Mechanics of Materials Philpot 3rd Edition Solutions Manual

P1.1 A stainless steel tube with an outside diameter of 60 mm and a wall thickness of 5 mm is used as a compression member. If the axial normal stress in the member must be limited to 200 MPa, determine the maximum load P that the member can support.

Solution The cross-sectional area of the stainless steel tube is   A  ( D 2  d 2 )  [(60 mm) 2  (50 mm) 2 ]  863.938 mm 2 4 4 The normal stress in the tube can be expressed as P  A The maximum normal stress in the tube must be limited to 200 MPa. Using 200 MPa as the allowable normal stress, rearrange this expression to solve for the maximum load P Ans. Pmax   allow A  (200 N/mm2 )(863.938 mm2 )  172,788 N  172.8 kN

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

P1.2 A 2024-T4 aluminum tube with an outside diameter of 2.50 in. will be used to support a 27-kip load. If the axial normal stress in the member must be limited to 18 ksi, determine the wall thickness required for the tube.

Solution From the definition of normal stress, solve for the minimum area required to support a 27-kip load without exceeding a stress of 18 ksi P P 27 kips   Amin    1.500 in.2 A  18 ksi The cross-sectional area of the aluminum tube is given by  A  (D2  d 2 ) 4 Set this expression equal to the minimum area and solve for the maximum inside diameter d



4

[(2.50 in.) 2  d 2 ]  1.500 in.2

(2.50 in.) 2  d 2  (2.50 in.) 2 

4



4



(1.500 in.2 )

(1.500 in.2 )  d 2

 d max  2.08330 in. The outside diameter D, the inside diameter d, and the wall thickness t are related by D  d  2t Therefore, the minimum wall thickness required for the aluminum tube is D  d 2.50 in.  2.08330 in. tmin    0.20835 in.  0.208 in. 2 2

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P1.3 Two solid cylindrical rods (1) and (2) are joined together at flange B and loaded, as shown in Figure P1.3/4. If the normal stress in each rod must be limited to 40 ksi, determine the minimum diameter required for each rod.

FIGURE P1.3/4

Solution Cut a FBD through rod (1). The FBD should include the free end of the rod at A. As a matter of course, we will assume that the internal force in rod (1) is tension (even though it obviously will be in compression). From equilibrium, Fy   F1  15 kips  0

 F1  15 kips  15 kips (C) Next, cut a FBD through rod (2) that includes the free end of the rod at A. Again, we will assume that the internal force in rod (2) is tension. Equilibrium of this FBD reveals the internal force in rod (2): Fy   F2  30 kips  30 kips  15 kips  0

 F2  75 kips  75 kips (C) Notice that rods (1) and (2) are in compression. In this situation, we are concerned only with the stress magnitude; therefore, we will use the force magnitudes to determine the minimum required cross-sectional areas. If the normal stress in rod (1) must be limited to 40 ksi, then the minimum cross-sectional area that can be used for rod (1) is F 15 kips A1,min  1   0.375 in.2  40 ksi The minimum rod diameter is therefore  Ans. A1,min  d12  0.375 in.2  d1  0.69099 in.  0.691 in. 4 Similarly, the normal stress in rod (2) must be limited to 40 ksi, which requires a minimum area of F 75 kips A2,min  2   1.875 in.2  40 ksi The minimum diameter for rod (2) is therefore  Ans. A2,min  d 22  1.875 in.2  d 2  1.545097 in.  1.545 in. 4


P1.4 Two solid cylindrical rods (1) and (2) are joined together at flange B and loaded, as shown in Figure P1.3/4. The diameter of rod (1) is 1.75 in. and the diameter of rod (2) is 2.50 in. Determine the normal stresses in rods (1) and (2).

FIGURE P1.3/4

Solution Cut a FBD through rod (1). The FBD should include the free end of the rod at A. We will assume that the internal force in rod (1) is tension (even though it obviously will be in compression). From equilibrium, Fy   F1  15 kips  0

 F1  15 kips  15 kips (C) Next, cut a FBD through rod (2) that includes the free end of the rod at A. Again, we will assume that the internal force in rod (2) is tension. Equilibrium of this FBD reveals the internal force in rod (2):

Fy   F2  30 kips  30 kips  15 kips  0  F2  75 kips  75 kips (C) From the given diameter of rod (1), the cross-sectional area of rod (1) is  A1  (1.75 in.) 2  2.4053 in.2 4 and thus, the normal stress in rod (1) is F 15 kips 1  1   6.23627 ksi  6.24 ksi (C) Ans. A1 2.4053 in.2 From the given diameter of rod (2), the cross-sectional area of rod (2) is  A2  (2.50 in.) 2  4.9087 in.2 4 Accordingly, the normal stress in rod (2) is F 75 kips 2  2   15.2789 ksi  15.28 ksi (C) A2 2.4053 in.2

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P1.5 Axial loads are applied with rigid bearing plates to the solid cylindrical rods shown in Figure P1.5/6. The diameter of aluminum rod (1) is 2.00 in., the diameter of brass rod (2) is 1.50 in., and the diameter of steel rod (3) is 3.00 in. Determine the axial normal stress in each of the three rods.

FIGURE P1.5/6

Solution Cut a FBD through rod (1). The FBD should include the free end A. We will assume that the internal force in rod (1) is tension (even though it obviously will be in compression). From equilibrium, Fy   F1  8 kips  4 kips  4 kips  0  F1  16 kips  16 kips (C)

FBD through rod (1) FBD through rod (2)

FBD through rod (3) Next, cut a FBD through rod (2) that includes the free end A. Again, we will assume that the internal force in rod (2) is tension. Equilibrium of this FBD reveals the internal force in rod (2): Fy   F2  8 kips  4 kips  4 kips  15 kips  15 kips  0  F2  14 kips  14 kips (T) Similarly, cut a FBD through rod (3) that includes the free end A. From this FBD, the internal force in rod (3) is: Fy   F3  8 kips  4 kips  4 kips  15 kips  15 kips  20 kips  20 kips  0

 F3  26 kips  26 kips (C) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

From the given diameter of rod (1), the cross-sectional area of rod (1) is  A1  (2.00 in.) 2  3.1416 in.2 4 and thus, the normal stress in aluminum rod (1) is F 16 kips 1  1   5.0930 ksi  5.09 ksi (C) A1 3.1416 in.2 From the given diameter of rod (2), the cross-sectional area of rod (2) is  A2  (1.50 in.) 2  1.7671 in.2 4 Accordingly, the normal stress in brass rod (2) is F 14 kips 2  2   7.9224 ksi  7.92 ksi (T) A2 1.7671 in.2 Finally, the cross-sectional area of rod (3) is  A3  (3.00 in.) 2  7.0686 in.2 4 and the normal stress in the steel rod is F 26 kips 3  3   3.6782 ksi  3.68 ksi (C) A3 7.0686 in.2

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P1.6 Axial loads are applied with rigid bearing plates to the solid cylindrical rods shown in Figure P1.5/6. The normal stress in aluminum rod (1) must be limited to 18 ksi, the normal stress in brass rod (2) must be limited to 25 ksi, and the normal stress in steel rod (3) must be limited to 15 ksi. Determine the minimum diameter required for each of the three rods.

FIGURE P1.5/6

Solution The internal forces in the three rods must be determined. Begin with a FBD cut through rod (1) that includes the free end A. We will assume that the internal force in rod (1) is tension (even though it obviously will be in compression). From equilibrium, Fy   F1  8 kips  4 kips  4 kips  0  F1  16 kips  16 kips (C)

FBD through rod (1) FBD through rod (2)

FBD through rod (3) Next, cut a FBD through rod (2) that includes the free end A. Again, we will assume that the internal force in rod (2) is tension. Equilibrium of this FBD reveals the internal force in rod (2): Fy   F2  8 kips  4 kips  4 kips  15 kips  15 kips  0  F2  14 kips  14 kips (T) Similarly, cut a FBD through rod (3) that includes the free end A. From this FBD, the internal force in rod (3) is: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Fy   F3  8 kips  4 kips  4 kips  15 kips  15 kips  20 kips  20 kips  0  F3  26 kips  26 kips (C) Notice that two of the three rods are in compression. In these situations, we are concerned only with the stress magnitude; therefore, we will use the force magnitudes to determine the minimum required crosssectional areas, and in turn, the minimum rod diameters. The normal stress in aluminum rod (1) must be limited to 18 ksi; therefore, the minimum cross-sectional area required for rod (1) is F 16 kips A1,min  1   0.8889 in.2  1 18 ksi The minimum rod diameter is therefore  Ans. A1,min  d12  0.8889 in.2  d1  1.0638 in.  1.064 in. 4 The normal stress in brass rod (2) must be limited to 25 ksi, which requires a minimum area of F 14 kips A2,min  2   0.5600 in.2  2 25 ksi which requires a minimum diameter for rod (2) of  A2,min  d 22  0.5600 in.2  d 2  0.8444 in.  0.844 in. 4

Ans.

The normal stress in steel rod (3) must be limited to 15 ksi. The minimum cross-sectional area required for this rod is: F 26 kips A3,min  3   1.7333 in.2  3 15 ksi which requires a minimum diameter for rod (3) of  Ans. A3,min  d32  1.7333 in.2  d3  1.4856 in.  1.486 in. 4


P1.7 Two solid cylindrical rods support a load of P = 50 kN, as shown in Figure P1.7/8. If the normal stress in each rod must be limited to 130 MPa, determine the minimum diameter required for each rod.

FIGURE P1.7/8

Solution Consider a FBD of joint B. Determine the angle  between rod (1) and the horizontal axis: 4.0 m tan    1.600    57.9946 2.5 m and the angle  between rod (2) and the horizontal axis: 2.3 m tan    0.7188    35.7067 3.2 m Write equilibrium equations for the sum of forces in the horizontal and vertical directions. Note: Rods (1) and (2) are two-force members. Fx  F2 cos(35.7067)  F1 cos(57.9946)  0 Fy  F2 sin(35.7067)  F1 sin(57.9946)  P  0

(a) (b)

Unknown forces F1 and F2 can be found from the simultaneous solution of Eqs. (a) and (b). Using the substitution method, Eq. (b) can be solved for F2 in terms of F1: cos(57.9946) F2  F1 (c) cos(35.7067) Substituting Eq. (c) into Eq. (b) gives cos(57.9946) F1 sin(35.7067)  F1 sin(57.9946)  P cos(35.6553)

F1  cos(57.9946) tan(35.7067)  sin(57.9946)   P  F1 

P P  cos(57.9946) tan(35.7067)  sin(57.9946) 1.2289

For the given load of P = 50 kN, the internal force in rod (1) is therefore: 50 kN F1   40.6856 kN 1.2289


Backsubstituting this result into Eq. (c) gives force F2: cos(57.9946) cos(57.9946) F2  F1  (40.6856 kN)  26.5553 kN cos(35.7067) cos(35.7067) The normal stress in rod (1) must be limited to 130 MPa; therefore, the minimum cross-sectional area required for rod (1) is F (40.6856 kN)(1,000 N/kN) A1,min  1   312.9664 mm 2 2 1 130 N/mm The minimum rod diameter is therefore  Ans. A1,min  d12  312.9664 mm 2  d1  19.9620 mm  19.96 mm 4 The minimum area required for rod (2) is F (26.5553 kN)(1,000 N/kN) A2,min  2   204.2718 mm 2 2 2 130 N/mm which requires a minimum diameter for rod (2) of  A2,min  d 22  204.2718 mm2  d 2  16.1272 mm  16.13 mm 4

Ans.


P1.8 Two solid cylindrical rods support a load of P = 27 kN, as shown in Figure P1.7/8. Rod (1) has a diameter of 16 mm and the diameter of rod (2) is 12 mm. Determine the axial normal stress in each rod.

FIGURE P1.7/8

Solution Consider a FBD of joint B. Determine the angle  between rod (1) and the horizontal axis: 4.0 m tan    1.600    57.9946 2.5 m and the angle  between rod (2) and the horizontal axis: 2.3 m tan    0.7188    35.7067 3.2 m Write equilibrium equations for the sum of forces in the horizontal and vertical directions. Note: Rods (1) and (2) are two-force members. Fx  F2 cos(35.7067)  F1 cos(57.9946)  0 Fy  F2 sin(35.7067)  F1 sin(57.9946)  P  0

(a) (b)

Unknown forces F1 and F2 can be found from the simultaneous solution of Eqs. (a) and (b). Using the substitution method, Eq. (b) can be solved for F2 in terms of F1: cos(57.9946) F2  F1 (c) cos(35.7067) Substituting Eq. (c) into Eq. (b) gives cos(57.9946) F1 sin(35.7067)  F1 sin(57.9946)  P cos(35.6553)

F1  cos(57.9946) tan(35.7067)  sin(57.9946)   P  F1 

P P  cos(57.9946) tan(35.7067)  sin(57.9946) 1.2289

For the given load of P = 27 kN, the internal force in rod (1) is therefore: 27 kN F1   21.9702 kN 1.2289


Backsubstituting this result into Eq. (c) gives force F2: cos(57.9946) cos(57.9946) F2  F1  (21.9702 kN)  14.3399 kN cos(35.7067) cos(35.7067) The diameter of rod (1) is 16 mm; therefore, its cross-sectional area is:  A1  (16 mm) 2  201.0619 mm 2 4 and the normal stress in rod (1) is: F (21.9702 kN)(1,000 N/kN) 1  1   109.2710 N/mm 2  109.3 MPa (T) 2 A1 201.0619 mm The diameter of rod (2) is 12 mm; therefore, its cross-sectional area is:  A2  (12 mm)2  113.0973 mm2 4 and the normal stress in rod (2) is: F (14.3399 kN)(1,000 N/kN) 2  2   126.7924 N/mm 2  126.8 MPa (T) 2 A2 113.0973 mm

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P1.9 A simple pin-connected truss is loaded and supported as shown in Figure P1.9. All members of the truss are aluminum pipes that have an outside diameter of 4.00 in. and a wall thickness of 0.226 in. Determine the normal stress in each truss member.

FIGURE P1.9

Solution Overall equilibrium: Begin the solution by determining the external reaction forces acting on the truss at supports A and B. Write equilibrium equations that include all external forces. Note that only the external forces (i.e., loads and reaction forces) are considered at this time. The internal forces acting in the truss members will be considered after the external reactions have been computed. The free-body diagram (FBD) of the entire truss is shown. The following equilibrium equations can be written for this structure: Fx  Ax  2 kips  0



 Ax  2 kips M A  By (6 ft)  (5 kips)(14 ft)  (2 kips)(7 ft)  0  By  14 kips Fy  Ay  By  5 kips  0  Ay  9 kips Method of joints: Before beginning the process of determining the internal forces in the axial members, the geometry of the truss will be used to determine the magnitude of the inclination angles of members AC and BC. Use the definition of the tangent function to determine AC and BC: 7 ft tan  AC   0.50   AC  26.565 14 ft 7 ft tan  BC   0.875   BC  41.186 8 ft


Joint A: Begin the solution process by considering a FBD of joint A. Consider only those forces acting directly on joint A. In this instance, two axial members, AB and AC, are connected at joint A. Additionally, two reaction forces, Ax and Ay, act at joint A. Tension forces will be assumed in each truss member. Fx  FAC cos(26.565)  FAB  Ax  0 (a) Fy  FAC sin(26.565)  Ay  0 (b) Solve Eq. (b) for FAC: Ay 9 kips FAC     20.125 kips sin(26.565) sin(26.565) and then compute FAB using Eq. (a): FAB   FAC cos(26.565)  Ax   (20.125 kips) cos(26.565)  ( 2 kips)  16.000 kips

Joint B: Next, consider a FBD of joint B. In this instance, the equilibrium equations associated with joint B seem easier to solve than those that would pertain to joint C. As before, tension forces will be assumed in each truss member. Fx   FAB  FBC cos(41.186)  0 (c) Fy  FBC sin(41.186)  By  0 (d) Solve Eq. (d) for FBC: By 14 kips FBC     21.260 kips sin(41.186) sin(41.186) Eq. (c) can be used as a check on our calculations: Fx   FAB  FBC cos(41.186)   ( 16.000 kips)  ( 21.260 kips) cos(41.186)  0

Section properties: For each of the three truss members: d  4.00 in.  2(0.226 in.)  3.548 in.

A

Checks!



(4.00 in.) 2  (3.548 in.) 2   2.67954 in.2 4

Normal stress in each truss member: F 16.000 kips  AB  AB   5.971 ksi  5.97 ksi (C) AAB 2.67954 in.2

Ans.

 AC 

FAC 20.125 kips   7.510 ksi  7.51 ksi (T) AAC 2.67954 in.2

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 BC 

FBC 21.260 kips   7.934 ksi  7.93 ksi (C) ABC 2.67954 in.2

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P1.10 A simple pin-connected truss is loaded and supported as shown in Figure P1.10. All members of the truss are aluminum pipes that have an outside diameter of 60 mm and a wall thickness of 4 mm. Determine the normal stress in each truss member.

FIGURE P1.10

Solution Overall equilibrium: Begin the solution by determining the external reaction forces acting on the truss at supports A and B. Write equilibrium equations that include all external forces. Note that only the external forces (i.e., loads and reaction forces) are considered at this time. The internal forces acting in the truss members will be considered after the external reactions have been computed. The freebody diagram (FBD) of the entire truss is shown. The following equilibrium equations can be written for this structure: Fx  Ax  12 kN  0

 Ax  12 kN M A  By (1 m)  (15 kN)(4.3 m)  0  By  64.5 kN Fy  Ay  By  15 kN  0  Ay  49.5 kN Method of joints: Before beginning the process of determining the internal forces in the axial members, the geometry of the truss will be used to determine the magnitude of the inclination angles of members AB and BC. Use the definition of the tangent function to determine AB and BC: 1.5 m tan  AB   1.50   AB  56.310 1.0 m 1.5 m tan  BC   0.454545   BC  24.444 3.3 m




Joint A: Begin the solution process by considering a FBD of joint A. Consider only those forces acting directly on joint A. In this instance, two axial members, AB and AC, are connected at joint A. Additionally, two reaction forces, Ax and Ay, act at joint A. Tension forces will be assumed in each truss member. Fx  FAC  FAB cos(56.310)  Ax  0 (a) Fy  Ay  FAB sin(56.310)  0 (b) Solve Eq. (b) for FAB: Ay 49.5 kN FAB    59.492 kN sin(56.310) sin(56.310) and then compute FAC using Eq. (a): FAC   FAB cos(56.310)  Ax   ( 59.492 kN)cos(56.310)  ( 12 kN)  45.000 kN

Joint C: Next, consider a FBD of joint C. In this instance, the equilibrium equations associated with joint C seem easier to solve than those that would pertain to joint B. As before, tension forces will be assumed in each truss member. Fx   FAC  FBC cos(24.444)  12 kN  0 (c) Fy   FBC sin(24.444)  15 kN  0 (d) Solve Eq. (d) for FBC: 15 kN FBC   36.249 kN sin(24.444) Eq. (c) can be used as a check on our calculations: Fx   FAC  FBC cos(24.444)  12 kN  0   (45.000 kN)  ( 36.249 kN) cos(24.444)  12 kN  0

Section properties: For each of the three truss members: d  60 mm  2(4 mm)  52 mm

A

Checks!



(60 mm) 2  (52 mm) 2   703.7168 mm2 4

Normal stress in each truss member: F ( 59.492 kN)(1,000 N/kN)  AB  AB   84.539 MPa  84.5 MPa (C) AAB 703.7168 mm 2 F (45.000 kN)(1,000 N/kN)  AC  AC   63.946 MPa  63.9 MPa (T) AAC 703.7168 mm 2 F ( 36.249 kN)(1,000 N/kN)  BC  BC   51.511 MPa  51.5 MPa (C) ABC 703.7168 mm 2

Ans. Ans. Ans.


P1.11 A simple pin-connected truss is loaded and supported as shown in Figure P1.11. All members of the truss are aluminum pipes that have an outside diameter of 42 mm and a wall thickness of 3.5 mm. Determine the normal stress in each truss member.

FIGURE P1.11

Solution

Overall equilibrium: Begin the solution by determining the external reaction forces acting on the truss at supports A and B. Write equilibrium equations that include all external forces. Note that only the external forces (i.e., loads and reaction forces) are considered at this time. The internal forces acting in the truss members will be considered after the external reactions have been computed. The free-body diagram (FBD) of the entire truss is shown. The following equilibrium equations can be written for this structure: Fy  Ay  30 kN  0

 Ay  30 kN

M A  (30 kN)(4.5 m)  (15 kN)(1.6 m)  Bx (5.6 m)  0



 Bx  19.821 kN Fx  Ax  Bx  15 kN  0 Ax  15 kN  Bx  15 kN  (19.821 kN)

 Ax  34.821 kN

Method of joints: Before beginning the process of determining the internal forces in the axial members, the geometry of the truss will be used to determine the magnitude of the inclination angles of members AC and BC. Use the definition of the tangent function to determine AC and BC: 1.6 m tan  AC   0.355556   AC  19.573 4.5 m 4m tan  BC   0.888889   BC  41.634 4.5 m Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Joint A: Begin the solution process by considering a FBD of joint A. Consider only those forces acting directly on joint A. In this instance, two axial members, AB and AC, are connected at joint A. Additionally, two reaction forces, Ax and Ay, act at joint A. Tension forces will be assumed in each truss member. Fx  Ax  FAC cos(19.573)  0 (a) Fy  Ay  FAC sin(19.573)  FAB  0 (b) Solve Eq. (a) for FAC: Ax 34.821 kN FAC    36.957 kN cos(19.573) cos(19.573) and then compute FAB using Eq. (b): FAB  Ay  FAC sin(19.573)

 (30.000 kN)  (36.957 kN)sin(19.573)  17.619 kN Joint B: Next, consider a FBD of joint B. In this instance, the equilibrium equations associated with joint B seem easier to solve than those that would pertain to joint C. As before, tension forces will be assumed in each truss member. Fx  Bx  FBC cos(41.634)  0 (c) Fy  FBC sin(41.634)  FAB  0 (d) Solve Eq. (c) for FBC: Bx ( 19.821 kN) FBC    26.520 kN cos(41.634) cos(41.634) Eq. (d) can be used as a check on our calculations: Fy  FBC sin(41.634)  FAB

 (26.520 kN)sin(41.634)  (17.619 kN)  0 Section properties: For each of the three truss members: d  42 mm  2(3.5 mm)  35 mm

A

Checks!



(42 mm) 2  (35 mm) 2   423.3296 mm2 4

Normal stress in each truss member: F (17.619 kN)(1,000 N/kN)  AB  AB   41.620 MPa  41.6 MPa (T) AAB 423.3296 mm 2

Ans.

 AC 

FAC (36.957 kN)(1,000 N/kN)   87.301 MPa  87.3 MPa (T) AAC 423.3296 mm 2

Ans.

 BC 

FBC ( 26.520 kN)(1,000 N/kN)   62.647 MPa  62.6 MPa (C) ABC 423.3296 mm 2

Ans.


P1.12 The rigid beam BC shown in Figure P1.12 is supported by rods (1) and (2) that have cross-sectional areas of 175 mm2 and 300 mm2, respectively. For a uniformly distributed load of w = 15 kN/m, determine the normal stress in each rod. Assume L = 3 m and a = 1.8 m.

FIGURE P1.12

Solution Equilibrium: Calculate the internal forces in members (1) and (2).  1.8 m  M C   F1 (3 m)  (15 kN/m)(1.8 m)  0  2 

 F1  8.100 kN 1.8 m   M B  F2 (3 m)  (15 kN/m)(1.8 m)  3 m   0  2   F2  18.900 kN Stresses:

1 

F1 (8.100 kN)(1,000 N/kN)   46.286 N/mm 2  46.3 MPa 2 A1 175 mm

Ans.

2 

F2 (18.900 kN)(1,000 N/kN)   63.000 N/mm 2  63.0 MPa 2 A2 300 mm

Ans.


P1.13 Bar (1) in Figure P1.15 has a crosssectional area of 0.75 in.2. If the stress in bar (1) must be limited to 30 ksi, determine the maximum load P that may be supported by the structure.

FIGURE P1.13

Solution Given that the cross-sectional area of bar (1) is 0.75 in.2 and its normal stress must be limited to 30 ksi, the maximum force that may be carried by bar (1) is F1,max  1 A1  (30 ksi)(0.75 in.2 )  22.5 kips Consider a FBD of ABC. From the moment equilibrium equation about joint A, the relationship between the force in bar (1) and the load P is: M A  (6 ft)F1  (10 ft)P  0 P 

6 ft F1 10 ft

Substitute the maximum force F1,max = 22.5 kips into this relationship to obtain the maximum load that may be applied to the structure: 6 ft 6 ft Ans. P F1  (22.5 kips)  13.50 kips 10 ft 10 ft


P1.14 The rectangular bar shown in Figure P1.14 is subjected to a uniformly distributed axial loading of w = 13 kN/m and a concentrated force of P = 9 kN at B. Determine the magnitude of the maximum normal stress in the bar and its location x. Assume a = 0.5 m, b = 0.7 m, c = 15 mm, and d = 40 mm. FIGURE P1.14

Solution Equilibrium: Draw a FBD for the interval between A and B where 0  x  a . Write the following equilibrium equation: 

  Fx  (13 kN/m)(1.2 m  x)  (9 kN)  F  0  F  (13 kN/m)(1.2 m  x)  (9 kN) The largest force in this interval occurs at x = 0 where F = 6.6 kN.

In the interval between B and C where a  x  a  b , and write the following equilibrium equation: 

  Fx  (13 kN/m)(1.2 m  x)  F  0  F  (13 kN/m)(1.2 m  x) The largest force in this interval occurs at x = a where F = 9.1 kN.

Maximum Normal Stress: (9.1 kN)(1,000 N/kN)  max   15.17 MPa at x  0.5 m (15 mm)(40 mm)

Ans.


P1.15 The solid 1.25-in.-diameter rod shown in Figure P1.15 is subjected to a uniform axial distributed loading along its length of w = 750 lb/ft. Two concentrated loads also act on the rod: P = 2,000 lb and Q = 1,000 lb. Assume a = 16 in. and b = 32 in. Determine the normal stress in the rod at the following locations: (a) x = 10 in. (b) x = 30 in. FIGURE P1.15

Solution (a) x = 10 in. Equilibrium: Draw a FBD for the interval between A and B where 0  x  a , and write the following equilibrium equation: 

 Fx  (750 lb/ft)(1 ft/12 in.)(48 in.  x)  (2,000 lb)  (1,000 lb)  F  0  F  (62.5 lb/in.)(48 in.  x)  3,000 lb At x = 10 in., F = 5,375 lb. Stress: The normal stress at this location can be calculated as follows. A





4

(1.25 in.) 2  1.227185 in.2

5,375 lb  4,379.944 psi  4,380 psi 1.227185 in.2

Ans.

(b) x = 30 in. Equilibrium: Draw a FBD for the interval between B and C where a  x  a  b , and write the following equilibrium equation: 

 Fx  (750 lb/ft)(1 ft/12 in.)(48 in.  x)  (1,000 lb)  F  0  F  (62.5 lb/in.)(48 in.  x)  1,000 lb At x = 30 in., F = 2,125 lb. Stress: The normal stress at this location can be calculated as follows. 2,125 lb   1,731.606 psi  1,730 psi 1.227185 in.2

Ans.


P1.16 Two 6 in. wide wooden boards are to be joined by splice plates that will be fully glued on the contact surfaces. The glue to be used can safely provide a shear strength of 120 psi. Determine the smallest allowable length L that can be used for the splice plates for an applied load of P = 10,000 lb. Note that a gap of 0.5 in. is required between boards (1) and (2).

FIGURE P1.16

Solution Consider a FBD of board (2). The glue on the splice plates provides resistance to the 10,000 lb applied load on both the top and bottom surfaces of board (2). Denoting the shear resistance on a glue surface as V, equilibrium in the horizontal direction requires Fx  P  V  V  0 V 

10,000 lb  5,000 lb 2

In other words, each glue surface must be large enough so that 5,000 lb of shear resistance can be provided to board (2). Since the glue has a shear strength of 120 psi, the area of each glue surface on board (2) must be at least 5,000 lb Amin   41.6667 in.2 120 psi The boards are 6-in. wide; therefore, glue must be spread along board (2) for a length of at least 41.6667 in.2 Lglue joint   6.9444 in. 6 in. Although we’ve discussed only board (2), the same rationale applies to board (1). For both boards (1) and (2), the glue must be applied along a length of at least 6.9444 in. on both the top and bottom of the boards in order to resist the 10,000 lb applied load. The glue applied to boards (1) and (2) must be matched by glue applied to the splice plates. Therefore, the splice plates must be at least 6.9444 in. + 6.9444 in. = 13.8889 in. long. However, we are told that a 0.5-in. gap is required between boards (1) and (2); therefore, the splice plates must be 0.5-in. longer. Altogether, the length of the splice plates must be at least Ans. Lmin  6.9444 in.  6.9444 in.  0.5 in.  14.39 in.


P1.17 For the clevis connection shown in Figure P1.17, determine the maximum applied load P that can be supported by the 10-mm-diameter pin if the average shear stress in the pin must not exceed 95 MPa.

FIGURE P1.17

Solution Consider a FBD of the bar that is connected by the clevis, including a portion of the pin. If the shear force acting on each exposed surface of the pin is denoted by V, then the shear force on each pin surface is related to the load P by: Fx  P  V  V  0  P  2V The area of the pin surface exposed by the FBD is simply the cross-sectional area of the pin:  2  Apin  d pin  (10 mm) 2  78.539816 mm 2 4 4 If the average shear stress in the pin must be limited to 95 MPa, the maximum shear force V on a single cross-sectional surface must be limited to V   Abolt  (95 N/mm 2 )(78.539816 mm 2 )  7, 461.283 N Therefore, the maximum load P that may be applied to the connection is P  2V  2(7, 461.283 N)  14,922.565 N  14.92 kN

Ans.


P1.18 For the connection shown in Figure P1.18, determine the average shear stress produced in the 3/8in. diameter bolts if the applied load is P = 2,500 lb.

FIGURE P1.18

Solution There are four bolts, and it is assumed that each bolt supports an equal portion of the external load P. Therefore, the shear force carried by each bolt is 2,500 lb V  625 lb 4 bolts The bolts in this connection act in single shear. The cross-sectional area of a single bolt is  2   Abolt  d bolt  (3 / 8 in.) 2  (0.375 in.) 2  0.110447 in.2 4 4 4 Therefore, the average shear stress in each bolt is V 625 lb    5,658.8427 psi  5,660 psi Ans. Abolt 0.110447 in.2


P1.19 The five-bolt connection shown in Figure P1.19 must support an applied load of P = 265 kN. If the average shear stress in the bolts must be limited to 120 MPa, determine the minimum bolt diameter that may be used for this connection. FIGURE P1.19

Solution There are five bolts, and it is assumed that each bolt supports an equal portion of the external load P. Therefore, the shear force carried by each bolt is 265 kN V  53 kN  53,000 N 5 bolts Since the average shear stress must be limited to 120 MPa, each bolt must provide a shear area of at least: 53,000 N AV   441.6667 mm 2 2 120 N/mm Each bolt in this connection acts in double shear; therefore, two cross-sectional bolt surfaces are available to transmit shear stress in each bolt. AV 441.6667 mm 2 Abolt    220.8333 mm 2 per surface 2 surfaces per bolt 2 surfaces The minimum bolt diameter must be  2 Ans. d bolt  220.8333 mm 2  d bolt  16.7682 mm  16.77 mm 4


P1.20 A coupling is used to connect a 2 in. diameter plastic pipe (1) to a 1.5 in. diameter pipe (2), as shown in Figure P1.20. If the average shear stress in the adhesive must be limited to 400 psi, determine the minimum lengths L1 and L2 required for the joint if the applied load P is 5,000 lb.

FIGURE P1.24

Solution To resist a shear force of 5,000 lb, the area of adhesive required on each pipe is V 5,000 lb AV    12.5 in.2  adhesive 400 psi Consider the coupling on pipe (1). The adhesive is applied to the circumference of the pipe, and the circumference C1 of pipe (1) is C1   D1   (2.0 in.)  6.2832 in. The minimum length L1 is therefore AV 12.5 in.2 L1    1.9894 in.  1.989 in. Ans. C1 6.2832 in. Consider the coupling on pipe (2). The circumference C2 of pipe (2) is C2   D2   (1.5 in.)  4.7124 in. The minimum length L2 is therefore AV 12.5 in.2 L2    2.6526 in.  2.65 in. C2 4.7124 in.

Ans.


1.21 A hydraulic punch press is used to punch a slot in a 0.50-in. thick plate, as illustrated in Fig. P1.21. If the plate shears at a stress of 30 ksi, determine the minimum force P required to punch the slot.

FIGURE P1.21

Solution The shear stress associated with removal of the slug exists on its perimeter. The perimeter of the slug is given by perimeter  2(3.00 in.) +  (0.75 in.)  8.35619 in. Thus, the area subjected to shear stress is AV  perimeter  plate thickness  (8.35619 in.)(0.50 in.)  4.17810 in.2 Given that the plate shears at  = 30 ksi, the force required to remove the slug is therefore Ans. Pmin   AV  (30 ksi)(4.17810 in.2 )  125.343 kips  125.3 kips


P1.22 The handle shown in Figure P1.22 is attached to a 40-mm-diameter shaft with a square shear key. The forces applied to the lever are P = 1,300 N. If the average shear stress in the key must not exceed 150 MPa, determine the minimum dimension a that must be used if the key is 25 mm long. The overall length of the handle is L = 0.70 m. FIGURE P1.22

Solution To determine the shear force V that must be resisted by the shear key, sum moments about the center of the shaft (which will be denoted O):  700 mm   700 mm   40 mm  M O  (1,300 N)   (1,300 N)     V  0   2  2   2 

V  45,500 N Since the average shear stress in the key must not exceed 150 MPa, the shear area required is V 45,500 N AV    303.3333 mm2 2  150 N/mm The shear area in the key is given by the product of its length L (i.e., 25 mm) and its width a. Therefore, the minimum key width a is A 303.3333 mm 2 a V   12.1333 mm  12.13 mm Ans. L 25 mm


P1.23 An axial load P is supported by the short steel column shown in Figure P1.23. The column has a crosssectional area of 14,500 mm2. If the average normal stress in the steel column must not exceed 75 MPa, determine the minimum required dimension a so that the bearing stress between the base plate and the concrete slab does not exceed 8 MPa. Assume b = 420 mm.

FIGURE P1.23

Solution Since the normal stress in the steel column must not exceed 75 MPa, the maximum column load is Pmax   A  (75 N/mm 2 )(14,500 mm 2 )  1,087,500 N The maximum column load must be distributed over a large enough area so that the bearing stress between the base plate and the concrete slab does not exceed 8 MPa; therefore, the minimum plate area is P 1,087,500 N Amin    135,937.5 mm 2 2 b 8 N/mm The area of the plate is a ×b. Since b = 420, the minimum length of a must be Amin  135,937.5 mm2  a  b

a 

135,937.5 mm2  324 mm 420 mm

Ans.


P1.24 The two wooden boards shown in Figure P1.24 are connected by a 0.5-in.-diameter bolt. Washers are installed under the head of the bolt and under the nut. The washer dimensions are D = 2 in. and d = 5/8 in. The nut is tightened to cause a tensile stress of 9,000 psi in the bolt. Determine the bearing stress between the washer and the wood.

FIGURE P1.24

Solution The tensile stress in the bolt is 9,000 psi; therefore, the tension force that acts in the bolt is  Fbolt   bolt Abolt  (9,000 psi) (0.5 in.) 2  (9,000 psi)(0.196350 in.2 )  1,767.146 lb 4 The contact area between the washer and the wood is  Awasher  (2 in.) 2  (0.625 in.) 2   2.834796 in.2 4 Thus, the bearing stress between the washer and the wood is 1,767.146 lb b   623 psi 2.834796 in.2

Ans.


P1.25 For the beam shown in Figure P1.25, the allowable bearing stress for the material under the supports at A and B is b = 800 psi. Assume w = 2,100 lb/ft, P = 4,600 lb, a = 20 ft, and b = 8 ft. Determine the size of square bearing plates required to support the loading shown. Dimension the plates to the nearest ½ in. FIGURE P1.25

Solution Equilibrium: Using the FBD shown, calculate the beam reaction forces.

 20 ft  M A  By (20 ft)  (2,100 lb/ft)(20 ft)   (4,600 lb)(28 ft)  0  2   By  27,440 lb  20 ft  M B   Ay (20 ft)  (2,100 lb/ft)(20 ft)   (4,600 lb)(8 ft)  0  2   Ay  19,160 lb Bearing plate at A: The area of the bearing plate required for support A is 19,160 lb AA   23.950 in.2 800 psi Since the plate is to be square, its dimensions must be

width  23.950 in.2  4.894 in.

use 5 in.  5 in. bearing plate at A

Ans.

Bearing plate at B: The area of the bearing plate required for support B is 27, 440 lb AB   34.300 in.2 800 psi Since the plate is to be square, its dimensions must be

width  34.300 in.2  5.857 in.

use 6 in.  6 in. bearing plate at B

Ans.


P1.26 The d = 15-mm-diameter solid rod shown in Figure P1.26 passes through a D = 20-mm-diameter hole in the support plate. When a load P is applied to the rod, the rod head rests on the support plate. The support plate has a thickness of b = 12 mm. The rod head has a diameter of a = 30 mm and the head has a thickness of t = 10 mm. If the normal stress produced in the rod by load P is 225 MPa, determine: (a) the bearing stress acting between the support plate and the rod head. (b) the average shear stress produced in the rod head. (c) the punching shear stress produced in the support plate by the rod head.

FIGURE P1.26

Solution The cross-sectional area of the rod is:  Arod  (15 mm)2  176.715 mm 2 4 The tensile stress in the rod is 225 MPa; therefore, the tension force in the rod is Frod   rod Arod  (225 N/mm 2 )(176.715 mm 2 )  39,760.782 N (a) The contact area between the support plate and the rod head is  Acontact  (30 mm) 2  (20 mm) 2   392.699 mm 2 4 Thus, the bearing stress between the support plate and the rod head is 39,760.782 N b   101.3 MPa 392.699 mm 2 (b) In the rod head, the area subjected to shear stress is equal to the perimeter of the rod times the thickness of the head. AV   (15 mm)(10 mm)  471.239 mm 2 and therefore, the average shear stress in the rod head is 39,760.782 N   84.4 MPa 471.239 mm2 (c) In the support plate, the area subjected to shear stress is equal to the product of the rod head perimeter and the thickness of the plate. AV   (30 mm)(12 mm)  1,130.973 mm 2 and therefore, the average punching shear stress in the support plate is 39,760.782 N   35.2 MPa 1,130.973 mm 2

Ans.

Ans.

Ans.


P1.27 The rectangular bar is connected to the support bracket with a circular pin, as shown in Figure P1.27. The bar width is w = 1.75 in. and the bar thickness is 0.375 in. For an applied load of P = 5,600 lb, determine the average bearing stress produced in the bar by the 0.625-in.-diameter pin. FIGURE P1.27

Solution The average bearing stress produced in the bar by the pin is based on the projected area of the pin. The projected area is equal to the pin diameter times the bar thickness. Therefore, the average bearing stress in the bar is 5,600 lb b   23,893.33 psi  23, 900 psi Ans. (0.625 in.)(0.375 in.)


P1.28 The steel pipe column shown in Figure P1.28 has an outside diameter of 8.625 in. and a wall thickness of 0.25 in. The timber beam is 10.75 in wide, and the upper plate has the same width. The load imposed on the column by the timber beam is 80 kips. Determine (a) The average bearing stress at the surfaces between the pipe column and the upper and lower steel bearing plates. (b) The length L of the rectangular upper bearing plate if its width is 10.75 in. and the average bearing stress between the steel plate and the wood beam is not to exceed 500 psi. (c) The dimension “a” of the square lower bearing plate if the average bearing stress between the lower bearing plate and the concrete slab is not to exceed 900 psi.

Figure P1.28

Solution (a) The area of contact between the pipe column and one of the bearing plates is simply the crosssectional area of the pipe. To calculate the pipe area, we must first calculate the pipe inside diameter d: D  d  2t  d  D  2t  8.625 in.  2(0.25 in.)  8.125 in. The pipe cross-sectional area is   Apipe   D 2  d 2   (8.625 in.)2  (8.125 in.) 2   6.5777 in.2 4 4 Therefore, the bearing stress between the pipe and one of the bearing plates is P 80 kips b    12.1623 ksi  12.16 ksi Ans. Ab 6.5777 in.2 (b) The bearing stress between the timber beam and the upper bearing plate must not exceed 500 psi (i.e., 0.5 ksi). To support a load of 80 kips, the contact area must be at least P 80 kips Ab    160 in.2  b 0.5 ksi If the width of the timber beam is 10.75 in., then the length L of the upper bearing plate must be Ab 160 in.2 L   14.8837 in.  14.88 in. Ans. beam width 10.75 in. (c) The bearing stress between the concrete slab and the lower bearing plate must not exceed 900 psi (i.e., 0.9 ksi). To support the 80-kip pipe load, the contact area must be at least P 80 kips Ab    88.8889 in.2  b 0.9 ksi Since the lower bearing plate is square, its dimension a must be Ans. Ab  a  a  88.8889 in.2  a  9.4281 in.  9.43 in.


P1.29 A clevis-type pipe hanger supports an 8-indiameter pipe as shown in Figure P1.29. The hanger rod has a diameter of 1/2 in. The bolt connecting the top yoke and the bottom strap has a diameter of 5/8 in. The bottom strap is 3/16-in.-thick by 1.75-in.-wide by 36-in.-long. The weight of the pipe is 2,000 lb. Determine the following: (a) the normal stress in the hanger rod (b) the shear stress in the bolt (c) the bearing stress in the bottom strap

FIGURE P1.29

Solution (a) The normal stress in the hanger rod is  Arod  (0.5 in.) 2  0.196350 in.2 4 2,000 lb  rod   10,185.917 psi  10,190 psi 0.196350 in.2

Ans.

(b) The cross-sectional area of the bolt is:

Abolt 



(0.625 in.) 2  0.306796 in.2

4 The bolt acts in double shear; therefore, its average shear stress is 2,000 lb  bolt   3, 259.493 psi  3, 260 psi 2(0.306796 in.2 )

Ans.

(c) The bearing stress in the bottom strap is based on the projected area of the bolt in contact with the strap. Also, keep in mind that there are two ends of the strap that contact the bolt. The bearing stress is thus 2,000 lb b   8,533.334 psi  8,530 psi Ans. 2(0.625 in.)(3/16 in.)


P1.30 Rigid bar ABC shown in Figure P1.30 is supported by a pin at bracket A and by tie rod (1). Tie rod (1) has a diameter of 5 mm, and it is supported by double-shear pin connections at B and D. The pin at bracket A is a single-shear connection. All pins are 7 mm in diameter. Assume a = 600 mm, b = 300 mm, h = 450 mm, P = 900 N, and  = 55°. Determine the following: (a) the normal stress in rod (1) (b) the shear stress in pin B (c) the shear stress in pin A FIGURE P1.30

Solution Equilibrium: Using the FBD shown, calculate the reaction forces that act on rigid bar ABC. M A  F1 sin(36.87)(600 mm)

(900 N)sin (55)(900 mm)  0  F1  1,843.092 N

Fx  Ax  (1,843.092 N)cos(36.87)  (900 N)cos(55)  0  Ax  958.255 N Fy  Ay  (1,843.092 N)sin (36.87)  (900 N)sin (55)  0  Ay  368.618 N The resultant force at A is

A  (958.255 N)2  (368.618 N)2  1,026.709 N (a) Normal stress in rod (1).  Arod  (5 mm) 2  19.635 mm 2 4 1,843.092 N  rod   93.9 MPa 19.635 mm 2

Ans.

(b) Shear stress in pin B. The cross-sectional area of a 7-mm-diameter pin is:

Apin 



(7 mm)2  38.485 mm 2

4 Pin B is a double shear connection; therefore, its average shear stress is 1,843.092 N  pin B   23.9 MPa 2(38.485 mm 2 ) (c) Shear stress in pin A. Pin A is a single shear connection; therefore, its average shear stress is 1,026.709 N  pin A   26.7 MPa 38.485 mm2

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P1.31 The bell crank shown in Figure P1.31 is in equilibrium for the forces acting in rods (1) and (2). The bell crank is supported by a 10-mm-diameter pin at B that acts in single shear. The thickness of the bell crank is 5 mm. Assume a = 65 mm, b = 150 mm, F1 = 1,100 N, and  = 50°. Determine the following: (a) the shear stress in pin B (b) the bearing stress in the bell crank at B

FIGURE P1.31

Solution Equilibrium: Using the FBD shown, calculate the reaction forces that act on the bell crank. M B  (1,100 N)sin(50)(65 mm)

 F2 (150 mm)  0  F2  365.148 N Fx  Bx  (1,100 N)cos(50) 365.148 N  0  Bx  341.919 N

Fy  By  (1,100 N)sin(50)  0  By  842.649 N The resultant force at B is

B  (341.919 N)2  (842.649 N)2  909.376 N (a) Shear stress in pin B. The cross-sectional area of the 10-mm-diameter pin is:

Apin 



(10 mm)2  78.540 mm 2

4 Pin B is a single shear connection; therefore, its average shear stress is 909.376 N  pin B   11.58 MPa 78.540 mm 2

Ans.

(b) Bearing stress in the bell crank at B. The average bearing stress produced in the bell crank by the pin is based on the projected area of the pin. The projected area is equal to the pin diameter times the bell crank thickness. Therefore, the average bearing stress in the bell crank is 909.376 N b   18.19 MPa Ans. (10 mm)(5 mm)


P1.32 The beam shown in Figure P1.32 is supported by a pin at C and by a short link AB. If w = 30 kN/m, determine the average shear stress in the pins at A and C. Each pin has a diameter of 25 mm. Assume L = 1.8 m and  = 35°.

FIGURE P1.32

Solution Equilibrium: Using the FBD shown, calculate the reaction forces that act on the beam.

 1.8 m  M C   F1 sin(35)(1.8 m)  (30 kN/m)(1.8 m)  0  2   F1  47.0731 kN

Fx  Cx  (47.0731 kN)cos(35)  0  Cx  38.5600 kN  1.8 m  M B  C y (1.8 m)  (30 kN/m)(1.8 m)  0  2   C y  27.0000 kN The resultant force at C is

C  (38.5600 kN)2  (27.0000 kN)2  47.0731 kN Shear stress in pin A. The cross-sectional area of a 25-mm-diameter pin is:

Apin 



(25 mm) 2  490.8739 mm 2

4 Pin A is a single shear connection; therefore, its average shear stress is 47,073.1 N  pin A   95.9 MPa 490.8739 mm 2 Shear stress in pin C. Pin C is a double shear connection; therefore, its average shear stress is 47,073.1 N  pin C   47.9 MPa 2(490.8739 mm 2 )

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Ans.


P1.33 The bell-crank mechanism shown in Figure P1.33 is in equilibrium for an applied load of P = 7 kN applied at A. Assume a = 200 mm, b = 150 mm, and  = 65°. Determine the minimum diameter d required for pin B for each of the following conditions: (a) The average shear stress in the pin may not exceed 40 MPa. (b) The bearing stress in the bell crank may not exceed 100 MPa. (c) The bearing stress in the support bracket may not exceed 165 MPa. FIGURE P1.33

Solution Equilibrium: Using the FBD shown, calculate the reaction forces that act on the bell crank. M B  (7,000 N)sin(65)(200 mm)

 F2 (150 mm)  0  F2  8, 458.873 N Fx  Bx  (7,000 N)cos(65) 8,458.873 N  0  Bx  11,417.201 N

Fy  By  (7,000 N)sin(65)  0  By  6,344.155 N The resultant force at B is

B  (11,417.201 N)2  (6,344.155 N)2  13,061.423 N (a) The average shear stress in the pin may not exceed 40 MPa. The shear area required for the pin at B is 13,061.423 N AV   326.536 mm2 2 40 N/mm Since the pin at B is supported in a double shear connection, the required cross-sectional area for the pin is A Apin  V  163.268 mm 2 2 and therefore, the pin must have a diameter of 4 d (163.268 mm 2 )  14.42 mm Ans. 


(b) The bearing stress in the bell crank may not exceed 100 MPa. The projected area of pin B on the bell crank must equal or exceed 13,061.423 N Ab   130.614 mm 2 2 100 N/mm The bell crank thickness is 8 mm; therefore, the projected area of the pin is Ab = (8 mm)d. Calculate the required pin diameter d: 130.614 mm 2 Ans. d  16.33 mm 8 mm (c) The bearing stress in the support bracket may not exceed 165 MPa. The pin at B bears on two 6mm-thick support brackets. Thus, the minimum pin diameter required to satisfy the bearing stress limit on the support bracket is 13,061.423 N Ab   79.160 mm2 165 N/mm2 d

79.160 mm 2  6.60 mm 2(6 mm)

Ans.


P1.34 A structural steel bar with a 25 mm × 75 mm rectangular cross section is subjected to an axial load of 150 kN. Determine the maximum normal and shear stresses in the bar.

Solution The maximum normal stress in the steel bar is F (150 kN)(1,000 N/kN)  max    80 MPa A (25 mm)(75 mm) The maximum shear stress is one-half of the maximum normal stress   max  max  40 MPa 2

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P1.35 A steel rod of circular cross section will be used to carry an axial load of 92 kips. The maximum stresses in the rod must be limited to 30 ksi in tension and 12 ksi in shear. Determine the required diameter for the rod.

Solution Based on the allowable 30 ksi tension stress limit, the minimum cross-sectional area of the rod is F 92 kips Amin    3.0667 in.2  max 30 ksi For the 12-ksi shear stress limit, the minimum cross-sectional area of the rod must be F 92 kips Amin    3.8333 in.2 2 max 2(12 ksi) Therefore, the rod must have a cross-sectional area of at least 3.8333 in.2 in order to satisfy both the normal and shear stress limits. The minimum rod diameter D is therefore  2 d min  3.8333 in.2  d min  2.2092 in.  2.21 in. 4

Ans.


P1.36 An axial load P is applied to the rectangular bar shown in Figure P1.36. The cross-sectional area of the bar is 400 mm2. Determine the normal stress perpendicular to plane AB and the shear stress parallel to plane AB if the bar is subjected to an axial load of P = 70 kN. FIGURE P1.36

Solution The angle  for the inclined plane is 35°. The normal force N perpendicular to plane AB is found from N  P cos  (40 kN)cos35  57.3406 kN and the shear force V parallel to plane AB is V  P sin   (70 kN)sin35  40.1504 kN

The cross-sectional area of the bar is 400 mm2, but the area along inclined plane AB is A 400 mm 2 An    488.3098 mm 2 cos  cos35 The normal stress n perpendicular to plane AB is N (57.3406 kN)(1,000 N/kN) n    117.4268 MPa  117.4 MPa An 488.3098 mm 2 The shear stress nt parallel to plane AB is V (40.1504 kN)(1,000 N/kN)  nt    82.2231 MPa  82.2 MPa An 488.3098 mm 2

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


P1.37 An axial load P is applied to the 1.75 in. by 0.75 in. rectangular bar shown in Figure P1.37. Determine the normal stress perpendicular to plane AB and the shear stress parallel to plane AB if the bar is subjected to an axial load of P = 18 kips. FIGURE P1.37

Solution The angle  for the inclined plane is 60°. The normal force N perpendicular to plane AB is found from N  P cos  (18 kips)cos60  9.0 kips and the shear force V parallel to plane AB is V  P sin   (18 kips)sin 60  15.5885 kips

The cross-sectional area of the bar is (1.75 in.)(0.75 in.) = 1.3125 in.2, but the area along inclined plane AB is 1.3125 in.2 An  A / cos    2.6250 in.2 cos 60 The normal stress n perpendicular to plane AB is N 9.0 kips n    3.4286 ksi  3.43 ksi An 2.6250 in.2 The shear stress nt parallel to plane AB is V 15.5885 kips  nt    5.9385 ksi  5.94 ksi An 2.6250 in.2

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


P1.38 A compression load of P = 80 kips is applied to a 4 in. by 4 in. square post, as shown in Figure P1.38/39. Determine the normal stress perpendicular to plane AB and the shear stress parallel to plane AB.

FIGURE P1.38/39

Solution The angle  for the inclined plane is 55°. The normal force N perpendicular to plane AB is found from N  P cos  (80 kips)cos55  45.8861 kips and the shear force V parallel to plane AB is V  P sin   (80 kips)sin55  65.5322 kips The cross-sectional area of the post is (4 in.)(4 in.) = 16 in.2, but the area along inclined plane AB is 16 in.2 An  A / cos    27.8951 in.2 cos55

The normal stress n perpendicular to plane AB is N 45.8861 kips n    1.6449 ksi  1.645 ksi An 27.8951 in.2 The shear stress nt parallel to plane AB is V 65.5322 kips  nt    2.3492 ksi  2.35 ksi An 27.8951 in.2

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P1.39 Specifications for the 50 mm × 50 mm square bar shown in Figure P1.38/39 require that the normal and shear stresses on plane AB not exceed 120 MPa and 90 MPa, respectively. Determine the maximum load P that can be applied without exceeding the specifications.

FIGURE P1.38/39

Solution The general equations for normal and shear stresses on an inclined plane in terms of the angle  are P (a) n  (1  cos 2 ) 2A and P (b)  nt  sin 2 2A The cross-sectional area of the square bar is A = (50 mm)2 = 2,500 mm2, and the angle  for plane AB is 55°. The normal stress on plane AB is limited to 120 MPa; therefore, the maximum load P that can be supported by the square bar is found from Eq. (a): 2 A n 2(2,500 mm 2 )(120 N/mm 2 ) P   911,882 N 1  cos 2 1  cos 2(55) The shear stress on plane AB is limited to 90 MPa. From Eq. (b), the maximum load P based the shear stress limit is 2 A nt 2(2,500 mm 2 )(90 N/mm 2 ) P   478,880 N sin 2 sin 2(55) Thus, the maximum load that can be supported by the bar is Pmax  479 kN

Ans.


P1.40 Specifications for the 6 in. × 6 in. square post shown in Figure P1.40 require that the normal and shear stresses on plane AB not exceed 800 psi and 400 psi, respectively. Determine the maximum load P that can be applied without exceeding the specifications.

FIGURE P1.40

Solution The general equations for normal and shear stresses on an inclined plane in terms of the angle  are P (a) n  (1  cos 2 ) 2A and P (b)  nt  sin 2 2A The cross-sectional area of the square post is A = (6 in.)2 = 36 in.2, and the angle  for plane AB is 40°. The normal stress on plane AB is limited to 800 psi; therefore, the maximum load P that can be supported by the square post is found from Eq. (a): 2 A n 2(36 in.2 )(800 psi) P   49,078 lb 1  cos 2 1  cos 2(40) The shear stress on plane AB is limited to 400 psi. From Eq. (b), the maximum load P based the shear stress limit is 2 A nt 2(36 in.2 )(400 psi) P   29, 244 lb sin 2 sin 2(40) Thus, the maximum load that can be supported by the post is Pmax  29,200 lb  29.2 kips

Ans.


P1.41 A 90 mm wide bar will be used to carry an axial tension load of 280 kN as shown in Figure P1.41. The normal and shear stresses on plane AB must be limited to 150 MPa and 100 MPa, respectively. Determine the minimum thickness t required for the bar.

FIGURE P1.41

Solution The general equations for normal and shear stresses on an inclined plane in terms of the angle  are P (a) n  (1  cos 2 ) 2A and P (b)  nt  sin 2 2A The angle  for plane AB is 50°. The normal stress on plane AB is limited to 150 MPa; therefore, the minimum cross-sectional area A required to support P = 280 kN can be found from Eq. (a): P (280 kN)(1,000 N/kN) A (1  cos 2 )  (1  cos 2(50))  771.2617 mm 2 2 n 2(150 N/mm 2 ) The shear stress on plane AB is limited to 100 MPa; therefore, the minimum cross-sectional area A required to support P = 280 kN can be found from Eq. (b): P (280 kN)(1,000 N/kN) A sin 2  sin 2(50)  1,378.7309 mm 2 2 nt 2(100 N/mm 2 ) To satisfy both the normal and shear stress requirements, the cross-sectional area must be at least Amin = 1,379.7309 mm2. Since the bar width is 90 mm, the minimum bar thickness t must be 1,378.7309 mm 2 tmin   15.3192 mm  15.32 mm Ans. 90 mm


P1.42 A rectangular bar having width w = 6.00 in. and thickness t = 1.50 in. is subjected to a tension load P as shown in Figure P1.42/43. The normal and shear stresses on plane AB must not exceed 16 ksi and 8 ksi, respectively. Determine the maximum load P that can be applied without exceeding either stress limit. FIGURE P1.42/43

Solution The general equations for normal and shear stresses on an inclined plane in terms of the angle  are P (a) n  (1  cos 2 ) 2A and P (b)  nt  sin 2 2A The angle  for inclined plane AB is calculated from 3 tan    3   71.5651 1 The cross-sectional area of the bar is A = w×t = (6.00 in.)(1.50 in.) = 9.0 in.2. The normal stress on plane AB is limited to 16 ksi; therefore, the maximum load P can be found from Eq. (a): 2 A n 2(9.0 in.2 )(16 ksi) P   1, 440 ksi 1  cos 2 1  cos 2(71.5651) The shear stress on plane AB is limited to 8 ksi. From Eq. (b), the maximum load P based the shear stress limit is 2 A nt 2(9.0 in.2 )(8 ksi) P   240 kips sin 2 sin 2(71.5651) Thus, the maximum load that can be supported by the bar is Pmax  240 kips

Ans.


P1.43 In Figure P1.42/43, a rectangular bar having width w = 1.25 in. and thickness t is subjected to a tension load of P = 30 kips. The normal and shear stresses on plane AB must not exceed 12 ksi and 8 ksi, respectively. Determine the minimum bar thickness t required for the bar. FIGURE P1.42/43

Solution The general equations for normal and shear stresses on an inclined plane in terms of the angle  are P (a) n  (1  cos 2 ) 2A and P (b)  nt  sin 2 2A The angle  for inclined plane AB is calculated from 3 tan    3   71.5651 1 The normal stress on plane AB is limited to 12 ksi; therefore, the minimum cross-sectional area A required to support P = 30 kips can be found from Eq. (a): P 30 kips A (1  cos 2 )  (1  cos 2(71.5651))  0.2500 in.2 2 n 2(12 ksi) The shear stress on plane AB is limited to 8 ksi; therefore, the minimum cross-sectional area A required to support P = 30 kips can be found from Eq. (b): P 30 kips A sin 2  sin 2(71.5651)  1.1250 in.2 2 nt 2(8 ksi) To satisfy both the normal and shear stress requirements, the cross-sectional area must be at least Amin = 1.1250 in.2. Since the bar width is 1.25 in., the minimum bar thickness t must be 1.1250 in.2 tmin   0.900 in.  0.900 in. Ans. 1.25 in.


P1.44 The rectangular bar has a width of w = 3.00 in. and a thickness of t = 2.00 in. The normal stress on plane AB of the rectangular block shown in Figure P1.44/45 is 6 ksi (C) when the load P is applied. Determine: (a) the magnitude of load P. (b) the shear stress on plane AB. (c) the maximum normal and shear stresses in the block at any possible orientation. FIGURE P1.44/45

Solution The general equation for normal stress on an inclined plane in terms of the angle  is P n  (1  cos 2 ) 2A and the angle  for inclined plane AB is 3 tan    0.75   36.8699 4 The cross-sectional area of the rectangular bar is A = (3.00 in.)(2.00 in.) = 6.00 in.2.

(a)

(a) Since the normal stress on plane AB is given as 6 ksi, the magnitude of load P can be calculated from Eq. (a): 2 A n 2(6.0 in.2 )(6 ksi) P   56.25 kips  56.3 kips Ans. 1  cos 2 1  cos 2(36.8699) (b) The general equation for shear stress on an inclined plane in terms of the angle  is P  nt  sin 2 2A therefore, the shear stress on plane AB is 56.25 kips  nt  sin 2(36.8699)  4.50 ksi 2(6.00 in.2 ) (c) The maximum normal stress at any possible orientation is P 56.25 kips  max    9.3750 ksi  9.38 ksi A 6.00 in.2 and the maximum shear stress at any possible orientation in the block is P 56.25 kips  max    4.6875 ksi  4.69 ksi 2 A 2(6.00 in.2 )

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Ans.


P1.45 The rectangular bar has a width of w = 100 mm and a thickness of t = 75 mm. The shear stress on plane AB of the rectangular block shown in Figure P1.44/45 is 12 MPa when the load P is applied. Determine: (a) the magnitude of load P. (b) the normal stress on plane AB. (c) the maximum normal and shear stresses in the block at any possible orientation. FIGURE P1.44/45

Solution The general equation for shear stress on an inclined plane in terms of the angle  is P  nt  sin 2 2A and the angle  for inclined plane AB is 3 tan    0.75   36.8699 4 The cross-sectional area of the rectangular bar is A = (100 mm)(75 mm) = 7,500 mm2.

(a)

(a) Since the shear stress on plane AB is given as 12 MPa, the magnitude of load P can be calculated from Eq. (a): 2 A nt 2(7,500 mm 2 )(12 N/mm 2 ) P   187,500 N  187.5 kN Ans. sin 2 sin 2(36.8699) (b) The general equation for normal stress on an inclined plane in terms of the angle  is P n  (1  cos 2 ) 2A therefore, the normal stress on plane AB is 187,500 N n  (1  cos 2(36.8699))  16.00 MPa 2(7,500 mm 2 ) (c) The maximum normal stress at any possible orientation is P 187,500 N  max    25.0 MPa A 7,500 mm 2 and the maximum shear stress at any possible orientation in the block is P 187,500 N  max    12.50 MPa 2 A 2(7,500 mm 2 )

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Mechanics of Materials Philpot 3rd Edition Solutions Manual

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