frame and solve for the support reactions. Note:

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 6.3 Define a free-body diagram for member BCD. The force exerted by the link DE has a known line of action but unknown magnitude. It is determined by summing moments about C. Sum of forces in the x and y directions may be used to find the force components at C.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Sample Problem 6.3 With member ACE as a free-body, check the solution by summing moments about A. (checks)

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: Statics EighthEdition Machines Machines are structures designed to transmit and modify forces. Their main purpose is to transform input forces into output forces. Given the magnitude of P, determine the magnitude of Q. Create a free-body diagram of the complete machine, including the reaction that the wire exerts. The machine is a nonrigid structure. Use one of the components as a free-body. Taking moments about A,

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29 Problem 6.4 The pin at B is attached to member ABCD and can slide along a slot cut in member BE. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium. A B C D E 15 in 80 lb 40 lb 36 in M

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30 Solving Problems on Your Own The pin at B is attached to member ABCD and can slide along a slot cut in member BE. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium. A B C D E 15 in 80 lb 40 lb 36 in M For this problem we note that there are no two- force members. In solving this problem, we 1. Dismember the frame, and draw a free-body diagram for each member. Problem To simplify the solution, seek a way to write an equation involving a single unknown.

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31 + Dismember the frame, and draw a free-body diagram for each member. B E 15 in 36 in M ExEx EyEy Free-Body: Member BE BE = ( ) 1/2 = 39 in To simplify the solution, seek a way to write an equation involving a single unknown. M E = 0: M - B (39 in) = 0 M = B (39 in) (1) Problem 6.4 Solution

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32 Problem 6.4 Solution Dismember the frame, and draw a free-body diagram for each member. B Free-Body: Member BE A C D 15 in 80 lb 40 lb DxDx DyDy + To simplify the solution, seek a way to write an equation involving a single unknown. M D = 0: (40 lb)(45 in) + (80 lb)(15 in) - B(30 in) = B = 260 lb

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33 Problem 6.4 Solution B A C D 15 in 80 lb 40 lb DxDx DyDy B = 260 lb B E 15 in 36 in M ExEx EyEy M = B (39 in) From EQ (1) M = B (39 in) = (260 lb) (39 in) = 10,140 lb-in M = kip-in

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34 Problem 6.5 For the frame shown and neglecting the effect of friction at the horizontal and vertical surfaces, determine the forces exerted at B and C on member BCE. 12 in A B C D E 6 in 4 in 50 lb 6 in 2 in 6 in

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35 Solving Problems on Your Own For this problem we note that there are no two-force members. In solving this problem, we 1. Dismember the frame, and draw a free-body diagram for each member. For the frame shown and neglecting the effect of friction at the horizontal and vertical surfaces, determine the forces exerted at B and C on member BCE. 12 in A B C D E 6 in 4 in 50 lb 6 in 2 in 6 in Problem To simplify the solution, seek a way to write an equation involving a single unknown.

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36 Problem 6.5 Solution Dismember the frame, and draw a free-body diagram for each member. A C 6 in 50 lb 6 in 2 in 18 in A D D CxCx CyCy H CxCx CyCy BxBx ByBy 6 in

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37 Problem 6.5 Solution A C 6 in 50 lb 6 in 2 in 18 in A D D CxCx CyCy H CxCx CyCy BxBx ByBy 6 in Free-Body: Member ACD To simplify the solution, seek a way to write an equation involving a single unknown. + M H = 0: C x (2 in) - C y (18 in) = 0 C x = 9C y

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38 Problem 6.5 Solution + A C 6 in 50 lb 6 in 2 in 18 in A D D CxCx CyCy H CxCx CyCy BxBx ByBy 6 in To simplify the solution, seek a way to write an equation involving a single unknown. M B = 0: C x (6 in) + C y (6 in) - (50 lb)(12 in) = 0 Free-Body: Member BCE C x = 9C y Substitute C x = 9C y : 9C y (6 in) + C y (6 in) = 0 C y = + 10 lb; C x = 9C y = 9(10) = 90 lb

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39 Problem 6.5 Solution C 6 in 50 lb CxCx CyCy BxBx ByBy 6 in Free-Body: Member BCE C C 90 lb 10 lb C = 90.6 lb 6.3 o F x = 0: B x - 90 lb = 0 B x = 90 lb F y = 0: B y + 10 lb - 50 lb = 0 B y = 40 lb + + B B 90 lb 40 lb B = 98.5 lb 24.0 o

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40 Problem 6.6 Using the method of joints, determine the force in each member of the truss shown. A B C D E F G 2 m 12.5 kN 2.5 m 2 m 12.5 kN

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41 Solving Problems on Your Own 1. Draw the free-body diagram of the entire truss, and use this diagram to determine the reactions at the supports. Using the method of joints, determine the force in each member of the truss shown. A B C D E F G 2 m 12.5 kN 2.5 m 2 m 12.5 kN 2. Locate a joint connecting only two members, and draw the free-body diagram for its pin. Use this free-body diagram to determine the unknown forces in each of the two members. Assuming all members are in tension, if the answer obtained from F x = 0 and F y = 0 is positive, the member is in tension. A negative answer means the member is in compression. Problem 6.6

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42 Solving Problems on Your Own 3. Next, locate a joint where the forces in only two of the connected members are still unknown. Draw the free-body diagram of the pin and use it as indicated in step 2 to determine the two unknown forces. Using the method of joints, determine the force in each member of the truss shown. A B C D E F G 2 m 12.5 kN 2.5 m 2 m 12.5 kN 4. Repeat this procedure until the forces in all the members of the truss have been determined. Problem 6.6

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43 + Problem 6.6 Solution Draw the free-body diagram of the entire truss, and use it to determine reactions at the supports. A B C D E F G 2 m 12.5 kN 2 m 12.5 kN 2.5 m AxAx AyAy E M A = 0: E(2.5 m) - (12.5 kN)(2 m) - (12.5 kN)(4 m) - (12.5 kN)(6 m) = 0 E = 60 kN F y = 0: A y - (4)(12.5 kN) = 0 A y = 50 kN F x = 0: A x - E = 0 A x = 60 kN + +

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44 Problem 6.6 Solution Locate a joint connecting only two members, and draw the free-body diagram for its pin. Use the free- body diagram to determine the unknown forces in each of the two members kN F CD F GD F y = 0: F GD kN = F GD = 32.5 kN C F x = 0: F GD - F CD = 0 F CD = 30 kN T Joint D A B C D F G 2 m 12.5 kN 2 m 12.5 kN 2.5 m E 60 kN 50 kN

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45 Problem 6.6 Solution A B C D F G 2 m 12.5 kN 2 m 12.5 kN 2.5 m E 60 kN 50 kN F CG F FG F = 0: F FG kN = 0 F FG = 32.5 kN C Joint G Next, locate a joint where the forces in only two of the connected members are still unknown. Draw the free-body diagram of the pin and use it to determine the two unknown forces kN F = 0: F CG = 0

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46 Problem 6.6 Solution F BC F CF F y = 0: kN - F CF sin = kN - F CF sin o = 0 F CF = kN C Joint C F CD = 30 kN Repeat this procedure until the forces in all the members of the truss have been determined kN A B C D F G 2 m 12.5 kN 2 m 12.5 kN 2.5 m E 60 kN 50 kN = BCF = tan -1 = o 2323 BF 2 BF = (2.5 m) = m F x = 0: 30 kN - F CF cos - F BC = 0 30 kN - (-19.53) cos o - F BC = 0 F BC = 45.0 kN T + +

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47 Problem 6.6 Solution F EF F y = 0: F BF - F EF - (32.5 kN) - (19.53) sin = 0 Joint F =39.81 o A B C D F G 2 m 12.5 kN 2 m 12.5 kN 2.5 m E 60 kN 50 kN F x = 0: - F EF - (32.5 kN) - F CF cos = F BF F FG = 32.5 kN F CF = 19.53kN F EF = kN - ( ) (19.53) cos o F EF = 48.8 kN C F BF - (-48.8 kN) kN kN = F BF = 6.25 kN T

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48 Problem 6.6 Solution F BE F y = 0: kN kN - F BE sin o = 0 F BE = kN Joint B A B C D F G 2 m 12.5 kN 2 m 12.5 kN 2.5 m E 60 kN 50 kN + + F AB F BC = 45.0 kN F BF = 6.25kN 12.5 kN tan = ; = o 2.5 m 2 m F x = 0: 45.0 kN - F AB + (24.0 kN) cos o = 0 F AB = 60.0 kN F BE = 24.0 kN C F AB = 60.0 kN T

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49 Problem 6.6 Solution F y = 0: F AE - (24 kN) sin o - (48.75 kN) = 0 Joint E A B C D F G 2 m 12.5 kN 2 m 12.5 kN 2.5 m E 60 kN 50 kN + F AE F EF = kN F BE = 24 kN F AE = 37.5 kN F AE = 37.5 kN T kN = o

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50 Problem 6.7 A Pratt roof truss is loaded as shown. Using the method of sections, determine the force in members FH, FI, and GI. D E F G 1.5 kN 3 m A B C H J I K L 3 kN 6.75 m 3 kN 1.5 kN

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51 Solving Problems on Your Own 1. Draw the free-body diagram of the entire truss, and use this diagram to determine the reactions at the supports. 2. Pass a section through three members of the truss, one of which is the desired member. A Pratt roof truss is loaded as shown. Using the method of sections, determine the force in members FH, FI, and GI. D E F G 3 m A B C H J I K L 3 kN 6.75 m 3 kN 1.5 kN Problem 6.7

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52 Solving Problems on Your Own 3. Select one of the two portions of the truss you have obtained, and draw its free-body diagram. This diagram should include the external forces applied to the selected portion as well as the forces exerted on it by the intersected members before these were removed. 4. Now write three equilibrium equations which can be solved for the forces in the three intersected members. A Pratt roof truss is loaded as shown. Using the method of sections, determine the force in members FH, FI, and GI. D E F G 3 m A B C H J I K L 3 kN 6.75 m 3 kN 1.5 kN Problem 6.7

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53 Problem 6.7 Solution Draw the free-body diagram of the entire truss, and use it to determine reactions at the supports. A y = L = (18kN) = 9 kN + D E F G A B C H J I K L 3 kN 1.5 kN 3 m AyAy AxAx L F x = 0: A x = 0 Total load = 5(3 kN) + 2(1.5 kN) = 18 kN By symmetry 1212

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54 Problem 6.7 Solution D E F G A B C H J I K L 3 kN 1.5 kN 3 m Pass a section through three members of the truss, one of which is the desired member. 9 kN

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55 Problem 6.7 Solution H J I K L 3 kN 1.5 kN 3 m 9 kN Select one of the two portions of the truss you have obtained, and draw its free-body diagram. G F F FH F GI F FI a 6.75 m Slope of FHJL Free Body: Portion HIL = tan = = = o FG GI Now write three equilibrium equations which can be solved for the forces in the three intersected members.

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56 Problem 6.7 Solution + H J I K L 3 kN 1.5 kN 3 m 9 kN G F F FH F GI F FI 6.75 m Free Body: Portion HIL Write the three equilibrium equations o Force in FH: M I = 0: F FH ( x 6.75 m) + (9 kN)(6 m) - (1.5 kN)(6m) - (3 kN)(3m) = F FH (4.5 m) + 36 kN-m = F FH = kN F FH = 10.0 kN C

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57 Problem 6.7 Solution + H J I K L 3 kN 1.5 kN 3 m 9 kN G F F FH F GI F FI 6.75 m Free Body: Portion HIL Write the three equilibrium equations o Force in FI: M L = 0: -F FI sin o (6 m) + (3 kN)(6 m) + (3 kN)(3 m) = 0 F FH = 10.0 kN C F FI sin o (6 m) = 27 kN-m F FI = kN F FI = 4.92 kN T

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58 Problem 6.7 Solution + H J I K L 3 kN 1.5 kN 3 m 9 kN G F F FH F GI F FI 6.75 m Free Body: Portion HIL Write the three equilibrium equations o Force in GI: M F = 0: -F GI (6.75 m) - (3 kN)(3 m) - (3 kN)(6 m) - (1.5 kN)(9 m) + (9 kN)(9 m) = 0 F FH = 10.0 kN C F FI = 4.92 kN T F GI (6.75 m) = kN-m F GI = kN F GI = 6.00 kN T

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59 Problem 6.8 A pipe of diameter 50 mm is gripped by the Stillson wrench shown. Portions AB and DE of the wrench are rigidly attached to each other and portion CF is connected by a pin at D. Assuming that no slipping occurs between the pipe and the wrench, determine the components of the forces exerted on the pipe at A and at C. 50 mm 40 mm 20 mm 400 mm 500 N A B C D E F

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60 Solving Problems on Your Own 1. Dismember the machine, and draw a free-body diagram of each member. 2. First consider the two-force members. Apply equal and opposite forces to each two-force member where it is connected to another member. A pipe of diameter 50 mm is gripped by the Stillson wrench shown. Portions AB and DE of the wrench are rigidly attached to each other and portion CF is connected by a pin at D. Assuming that no slipping occurs between the pipe and the wrench, determine the components of the forces exerted on the pipe at A and at C. 50 mm 40 mm 20 mm 400 mm 500 N A B C D E F Problem 6.8

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61 Solving Problems on Your Own 3. Next consider the multi force members. 4. Equilibrium equations can be written after completing each free-body. A pipe of diameter 50 mm is gripped by the Stillson wrench shown. Portions AB and DE of the wrench are rigidly attached to each other and portion CF is connected by a pin at D. Assuming that no slipping occurs between the pipe and the wrench, determine the components of the forces exerted on the pipe at A and at C. 50 mm 40 mm 20 mm 400 mm 500 N A B C D E F Problem 6.8

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62 Problem 6.8 Solution 90 mm 20 mm Dismember the machine, and draw a free-body diagram of each member. First consider the two-force members. AxAx AyAy DyDy DxDx Free Body : Portion ABDE A B D E A y 20 mm = ; A x 90 mm D y = A y : D x = A x = 4.5 D y (1) A x = 4.5 A y 90 mm 20 mm A D

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63 Problem 6.8 Solution + 20 mm 400 mm 500 N F Next consider the multi force members. Equilibrium equations can be written after completing each free-body. DyDy DxDx CyCy CxCx Free Body : Portion CF D y = A y : D x = A x = 4.5 D y (1) M C = 0: D x (20 mm) - D y (40 mm) - (500 N)(440 mm) = 0 Substitute from (1) 4.5D y (20) - D y (40) x10 3 = 0 D y = 4400 N = 4.4 kN D x = 4.5 D y = 19.8 kN F x = 0: C x kN = 0 C x = 19.8 kN F y = 0: C y kN -0.5 kN = 0 C y = 4.9 kN + + Using (1) A x = D x = 19.8 kN A y = D y = 4.4 kN 40 mm

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64 Problem 6.8 Solution 20 mm 400 mm 500 N F A B D E 4.4 kN 19.8 kN 4.4 kN 19.8 kN 4.9 kN All components act in the directions shown. Components on the pipe are equal and opposite to those on the wrench. 4.4 kN 19.8 kN 4.9 kN A x = 19.8 kN A y = 4.4 kN C x = 19.8 kN C y = 4.9 kN 40 mm

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