ch01

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1.1 A stainless steel tube with an outside diameter of 60 mm and a wall thickness of 5 mm is used as acompression member. If the axial stress in the member must be limited to 340 MPa, determine themaximum load P that the member can support.

Solution The cross-sectional area of the stainless steel tube is

2 2 2 2 2( ) [(60 mm) (50 mm) ] 863.938 mm4 4

A D dπ π= − = − =

The normal stress in the tube can be expressed as PA

σ =

The maximum normal stress in the tube must be limited to 340 MPa. Using 340 MPa as the allowable normal stress, rearrange this expression to solve for the maximum load P 2 2

max allow (340 N/mm )(863.938 mm ) 293,739 N 294 kNP Aσ≤ = = = Ans.


1.2 A 2024-T4 aluminum tube with an outside diameter of 2.50 in. will be used to support a 12 kipload. If the axial stress in the member must be limited to 25 ksi, determine the wall thickness requiredfor the tube.

Solution From the definition of normal stress, solve for the minimum area required to support a 12-kip load without exceeding a stress of 25 ksi

2min

12 kips 0.480 in.25 ksi

P PAA

σσ

= ∴ ≥ = =

The cross-sectional area of the aluminum tube is given by

2 2( )4

A D dπ= −

Set this expression equal to the minimum area and solve for the maximum inside diameter d

2 2 2

2 2 2

2 2 2

max

[(2.50 in.) ] 0.480 in.4

4(2.50 in.) (0.480 in. )

4(2.50 in.) (0.480 in. )

2.374625 in.

d

d

d

d

π

π

π

− ≥

− ≥

− ≥

∴ ≤

The outside diameter D, the inside diameter d, and the wall thickness t are related by 2D d t= + Therefore, the minimum wall thickness required for the aluminum tube is

min2.50 in. 2.374525 in. 0.062738 in. 0.0627 in.

2 2D dt − −≥ = = = Ans.


1.3 Two solid cylindrical rods (1) and (2) arejoined together at flange B and loaded, asshown in Fig. P1.3. The diameter of rod (1) isD1 = 24 mm and the diameter of rod (2) is D2= 42 mm. Determine the normal stresses inrods (1) and (2). Fig. P1.3

Solution Cut a FBD through rod (1) that includes the free end of the rod at A. Assume that the internal force in rod (1) is tension. From equilibrium, 1 180 kN 0 80 kN (T)xF F FΣ = − = ∴ = Next, cut a FBD through rod (2) that includes the free end of the rod A. Assume that the internal force in rod (2) is tension. Equilibrium of this FBD reveals the internal force in rod (2): 2 2140 kN 140 kN 80 kN 0 200 kN 200 kN (C)xF F FΣ = + + − = ∴ = − = From the given diameter of rod (1), the cross-sectional area of rod (1) is

2 21 (24 mm) 452.3893 mm

4A π= =

and thus, the normal stress in rod (1) is

11 2

1

(80 kN)(1,000 N/kN) 176.8388 MPa 176.8 MPa (T)452.3893 mm

FA

σ = = = = Ans.

From the given diameter of rod (2), the cross-sectional area of rod (2) is

2 22 (42 mm) 1,385.4424 mm

4A π= =

Accordingly, the normal stress in rod (2) is

22 2

2

( 200 kN)(1,000 N/kN) 144.3582 MPa 144.4 MPa (C)1,385.4424 mm

FA

σ −= = = − = Ans.


1.4 Two solid cylindrical rods (1) and (2) arejoined together at flange B and loaded, asshown in Fig. P1.4. If the normal stress ineach rod must be limited to 120 MPa,determine the minimum diameter D requiredfor each rod. Fig. P1.4

Solution Cut a FBD through rod (1) that includes the free end of the rod at A. Assume that the internal force in rod (1) is tension. From equilibrium, 1 180 kN 0 80 kN (T)xF F FΣ = − = ∴ = Next, cut a FBD through rod (2) that includes the free end of the rod A. Assume that the internal force in rod (2) is tension. Equilibrium of this FBD reveals the internal force in rod (2): 2 2140 kN 140 kN 80 kN 0 200 kN 200 kN (C)xF F FΣ = + + − = ∴ = − = If the normal stress in rod (1) must be limited to 120 MPa, then the minimum cross-sectional area that can be used for rod (1) is

211,min 2

(80 kN)(1,000 N/kN) 666.6667 mm120 N/mm

FAσ

≥ = =

The minimum rod diameter is therefore

2 21,min 1,min 1,min666.6667 mm 29.1346 mm 29.1 mm

4A D Dπ= ≥ ∴ ≥ = Ans.

Similarly, the normal stress in rod (2) must be limited to 120 MPa. Notice that rod (2) is in compression. In this situation, we are concerned only with the magnitude of the stress; therefore, we will use the magnitude of F2 in the calculations for the minimum required cross-sectional area.

222,min 2

(200 kN)(1,000 N/kN) 1,666.6667 mm120 N/mm

FAσ

≥ = =

The minimum diameter for rod (2) is therefore

2 22,min 2,min 2,min1,666.6667 mm 46.0659 mm 46.1 mm

4A D Dπ= ≥ ∴ ≥ = Ans.


1.5 Two solid cylindrical rods (1) and (2) are joinedtogether at flange B and loaded, as shown in Fig.P1.5. The diameter of rod (1) is 1.25 in. and thediameter of rod (2) is 2.00 in. Determine the normalstresses in rods (1) and (2).

Fig. P1.5

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,

1

1

15 kips 0

15 kips 15 kips (C)yF F

FΣ = − − =

∴ = − = 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):

2

2

30 kips 30 kips 15 kips 0


FΣ = − − − − =

∴ = − = From the given diameter of rod (1), the cross-sectional area of rod (1) is

2 21 (1.25 in.) 1.2272 in.

4A π= =

and thus, the normal stress in rod (1) is

11 2

1

15 kips 12.2231 ksi 12.22 ksi (C)1.2272 in.

FA

σ −= = = − = Ans.


2 22 (2.00 in.) 3.1416 in.

4A π= =

Accordingly, the normal stress in rod (2) is

22 2

2

75 kips 23.8732 ksi 23.9 ksi (C)3.1416 in.

FA

σ −= = = − = Ans.


1.6 Two solid cylindrical rods (1) and (2) arejoined together at flange B and loaded, asshown in Fig. P1.6. If the normal stress ineach rod must be limited to 18 ksi,determine the minimum diameter D requiredfor each rod.

Fig. P1.6

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,

1

1

15 kips 0


FΣ = − − =

∴ = − = 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):

2

2

30 kips 30 kips 15 kips 0


FΣ = − − − − =

∴ = − = 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 18 ksi, then the minimum cross-sectional area that can be used for rod (1) is

211,min

15 kips 0.8333 in.18 ksi

FAσ

≥ = =


2 21,min 1,min 1,min0.8333 in. 1.0301 in. 1.030 in.

4A D Dπ= ≥ ∴ ≥ = Ans.

Similarly, the normal stress in rod (2) must be limited to 18 ksi, which requires a minimum area of

222,min

75 kips 4.1667 in.18 ksi

FAσ

≥ = =

The minimum diameter for rod (2) is therefore


4A D Dπ= ≥ ∴ ≥ = Ans.


1.7 Axial loads are applied with rigid bearing plates to the solid cylindrical rods shown in Fig. P1.7. The diameter ofaluminum rod (1) is 2.00 in., the diameter of brass rod (2) is1.50 in., and the diameter of steel rod (3) is 3.00 in.Determine the axial stress in each of the three rods.

Fig. P1.7

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, 1 130 kips 15 kips 15 kips 0 60 kips 60 kips (C)yF F FΣ = − − − − = ∴ = − =

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): 2 230 kips 15 kips 15 kips 20 kips 20 kips 0 20 kips 20 kips (C)yF F FΣ = − − − − + + = ∴ = − = Similarly, cut a FBD through rod (3) that includes the free end A. From this FBD, the internal force in rod (3) is:


3

3

30 kips 15 kips 15 kips 20 kips 20 kips 35 kips 35 kips 0


FΣ = − − − − + + − − =

∴ = − =


2 21 (2.00 in.) 3.1416 in.

4A π= =

and thus, the normal stress in aluminum rod (1) is

11 2

1

60 kips 19.0986 ksi 19.10 ksi (C)3.1416 in.

FA

σ −= = = − = Ans.


2 22 (1.50 in.) 1.7671 in.

4A π= =

Accordingly, the normal stress in brass rod (2) is

22 2

2

20 kips 11.3177 ksi 11.32 ksi (C)1.7671 in.

FA

σ −= = = − = Ans.

Finally, the cross-sectional area of rod (3) is

2 23 (3.00 in.) 7.0686 in.

4A π= =

and the normal stress in the steel rod is

33 2

3

90 kips 12.7324 ksi 12.73 ksi (C)7.0686 in.

FA

σ −= = = − = Ans.


1.8 Axial loads are applied with rigid bearing plates to the solidcylindrical rods shown in Fig. P1.8. The normal stress inaluminum rod (1) must be limited to 25 ksi, the normal stress inbrass rod (2) must be limited to 15 ksi, and the normal stress insteel rod (3) must be limited to 10 ksi. Determine the minimumdiameter D required for each of the three rods.

Fig. P1.8

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, 1 130 kips 15 kips 15 kips 0 60 kips 60 kips (C)yF F FΣ = − − − − = ∴ = − =

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): 2 230 kips 15 kips 15 kips 20 kips 20 kips 0 20 kips 20 kips (C)yF F FΣ = − − − − + + = ∴ = − =


Similarly, cut a FBD through rod (3) that includes the free end A. From this FBD, the internal force in rod (3) is:

3

3

30 kips 15 kips 15 kips 20 kips 20 kips 35 kips 35 kips 0


FΣ = − − − − + + − − =

∴ = − =

Notice that all three rods 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, and in turn, the minimum rod diameters. The normal stress in aluminum rod (1) must be limited to 25 ksi; therefore, the minimum cross-sectional area required for rod (1) is

211,min

1

60 kips 2.40 in.25 ksi

FAσ

≥ = =



4A D Dπ= ≥ ∴ ≥ = Ans.

The normal stress in brass rod (2) must be limited to 15 ksi, which requires a minimum area of

222,min

2

20 kips 1.3333 in.15 ksi

FAσ

≥ = =

which requires a minimum diameter for rod (2) of


4A D Dπ= ≥ ∴ ≥ = Ans.

The normal stress in steel rod (3) must be limited to 10 ksi. The minimum cross-sectional area required for this rod is:

233,min

3

90 kips 9.0 in.10 ksi

FAσ

≥ = =



4A D Dπ= ≥ ∴ ≥ = Ans.


1.9 Two solid cylindrical rodssupport a load of P = 32 kN, as shown in Fig. P1.9. Rod (1) has adiameter of 16 mm and the diameterof rod (2) is 12 mm. Determine theaxial stress in each rod.

Fig. P1.9

Solution Consider a FBD of joint B. Determine the angle α between rod (1) and the horizontal axis:

5.6 mtan 1.4737 55.84033.8 m

α α= = ∴ = °

and the angle β between rod (2) and the horizontal axis:

3.3 mtan 0.7174 35.65534.6 m

β β= = ∴ = °

Write equilibrium equations for the sum of forces in the horizontal and vertical directions. Note: Rods (1) and (2) are two-force members. 2 1cos(35.6553 ) cos(55.8403 ) 0xF F FΣ = ° − ° = (a) 2 1sin(35.6553 ) sin(55.8403 ) 0yF F F PΣ = ° + ° − = (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:

2 1cos(55.8403 )cos(35.6553 )

F F °=°

(c)

Substituting Eq. (c) into Eq. (b) gives

[ ]1 1

1

1

cos(55.8403 ) sin(35.6553 ) sin(55.8403 )cos(35.6553 )cos(55.8403 ) tan(35.6553 ) sin(55.8403 )

cos(55.8403 ) tan(35.6553 ) sin(55.8403 ) 1.2303

F F P

F PP PF

° ° + ° =°° ° + ° =

∴ = =° ° + °

For the given load of P = 32 kN, the internal force in rod (1) is therefore:


132 kN 26.0101 kN1.2303

F = =

Backsubstituting this result into Eq. (c) gives force F2:

2 1cos(55.8403 ) cos(55.8403 )(26.0101 kN) 17.9742 kNcos(35.6553 ) cos(35.6553 )

F F ° °= = =° °

The diameter of rod (1) is 16 mm; therefore, its cross-sectional area is:

2 21 (16 mm) 201.0619 mm

4A π= =

and the normal stress in rod (1) is:

211 2

1

(26.0101 kN)(1,000 N/kN) 129.3636 N/mm 129.4 MPa (T)201.0619 mm

FA

σ = = = = Ans.

The diameter of rod (2) is 12 mm; therefore, its cross-sectional area is:

2 22 (12 mm) 113.0973 mm

4A π= =

and the normal stress in rod (2) is:

222 2

2

(17.9742 kN)(1,000 N/kN) 158.9269 N/mm 158.9 MPa (T)113.0973 mm

FA

σ = = = = Ans.


1.10 Two solid cylindrical rods supporta load of P = 70 kN, as shown in Fig.P1.10. If the normal stress in each rodmust be limited to 165 MPa, determinethe minimum diameter D required foreach rod.

Fig. P1.10

Solution Consider a FBD of joint B. Determine the angle α between rod (1) and the horizontal axis:

5.6 mtan 1.4737 55.84033.8 m

α α= = ∴ = °

and the angle β between rod (2) and the horizontal axis:

3.3 mtan 0.7174 35.65534.6 m

β β= = ∴ = °

Write equilibrium equations for the sum of forces in the horizontal and vertical directions. Note: Rods (1) and (2) are two-force members. 2 1cos(35.6553 ) cos(55.8403 ) 0xF F FΣ = ° − ° = (a) 2 1sin(35.6553 ) sin(55.8403 ) 0yF F F PΣ = ° + ° − = (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:

2 1cos(55.8403 )cos(35.6553 )

F F °=°

(c)

Substituting Eq. (c) into Eq. (b) gives

[ ]1 1

1

1

cos(55.8403 ) sin(35.6553 ) sin(55.8403 )cos(35.6553 )cos(55.8403 ) tan(35.6553 ) sin(55.8403 )

cos(55.8403 ) tan(35.6553 ) sin(55.8403 ) 1.2303

F F P

F PP PF

° ° + ° =°° ° + ° =

∴ = =° ° + °

For the given load of P = 70 kN, the internal force in rod (1) is therefore:


170 kN 56.8967 kN1.2303

F = =

Backsubstituting this result into Eq. (c) gives force F2:

2 1cos(55.8403 ) cos(55.8403 )(56.8967 kN) 39.3182 kNcos(35.6553 ) cos(35.6553 )

F F ° °= = =° °

The normal stress in rod (1) must be limited to 165 MPa; therefore, the minimum cross-sectional area required for rod (1) is

211,min 2

1

(56.8967 kN)(1,000 N/kN) 344.8285 mm165 N/mm

FAσ

≥ = =



4A D Dπ= ≥ ∴ ≥ = Ans.

The minimum area required for rod (2) is

222,min 2

2

(39.3182 kN)(1,000 N/kN) 238.2921 mm165 N/mm

FAσ

≥ = =



4A D Dπ= ≥ ∴ ≥ = Ans.


1.11 Bar (1) in Fig. P1.11 has a cross-sectional area of 0.75 in.2. If the stress in bar(1) must be limited to 30 ksi, determine themaximum load P that may be supported bythe structure.

Fig. P1.11

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 2

1,max 1 1 (30 ksi)(0.75 in. ) 22.5 kipsF Aσ= = = 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:

1

1

(6 ft) (10 ft) 06 ft

10 ft

AM F P

P F

Σ = − =

∴ =

Substitute the maximum force F1,max = 22.5 kips into this relationship to obtain the maximum load that may be applied to the structure:

16 ft 6 ft (22.5 kips) 13.50 kips

10 ft 10 ftP F= = = Ans.


1.12 Two 6 in. wide wooden boards are tobe joined by splice plates that will be fullyglued on the contact surfaces. The glue tobe used can safely provide a shear strengthof 120 psi. Determine the smallestallowable length L that can be used for thesplice plates for an applied load of P = 10,000 lb. Note that a gap of 0.5 in. isrequired between boards (1) and (2).

Fig. P1.12

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

010,000 lb 5,000 lb

2

xF P V V

V

Σ = − − =

∴ = =

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

2min

5,000 lb 41.6667 in.120 psi

A ≥ =

The boards are 6-in. wide; therefore, glue must be spread along board (2) for a length of at least

2

glue joint41.6667 in. 6.9444 in.

6 in.L ≥ =

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 min 6.9444 in. 6.9444 in. 0.5 in. 14.39 in.L = + + = Ans.


1.13 For the clevis connection shown in Fig. P1.13,determine the shear stress in the 24 mm diameter boltfor an applied load of P = 175 kN.

Fig. P1.13

Solution Consider a FBD of the bar that is connected by the clevis, including a portion of the bolt. If the shear force acting on each exposed surface of the bolt is denoted by V, then the shear force on each bolt surface is

175 kN0 87.5 kN2xF P V V VΣ = − − = ∴ = =

The area of the bolt surface exposed by the FBD is simply the cross-sectional area of the bolt:

2 2 2bolt bolt (24 mm) 452.3893 mm

4 4A Dπ π= = =

Therefore, the shear stress in the bolt is

22

bolt

(87.5 kN)(1,000 N/kN) 193.4175 N/mm 193.4 MPa452.3893 mm

VA

τ = = = = Ans.


1.14 For the clevis connection shown in Fig.P1.14, the shear stress in the 5/16 in.diameter bolt must be limited to 40 ksi.Determine the maximum load P that may beapplied to the connection. Fig. P1.14

Solution Consider a FBD of the bar that is connected by the clevis, including a portion of the bolt. If the shear force acting on each exposed surface of the bolt is denoted by V, then the shear force on each bolt surface is related to the load P by: 0 2xF P V V P VΣ = − − = ∴ =

The area of the bolt surface exposed by the FBD is simply the cross-sectional area of the bolt:

2 2 2 2bolt bolt (5 /16 in.) (0.3125 in.) 0.076699 in.

4 4 4A Dπ π π= = = =

If the shear stress in the bolt must be limited to 40 ksi, the maximum shear force V on a single cross-sectional surface must be limited to 2

bolt (40 ksi)(0.076699 in. ) 3.067962 kipsV Aτ= = = Therefore, the maximum load P that may be applied to the connection is 2 2(3.067962 kips) 6.135923 kips 6.14 kipsP V= = = = Ans.


1.15 For the connection shown in Fig. P1.15,determine the average shear stress in the 0.75 in.diameter bolts if the load is P = 60 kips.

Fig. P1.15

Solution The bolts in this connection act in single shear. The cross-sectional area of a single bolt is

2 2 2bolt bolt (0.75 in.) 0.4418 in.

4 4A Dπ π= = =

Since there are five bolts, the total area that carries shear stress is 2 2

bolt5 5(0.4418 in. ) 2.2089 in.VA A= = = Therefore, the shear stress in each bolt is

2

60 kips 27.1624 ksi 27.2 ksi2.2089 in.V

PA

τ = = = = Ans.


1.16 The five-bolt connection shown in Fig.P1.16 must support an applied load of P = 550 kN. If the average shear stress in the bolts mustbe limited to 270 MPa, determine the minimumbolt diameter that may be used in the connection.

Fig P1.16

Solution To support a load of 550 kN while not exceeding an average shear stress of 270 MPa, the total shear area provided by the bolts must be at least

22

(550 kN)(1,000 N/kN) 2,037.0370 mm270 N/mmV

PAτ

≥ = =

Since there are five single-shear bolts in this connection, five cross-sectional surfaces carry shear stress. Consequently, each bolt must provide a minimum area of

2

2bolt

2,037.0370 mm 407.4074 mm5 5VAA ≥ = =

The minimum bolt diameter is therefore

2 2bolt bolt bolt407.4074 mm 22.7756 mm 22.8 mm

4A D Dπ≥ = ∴ ≥ = Ans.


1.17 For the connection shown in Fig. P1.17,the average shear stress in the 16 mm diameterbolts must be limited to 210 MPa. Determinethe maximum load P that may be applied to theconnection.

Fig. P1.17

Solution The cross-sectional area of a 16-mm-diameter bolt is

2 2 2bolt bolt (16 mm) 201.0619 mm

4 4A Dπ π= = =

This is a double-shear connection. Therefore, the three bolts provide a total shear area of 2 2

bolt2(3 bolts) 2(3 bolts)(201.0619 mm ) 1,206.3716 mmVA A= = = Since the shear stress must be limited to 210 MPa, the total shear force that can be resisted by the three bolts is 2 2

max (210 N/mm )(1,206.3716 mm ) 253,338.0316 NVV Aτ= = = In this connection, the shear force in the bolts is equal to the applied load P; therefore, max 253 kNP = Ans.


1.18 The three-bolt connection shown in Fig. P1.18must support an applied load of P = 60 kips. If the average shear stress in the bolts must be limited to15 ksi, determine the minimum bolt diameter thatmay be used in the connection.

Fig. P1.18

Solution The shear force V that must be provided by the bolts equals the applied load of P = 60 kips. The total shear area required is thus

260 kips 4.0 in.15 ksiV

VAτ

≥ = =

The three bolts in this connection act in double shear; therefore, six cross-sectional bolt surfaces are available to transmit shear stress.

2

2bolt

4.0 in. 0.6667 in. per surface(2 surfaces per bolt)(3 bolts) 6 surfaces

VAA = = =

The minimum bolt diameter must be

2 2bolt bolt0.6667 in. 0.9213 in. 0.921 in.

4D Dπ ≥ ∴ ≥ = Ans.


1.19 A hydraulic punch press is used topunch a slot in a 10 mm thick plate, asillustrated in Fig. P1.19. If the plate shearsat a stress of 250 MPa, determine theminimum force P required to punch theslot.

Fig. P1.19

Solution The shear stress associated with removal of the slug exists on its perimeter. The perimeter of the slug is given by perimeter 2(75 mm) + (20 mm) 212.8319 mmπ= = Thus, the area subjected to shear stress is 2perimeter plate thickness (212.8319 mm)(10 mm) 2,128.319 mmVA = × = = Given that the plate shears at τ = 250 MPa, the force required to remove the slug is therefore 2 2

min (250 N/mm )(2,128.319 mm ) 532,080 N 532 kNVP Aτ= = = = Ans.


1.20 A coupling is used to connect a 2 in. diameterplastic pipe (1) to a 1.5 in. diameter pipe (2), asshown in Fig. P1.20. If the average shear stress in the adhesive must be limited to 400 psi, determinethe minimum lengths L1 and L2 required for the jointif the applied load P is 5,000 lb.

Fig. P1.20

Solution To resist a shear force of 5,000 lb, the area of adhesive required on each pipe is

2

adhesive

5,000 lb 12.5 in.400 psiV

VAτ

= = =

Consider the coupling on pipe (1). The adhesive is applied to the circumference of the pipe, and the circumference C1 of pipe (1) is 1 1 (2.0 in.) 6.2832 in.C Dπ π= = = The minimum length L1 is therefore

2

11

12.5 in. 1.9894 in. 1.989 in.6.2832 in.

VALC

≥ = = = Ans.

Consider the coupling on pipe (2). The circumference C2 of pipe (2) is 2 2 (1.5 in.) 4.7124 in.C Dπ π= = = The minimum length L2 is therefore

2

22

12.5 in. 2.6526 in. 2.65 in.4.7124 in.

VALC

≥ = = = Ans.


1.21 A lever is attached to a shaft with asquare shear key, as shown in Fig. P1.21. The force applied to the lever is P = 350 N.If the shear stress in the key must not exceed80 MPa, determine the minimum dimension“a” that must be used if the key is 25 mmlong.

Fig. P1.21

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):

42 mm(350 N)(700 mm) 0 11,666.6667 N2OM V V⎛ ⎞Σ = − + = ∴ =⎜ ⎟

⎝ ⎠

Since the shear stress in the key must not exceed 80 MPa, the shear area required is

22

11,666.6667 N 145.8333 mm80 N/mmV

VAτ

≥ = =

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

2145.8333 mm 5.8333 mm 5.83 mm

25 mmVAa

L≥ = = = Ans.


1.22 A common trailer hitch connection is shown inFig. P1.22. The shear stress in the pin must be limited to 30,000 psi. If the applied load is P = 4,000 lb, determine the minimum diameter that must be usedfor the pin.

Fig. P1.22

Solution The shear force V acting in the hitch pin is equal to the applied load; therefore, V = P = 4,000 lb. The shear area required to support a 4,000 lb shear force is

24,000 lb 0.1333 in.30,000 psiV

VAτ

≥ = =

The hitch pin is used in a double-shear connection; therefore, two cross-sectional areas of the pin are subjected to shear stress. Thus, the cross-sectional area of the pin is given by

2

2pin pin

0.1333 in.2 0.0667 in.2 2V

VAA A A= ∴ = = =

and the minimum pin diameter is

2 2pin pin0.0667 in. 0.2913 in. 0.291 in.

4D Dπ ≥ ∴ ≥ = Ans.


1.23 An axial load P is supported by a short steelcolumn, which has a cross-sectional area of11,400 mm2. If the average normal stress in the steel column must not exceed 110 MPa, determine theminimum required dimension “a” so that the bearingstress between the base plate and the concrete slab doesnot exceed 8 MPa.

Fig. P1.23

Solution Since the normal stress in the steel column must not exceed 110 MPa, the maximum column load is 2 2

max (110 N/mm )(11,400 mm ) 1,254,000 NP Aσ= = = 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

2min 2

1,254,000 N 156,750 mm8 N/mmb

PAσ

= = =

Since the plate is square, the minimum plate dimension a must be

2min 156,750 mm

395.9167 mm 396 mm

A a a

a

= = ×

∴ ≥ = Ans.


1.24 The steel pipe column shown in Fig. P1.24 has anoutside diameter of 8.625 in. and a wall thickness of 0.25 in. The timber beam is 10.75 in wide, and the upper platehas the same width. The load imposed on the column bythe 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 ifits width is 10.75 in. and the average bearing stressbetween the steel plate and the wood beam is not toexceed 500 psi.

(c) The dimension “a” of the square lower bearing plate if the average bearing stress between the lower bearingplate and the concrete slab is not to exceed 900 psi.

Fig. P1.24

Solution (a) The area of contact between the pipe column and one of the bearing plates is simply the cross-sectional area of the pipe. To calculate the pipe area, we must first calculate the pipe inside diameter d: 2 2 8.625 in. 2(0.25 in.) 8.125 in.D d t d D t= + ∴ = − = − = The pipe cross-sectional area is

2 2 2 2 2pipe (8.625 in.) (8.125 in.) 6.5777 in.

4 4A D dπ π⎡ ⎤ ⎡ ⎤= − = − =⎣ ⎦ ⎣ ⎦

Therefore, the bearing stress between the pipe and one of the bearing plates is

2

80 kips 12.1623 ksi 12.16 ksi6.5777 in.b

b

PA

σ = = = = Ans.

(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

280 kips 160 in.0.5 ksib

b

PAσ

≥ = =

If the width of the timber beam is 10.75 in., then the length L of the upper bearing plate must be

2160 in. 14.8837 in. 14.88 in.

beam width 10.75 in.bAL ≥ = = =

(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

280 kips 88.8889 in.0.9 ksib

b

PAσ

≥ = =

Since the lower bearing plate is square, its dimension a must be 288.8889 in. 9.4281 in. 9.43 in.bA a a a= × = ∴ ≥ = Ans.


1.25 A vertical shaft is supported by a thrustcollar and bearing plate, as shown in Fig.P1.25. The average shear stress in the collarmust be limited to 18 ksi. The average bearingstress between the collar and the plate must belimited to 24 ksi. Based on these limits,determine the maximum axial load P that canbe applied to the shaft.

Fig. P1.25

Solution Consider collar shear stress: The area subjected to shear stress in the collar is equal to the product of the shaft circumference and the collar thickness; therefore, 2shaft circumference collar thickness (1.0 in.)(0.5 in.) 1.5708 in.VA π= × = = If the shear stress must not exceed 18 ksi, the maximum load that can be supported by the vertical shaft is: 2(18 ksi)(1.5708 in. ) 28.2743 kipsVP Aτ≤ = = Consider collar bearing stress: We must determine the area of contact between the collar and the plate. The overall cross-sectional area of the collar is

2 2collar (1.5 in.) 1.7671 in.

4A π= =

is reduced by the area taken up by the shaft

2 2shaft (1.0 in.) 0.7854 in.

4A π= =

Therefore, the area of the collar that actually contacts the plate is 2 2 2

collar shaft 1.7671 in. 0.7854 in. 0.9817 in.bA A A= − = − = If the bearing stress must not exceed 24 ksi, the maximum load that can be supported by the vertical shaft is: 2(24 ksi)(0.9817 in. ) 23.5619 kipsb bP Aσ≤ = = Controlling P: Considering both shear stress in the collar and bearing stress between the collar and the plate, the maximum load that can be supported by the shaft is max 23.6 kipsP = Ans.


1.26 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

max(150 kN)(1,000 N/kN) 80 MPa

(25 mm)(75 mm)FA

σ = = = Ans.

The maximum shear stress is one-half of the maximum normal stress

maxmax 40 MPa

2στ = = Ans.


1.27 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 D for the rod.

Solution Based on the allowable 30 ksi tension stress limit, the minimum cross-sectional area of the rod is

2min

max

92 kips 3.0667 in.30 ksi

FAσ

= = =

For the 12-ksi shear stress limit, the minimum cross-sectional area of the rod must be

2min

max

92 kips 3.8333 in.2 2(12 ksi)

FAτ

= = =

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 2min min3.8333 in. 2.2092 in. 2.21 in.

4D Dπ ≥ ∴ = = Ans.


1.28 An axial load P is applied to therectangular bar shown in Fig. P1.28. Thecross-sectional area of the bar is 300 mm2. Determine the normal stress perpendicular toplane AB and the shear stress parallel toplane AB if the bar is subjected to an axialload of P = 25 kN.

Fig. P1.28

Solution The angle θ for the inclined plane is 35°. The normal force N perpendicular to plane AB is found from cos (25 kN)cos35 20.4788 kNN P θ= = ° = and the shear force V parallel to plane AB is sin (25 kN)sin 35 14.3394 kNV P θ= = ° =

The cross-sectional area of the bar is 300 mm2, but the area along inclined plane AB is

2

2300 mm 366.2324 mmcos cos35n

AAθ

= = =°

The normal stress σn perpendicular to plane AB is

2

(20.4788 kN)(1,000 N/kN) 55.9175 MPa 55.9 MPa366.2324 mmn

n

NA

σ = = = = Ans.

The shear stress τnt parallel to plane AB is

2

(14.3394 kN)(1,000 N/kN) 39.1539 MPa 39.2 MPa366.2324 mmnt

n

VA

τ = = = = Ans.


1.29 An axial load P is applied to the 1.25 in.by 0.75 in. rectangular bar shown in Fig.P1.29. Determine the normal stressperpendicular to plane AB and the shear stressparallel to plane AB if the bar is subjected toan axial load of P = 20 kips. Fig. P1.29

Solution The angle θ for the inclined plane is 60°. The normal force N perpendicular to plane AB is found from cos (20 kips)cos60 10.0 kipsN P θ= = ° = and the shear force V parallel to plane AB is sin (20 kips)sin 60 17.3205 kipsV P θ= = ° =

The cross-sectional area of the bar is (1.25 in.)(0.75 in.) = 0.9375 in.2, but the area along inclined plane AB is

2

20.9375 in./ cos 1.8750 in.cos60nA A θ= = =

°


2

10.0 kips 5.3333 ksi 5.33 ksi1.8750 in.n

n

NA

σ = = = = Ans.


2

17.3205 kips 9.2376 ksi 9.24 ksi1.8750 in.nt

n

VA

τ = = = = Ans.


1.30 A compression load of P = 80 kips is applied to a 4 in.by 4 in. square post, as shown in Fig. P1.30. Determine thenormal stress perpendicular to plane AB and the shear stress parallel to plane AB.

Fig. P1.30

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

2

216 in./ cos 27.8951 in.cos55nA A θ= = =

°


2

45.8861 kips 1.6449 ksi 1.645 ksi27.8951 in.n

n

NA

σ = = = = Ans.


2

65.5322 kips 2.3492 ksi 2.35 ksi27.8951 in.nt

n

VA

τ = = = = Ans.


1.31 Specifications for the 50 mm × 50 mm square barshown in Fig. P1.31 require that the normal and shear stresses on plane AB not exceed 100 MPa and 70 MPa,respectively. Determine the maximum load P that can be applied without exceeding the specifications.

Fig. P1.31

Solution The general equations for normal and shear stresses on an inclined plane in terms of the angle θ are

(1 cos 2 )2nPA

σ θ= + (a)

and

sin 22ntPA

τ θ= (b)

The cross-sectional area of the square bar is A = (50 mm)2 = 2,500 mm2, and the angle θ for plane AB is 55°. The normal stress on plane AB is limited to 100 MPa; therefore, the maximum load P that can be supported by the square bar is found from Eq. (a):

2 22 2(2,500 mm )(100 N/mm ) 759,902 N

1 cos 2 1 cos 2(55 )nAP σ

θ≤ = =

+ + °

The shear stress on plane AB is limited to 70 MPa. From Eq. (b), the maximum load P based the shear stress limit is

2 22 2(2,500 mm )(70 N/mm ) 372,462 N

sin 2 sin 2(55 )ntAP τθ

≤ = =°

Thus, the maximum load that can be supported by the bar is max 372.5 kNP = Ans.


1.32 Specifications for the 6 in. × 6 in. square post shown inFig. P1.32 require that the normal and shear stresses on planeAB not exceed 800 psi and 400 psi, respectively. Determinethe maximum load P that can be applied without exceedingthe specifications.

Fig. P1.32


(1 cos 2 )2nPA

σ θ= + (a)

and

sin 22ntPA

τ θ= (b)

The cross-sectional area of the square post is A = (6 in.)2 = 36 in.2, and the angle θ 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):

22 2(36 in. )(800 psi) 49,078 lb

1 cos 2 1 cos 2(40 )nAP σ

θ≤ = =

+ + °

The shear stress on plane AB is limited to 400 psi. From Eq. (b), the maximum load P based the shear stress limit is

22 2(36 in. )(400 psi) 29,244 lb

sin 2 sin 2(40 )ntAP τθ

≤ = =°

Thus, the maximum load that can be supported by the post is max 29,200 lb 29.2 kipsP = = Ans.


1.33 A 90 mm wide bar will be used to carry an axialtension load of 280 kN. The normal and shear stresseson plane AB must be limited to 150 MPa and 100 MPa,respectively. Determine the minimum thickness trequired for the bar.

Fig. P1.33


(1 cos 2 )2nPA

σ θ= + (a)

and

sin 22ntPA

τ θ= (b)

The angle θ 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):

22

(280 kN)(1,000 N/kN)(1 cos 2 ) (1 cos 2(50 )) 771.2617 mm2 2(150 N/mm )n

PA θσ

≥ + = + ° =

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):

22

(280 kN)(1,000 N/kN)sin 2 sin 2(50 ) 1,378.7309 mm2 2(100 N/mm )nt

PA θτ

≥ = ° =

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

2

min1,378.7309 mm 15.3192 mm 15.32 mm

90 mmt = = = Ans.


1.34 A rectangular bar having width w = 6.00in. and thickness t = 1.50 in. is subjected to atension load P. The normal and shear stresseson plane AB must not exceed 16 ksi and 8 ksi,respectively. Determine the maximum load Pthat can be applied without exceeding eitherstress limit. Fig. P1.34


(1 cos 2 )2nPA

σ θ= + (a)

and

sin 22ntPA

τ θ= (b)

The angle θ for inclined plane AB is calculated from

3tan 3 71.56511

θ θ= = ∴ = °

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):

22 2(9.0 in. )(16 ksi) 1,440 ksi

1 cos 2 1 cos 2(71.5651 )nAP σ

θ≤ = =

+ + °

The shear stress on plane AB is limited to 8 ksi. From Eq. (b), the maximum load P based the shear stress limit is

22 2(9.0 in. )(8 ksi) 240 kips

sin 2 sin 2(71.5651 )ntAP τθ

≤ = =°

Thus, the maximum load that can be supported by the bar is max 240 kipsP = Ans.


1.35 In Fig. P1.35, a rectangular bar having widthw = 2.50 in. and thickness t is subjected to atension load of P = 85 kips. The normal and shearstresses on plane AB must not exceed 16 ksi and 8ksi, respectively. Determine the minimum barthickness t required for the bar.

Fig. P1.35


(1 cos 2 )2nPA

σ θ= + (a)

and

sin 22ntPA

τ θ= (b)

The angle θ for inclined plane AB is calculated from

3tan 3 71.56511

θ θ= = ∴ = °

The normal stress on plane AB is limited to 16 ksi; therefore, the minimum cross-sectional area A required to support P = 85 kips can be found from Eq. (a):

285 kips(1 cos 2 ) (1 cos 2(71.5651 )) 0.5312 in.2 2(16 ksi)n

PA θσ

≥ + = + ° =

The shear stress on plane AB is limited to 8 ksi; therefore, the minimum cross-sectional area A required to support P = 85 kips can be found from Eq. (b):

285 kipssin 2 sin 2(71.5651 ) 3.1875 in.2 2(8 ksi)nt

PA θτ

≥ = ° =

To satisfy both the normal and shear stress requirements, the cross-sectional area must be at least Amin = 3.1875 in.2. Since the bar width is 2.50 in., the minimum bar thickness t must be

2

min3.1875 in. 1.2750 in. 1.275 in.

2.50 in.t = = = Ans.


1.36 The rectangular bar has a width of w = 3.00 in. and a thickness of t = 2.00 in. The normalstress on plane AB of the rectangular blockshown in Fig. P1.36 is 6 ksi (C) when the load Pis 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.

Fig. P1.36

Solution The general equation for normal stress on an inclined plane in terms of the angle θ is

(1 cos 2 )2nPA

σ θ= + (a)

and the angle θ for inclined plane AB is

3tan 0.75 36.86994

θ θ= = ∴ = °

The cross-sectional area of the rectangular bar is A = (3.00 in.)(2.00 in.) = 6.00 in.2. (a) Since the normal stress on plane AB is given as 6 ksi, the magnitude of load P can be calculated from Eq. (a):

22 2(6.0 in. )(6 ksi) 56.25 kips 56.3 kips

1 cos 2 1 cos 2(36.8699 )nAP σ

θ= = = =

+ + ° Ans.

(b) The general equation for shear stress on an inclined plane in terms of the angle θ is

sin 22ntPA

τ θ=

therefore, the shear stress on plane AB is

2

56.25 kips sin 2(36.8699 ) 4.50 ksi2(6.00 in. )ntτ = ° = Ans.

(c) The maximum normal stress at any possible orientation is

max 2

56.25 kips 9.3750 ksi 9.38 ksi6.00 in.

PA

σ = = = = Ans.

and the maximum shear stress at any possible orientation in the block is

max 2

56.25 kips 4.6875 ksi 4.69 ksi2 2(6.00 in. )PA

τ = = = = Ans.


1.37 The rectangular bar has a width of w = 100 mm and a thickness of t = 75 mm. The shear stresson plane AB of the rectangular block shown inFig. P1.37 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.

Fig. P1.37

Solution The general equation for shear stress on an inclined plane in terms of the angle θ is

sin 22ntPA

τ θ= (a)

and the angle θ for inclined plane AB is

3tan 0.75 36.86994

θ θ= = ∴ = °

The cross-sectional area of the rectangular bar is A = (100 mm)(75 mm) = 7,500 mm2. (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 22 2(7,500 mm )(12 N/mm ) 187,500 N 187.5 kN

sin 2 sin 2(36.8699 )ntAP τθ

= = = =°

Ans.

(b) The general equation for normal stress on an inclined plane in terms of the angle θ is

(1 cos 2 )2nPA

σ θ= +

therefore, the normal stress on plane AB is

2

187,500 N (1 cos 2(36.8699 )) 16.00 MPa2(7,500 mm )nσ = + ° = Ans.

(c) The maximum normal stress at any possible orientation is

max 2

187,500 N 25.0 MPa7,500 mm

PA

σ = = = Ans.

and the maximum shear stress at any possible orientation in the block is

max 2

187,500 N 12.50 MPa2 2(7,500 mm )PA

τ = = = Ans.

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