mechanics of material solution maunl

ISM for

Chapter1Stress1.1Introduction1.2NormalStressUnderAxialLoading1.3DirectShearStress1.4BearingStress1.5StressesonInclinedSections1.6EqualityofShearStressesonPerpendicularPlanesChapter2Strain2.1Displacement,Deformation,andtheConceptofStrain2.2NormalStrain2.3ShearStrain2.4ThermalStrainChapter3MechanicalPropertiesofMaterials3.1TheTensionTest3.2TheStressStrainDiagram3.3HookesLaw3.4PoissonsRatioChapter4DesignConcepts4.1Introduction4.2TypesofLoads4.3Safety4.4AllowableStressDesign4.5LoadandResistanceFactorDesignChapter5AxialDeformation5.1Introduction5.2SaintVenantsPrinciple5.3DeformationsinAxiallyLoadedBars5.4DeformationsinaSystemofAxiallyLoadedBars5.5StaticallyIndeterminateAxiallyLoadedMembers5.6ThermalEffectsonAxialDeformation5.7StressConcentrationsChapter6Torsion6.1Introduction6.2TorsionalShearStrain6.3TorsionalShearStress6.4StressesonObliquePlanes6.5TorsionalDeformations6.6TorsionSignConventions6.7GearsinTorsionAssemblies6.8PowerTransmission6.9StaticallyIndeterminateTorsionMembers6.10StressConcentrationsinCircularShaftsUnderTorsionalLoadings6.11TorsionofNoncircularSections6.12TorsionofThinWalledTubes:ShearFlow

Chapter7EquilibriumofBeams7.1Introduction7.2ShearandMomentinBeams7.3GraphicalMethodforConstructingShearandMomentDiagrams7.4DiscontinuityFunctionstoRepresentLoad,Shear,andMomentChapter8Bending8.1Introduction8.2FlexuralStrains8.3NormalStrainsinBeams8.4AnalysisofBendingStressesinBeams8.5IntroductoryBeamDesignforStrength8.6FlexuralStressesinBeamsofTwoMaterials8.7BendingDuetoEccentricAxialLoad8.8UnsymmetricBending8.9StressConcentrationsUnderFlexuralLoadingsChapter9ShearStressinBeams9.1Introduction9.2ResultantForcesProducedbyBendingStresses9.3TheShearStressFormula9.4TheFirstMomentofAreaQ9.5ShearStressesinBeamsofRectangularCrossSection9.6ShearStressesinBeamsofCircularCrossSection9.7ShearStressesinWebsofFlangedBeams9.8ShearFlowinBuiltUpMembersChapter10BeamDeflections10.1Introduction10.2MomentCurvatureRelationship10.3TheDifferentialEquationoftheElasticCurve10.4DeflectionsbyIntegrationofaMomentEquation10.5DeflectionsbyIntegrationofShearForceorLoadEquations10.6DeflectionsUsingDiscontinuityFunctions10.7MethodofSuperpositionChapter11StaticallyIndeterminateBeams11.1Introduction11.2TypesofStaticallyIndeterminateBeams11.3TheIntegrationMethod11.4UseofDiscontinuityFunctionsforStaticallyIndeterminateBeams11.5TheSuperpositionMethod

Chapter12StressTransformations12.1Introduction12.2StressataGeneralPointinanArbitrarilyLoadedBody12.3EquilibriumoftheStressElement12.4TwoDimensionalorPlaneStress12.5GeneratingtheStressElement12.6EquilibriumMethodforPlaneStressTransformation12.7GeneralEquationsofPlaneStressTransformation12.8PrincipalStressesandMaximumShearStress12.9PresentationofStressTransformationResults12.10MohrsCircleforPlaneStress12.11GeneralStateofStressataPointChapter13StrainTransformations13.1Introduction13.2TwoDimensionalorPlaneStrain13.3TransformationEquationsforPlaneStrain13.4PrincipalStrainsandMaximumShearingStrain13.5PresentationofStrainTransformationResults13.6MohrsCircleforPlaneStrain13.7StrainMeasurementandStrainRosettes13.8GeneralizedHookesLawforIsotropicMaterialsChapter14ThinWalledPressureVessels14.1Introduction14.2SphericalPressureVessels14.3CylindricalPressureVessels14.4StrainsinPressureVesselsChapter15CombinedLoads15.1Introduction15.2CombinedAxialandTorsionalLoads15.3PrincipalStressesinaFlexuralMember15.4GeneralCombinedLoadings15.5TheoriesofFailureChapter16Columns16.1Introduction16.2BucklingofPinEndedColumns16.3TheEffectofEndConditionsonColumnBuckling16.4TheSecantFormula16.5EmpiricalColumnFormulas&CentricLoading16.6EccentricallyLoadedColumns

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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 a

compression member. If the 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

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

A D d

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

2 2max allow (200 N/mm )(863.938 mm ) 172,788 172.8 kN NP A Ans.




1.2 A 2024-T4 aluminum tube with an outside diameter of 2.50 in. will be used to support a 27-kip

load. If the 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

2

min

27 kips1.500 in.

18 ksi

P PA

A

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

2 2( )4

A D d

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.) ] 1.500 in.4

4(2.50 in.) (1.500 in. )

4(2.50 in.) (1.500 in. )

2.08330 in.

d

d

d

d

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. 2.08330 in.

0.20835 in. 0.208 in.2 2

D dt

Ans.




1.3 Two solid cylindrical rods (1) and (2) are

joined together at flange B and loaded, as shown in

Fig. P1.3. The diameter of rod (1) is d1 = 24 mm

and the diameter of rod (2) is d2 = 42 mm.

Determine the normal stresses in rods (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 F

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 F

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

2 21 (24 mm) 452.3893 mm4

A

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

11 21

(80 kN)(1,000 N/kN)176.8388 MPa

452.389176

3 m.8 M )

mPa (T

F

A Ans.


2 22 (42 mm) 1,385.4424 mm4

A

Accordingly, the normal stress in rod (2) is

22 22

( 200 kN)(1,000 N/kN)144.3582 MPa

1,385.4144.4 MPa (

424C)

mm

F

A

Ans.





joined together at flange B and loaded, as shown in

Fig. P1.4. If the normal stress in each rod must be

limited to 120 MPa, determine the minimum

diameter required for 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 F

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 F

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

FA

The minimum rod diameter is therefore

2 21,min 1 1666.6667 mm 29.1346 mm 29. mm4

1 A d d

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

FA

The minimum diameter for rod (2) is therefore

2 22,min 2 21,666.6667 mm 46.0659 mm 46. mm4

1 A d d

Ans.





joined together at flange B and loaded, as

shown in Fig. P1.5. If the normal stress in

each rod must be limited to 40 ksi,

determine the minimum diameter required

for each rod.

Fig. P1.5

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

15 kips 15 kips (C)

yF F

F

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

75 kips 75 kips (C)

yF F

F

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

211,min15 kips

0.375 in.40 ksi

FA


2 21,min 1 10.375 in. 0.6909 0.691 9 i4

inn. .A d d

Ans.

Similarly, the normal stress in rod (2) must be limited to 40 ksi, which requires a minimum area of

222,min75 kips

1.875 in.40 ksi

FA

The minimum diameter for rod (2) is therefore

2 22,min 2 21.875 in. 1.54509 1.545 in.7 in.4

A d d

Ans.




1.6 Two solid cylindrical rods (1) and (2) are joined

together at flange B and loaded, as shown in Fig.

P1.6. 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).

Fig. P1.6

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

F

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

75 kips 75 kips (C)

yF F

F


2 21 (1.75 in.) 2.4053 in.4

A

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

11 21

15 kips6.23627 ksi

2.4053 in6.24 ksi )

.(C

F

A

Ans.


2 22 (2.50 in.) 4.9087 in.4

A

Accordingly, the normal stress in rod (2) is

22 22

75 kips15.2789 ksi

2.4053 in.15.28 ksi (C)

F

A

Ans.




1.7 Axial loads are applied with rigid bearing plates to the

solid cylindrical rods shown in Fig. P1.7. 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 normal 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 18 kips 4 kips 4 kips 0 16 kips 16 kips (C)yF F F

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 28 kips 4 kips 4 kips 15 kips 15 kips 0 14 kips 14 kips (T)yF F F

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

8 kips 4 kips 4 kips 15 kips 15 kips 20 kips 20 kips 0

26 kips 26 kips (C)

yF F

F





2 21 (2.00 in.) 3.1416 in.

4A

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

11 21

16 kips5.0930 ksi

3.1416 in5.09 ksi (C)

.

F

A

Ans.


2 22 (1.50 in.) 1.7671 in.

4A

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

22 22

14 kips7.9224 ksi

1.7671 in.7.92 ksi (T)

F

A Ans.

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

2 23 (3.00 in.) 7.0686 in.4

A

and the normal stress in the steel rod is

33 23

26 kips3.6782 ksi

7.0686 in3.68 ksi (C)

.

F

A

Ans.




1.8 Axial loads are applied with rigid bearing plates to the solid

cylindrical rods shown in Fig. P1.8. 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.

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 18 kips 4 kips 4 kips 0 16 kips 16 kips (C)yF F F

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 28 kips 4 kips 4 kips 15 kips 15 kips 0 14 kips 14 kips (T)yF F F

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

8 kips 4 kips 4 kips 15 kips 15 kips 20 kips 20 kips 0

26 kips 26 kips (C)

yF F

F

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 cross-

sectional 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

211,min1

16 kips0.8889 in.

18 ksi

FA


2 21,min 1 10.8889 in. 1.0638 in 1.064 in..4

A d d

Ans.

The normal stress in brass rod (2) must be limited to 25 ksi, which requires a minimum area of

22

2,min

2

14 kips0.5600 in.

25 ksi

FA

which requires a minimum diameter for rod (2) of

2 22,min 2 20.5600 in. 0.8444 in 0.844 in..4

A d d

Ans.

The normal stress in steel rod (3) must be limited to 15 ksi. The minimum cross-sectional area required

for this rod is:

23

3,min

3

26 kips1.7333 in.

15 ksi

FA


2 23,min 3 31.7333 in. 1.4856 in 1.486 in..4

A d d

Ans.




1.9 Two solid cylindrical rods support

a load of P = 50 kN, as shown in Fig.

P1.9. If the normal stress in each rod

must be limited to 130 MPa, determine

the minimum diameter required for

each rod.

Fig. P1.10

Solution

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

4.0 m

tan 1.600 57.99462.5 m

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

2.3 m

tan 0.7188 35.70673.2 m

Write equilibrium equations for the sum of forces in the

horizontal and vertical directions. Note: Rods (1) and (2)

are two-force members.

2 1cos(35.7067 ) cos(57.9946 ) 0xF F F (a)

2 1sin(35.7067 ) sin(57.9946 ) 0yF F F P (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(57.9946 )

cos(35.7067 )F F

(c)

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

1 1

1

1

cos(57.9946 )sin(35.7067 ) sin(57.9946 )

cos(35.6553 )

cos(57.9946 ) tan(35.7067 ) sin(57.9946 )

cos(57.9946 ) tan(35.7067 ) sin(57.9946 ) 1.2289

F F P

F P

P PF

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

150 kN

40.6856 kN1.2289

F




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

2 1

cos(57.9946 ) cos(57.9946 )(40.6856 kN) 26.5553 kN

cos(35.7067 ) cos(35.7067 )F F

The normal stress in rod (1) must be limited to 130 MPa; therefore, the minimum cross-sectional area

required for rod (1) is

21

1,min 2

1

(40.6856 kN)(1,000 N/kN)312.9664 mm

130 N/mm

FA


2 21,min 1 1312.9664 mm 19.9620 19.4

96 mmmmA d d

Ans.

The minimum area required for rod (2) is

22

2,min 2

2

(26.5553 kN)(1,000 N/kN)204.2718 mm

130 N/mm

FA


2 22,min 2 2204.2718 mm 16.1272 16.4

13 mmmmA d d

Ans.




1.10 Two solid cylindrical rods

support a load of P = 27 kN, as

shown in Fig. P1.10. Rod (1) has a

diameter of 16 mm and the diameter

of rod (2) is 12 mm. Determine the

normal stress in each rod.

Fig. P1.10

Solution

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

4.0 m

tan 1.600 57.99462.5 m

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

2.3 m

tan 0.7188 35.70673.2 m

Write equilibrium equations for the sum of forces in the

horizontal and vertical directions. Note: Rods (1) and (2)

are two-force members.

2 1cos(35.7067 ) cos(57.9946 ) 0xF F F (a)

2 1sin(35.7067 ) sin(57.9946 ) 0yF F F P (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(57.9946 )

cos(35.7067 )F F

(c)

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

1 1

1

1

cos(57.9946 )sin(35.7067 ) sin(57.9946 )

cos(35.6553 )

cos(57.9946 ) tan(35.7067 ) sin(57.9946 )

cos(57.9946 ) tan(35.7067 ) sin(57.9946 ) 1.2289

F F P

F P

P PF

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

127 kN

21.9702 kN1.2289

F




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

2 1

cos(57.9946 ) cos(57.9946 )(21.9702 kN) 14.3399 kN

cos(35.7067 ) cos(35.7067 )F F

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

2 21 (16 mm) 201.0619 mm4

A

and the normal stress in rod (1) is:

211 21

(21.9702 kN)(1,000 N/kN)109.2710 N/mm

201.0109.3 MPa (T)

619 mm

F

A Ans.

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

2 22 (12 mm) 113.0973 mm4

A

and the normal stress in rod (2) is:

22

2 2

2

(14.3399 kN)(1,000 N/kN)126.7924 N/mm

113.0126.8 MPa (T)

973 mm

F

A Ans.




1.11 A simple pin-connected truss is loaded

and supported as shown in Fig. P1.11. 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.

Fig. 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:

2 kips

2 ki

0

ps

x

x

xF A

A

(6 ft) (5 kips)(14 ft) (2 kips)(7 ft)

14 kips

0

y

A yB

B

M

5 kips 0

9 kips

y y y

y

F A B

A

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 fttan 0.50 26.565

14 ft

7 fttan 0.875 41.186

8 ft

AC AC

BC BC




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.

cos(26.565 ) 0x AC AB xF F F A (a)

sin(26.565 ) 0y AC yF F A (b)

Solve Eq. (b) for FAC:

9 kips

sin(26.565 ) sin(26.520.125 kip

65 )s

y

AC

AF

and then compute FAB using Eq. (a):

cos(26.565 )

(20.125 kips)cos(26.5 16.000 kips65 ) ( 2 kips)

AB AC xF F A

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.

cos(41.186 ) 0x AB BCF F F (c)

sin(41.186 ) 0y BC yF F B (d)

Solve Eq. (d) for FBC:

14 kips

sin(41.186 ) sin(41.1821.260 kip

6s

)

y

BC

BF

Eq. (c) can be used as a check on our calculations:

cos(41.186 )

( 16.000 kips) ( 21.260 kips)cos(41.186 ) 0

x AB BCF F F

Checks!

Section properties:

For each of the three truss members:

2 2 24.00 in. 2(0.226 in.) 3.548 in. (4.00 in.) (3.548 in.) 2.67954 in.4

d A

Normal stress in each truss member:

2

16.000 kips5.971 ksi

2.679545.97 ksi (C)

in.

ABAB

AB

F

A

Ans.

2

20.125 kips7.510 ksi

2.679547.51 ksi (T)

in.

ACAC

AC

F

A Ans.

2

21.260 kips7.934 ksi

2.679547.93 ksi (C)

in.

BCBC

BC

F

A

Ans.







have an outside diameter of 60 mm and a wall

thickness of 4 mm. Determine the normal


Fig. P1.12

Solution



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:

12 k

12

N 0

kNx

x xF A

A

(1 m) (15 kN)(4.3 m) 0

64.5 kNy

A yB

B

M

15 kN

49.5 kN

0

y

y y yF

A

A B

Method of joints:


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 mtan 1.50 56.310

1.0 m

1.5 mtan 0.454545 24.444

3.3 m

AB AB

BC BC




Joint A:




reaction forces, Ax and Ay, act at joint A. Tension forces will be assumed

in each truss member.

cos(56.310 ) 0x AC AB xF F F A (a)

sin(56.310 ) 0y y ABF A F (b)

Solve Eq. (b) for FAB:

49.5 kN

sin(56.310 ) sin(56.310 )59.492 kN

y

AB

AF

and then compute FAC using Eq. (a):

cos(56.310 )

( 59.492 kN)cos(56.3 45.000 10 ) ( 12 kN) kN

AC AB xF F A

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.

cos(24.444 ) 12 kN 0x AC BCF F F (c)

sin(24.444 ) 15 kN 0y BCF F (d)

Solve Eq. (d) for FBC:

15 kN

sin(24.444 )36.249 kNBCF

Eq. (c) can be used as a check on our calculations:

cos(24.444 ) 12 kN 0

(45.000 kN) ( 36.249 kN)cos(24.444 ) 12 kN 0

x AC BCF F F

Checks!

Section properties:


2 2 260 mm 2(4 mm) 52 mm (60 mm) (52 mm) 703.7168 mm4

d A


2

( 59.492 kN)(1,000 N/kN)84.539 MPa

7084.5 MPa (C)

3.7168 mm

ABAB

AB

F

A

Ans.

2

(45.000 kN)(1,000 N/kN)63.946 MPa

7063.9 MPa

3.7168)

mm(TACAC

AC

F

A Ans.

2

( 36.249 kN)(1,000 N/kN)51.511 MPa

7051.5 MPa (C)

3.7168 mm

BCBC

BC

F

A

Ans.







have an outside diameter of 42 mm and a wall

thickness of 3.5 mm. Determine the normal


Fig. P1.13

Solution


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:

30 kN

30 kN

0y

y

yF A

A

(30 kN)(4.5 m) (15 kN)(1.6 m) (5.6 m)

19.821 kN

0

x

A x

B

M B

15 kN 0

15 kN 15 kN ( 19.821 kN 34.821 ) kN

x x

x x

x

x

F A B

A AB

Method of joints:


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 mtan 0.355556 19.573

4.5 m

4 mtan 0.888889 41.634

4.5 m

AC AC

BC BC




Joint A:






cos(19.573 ) 0x x ACF A F (a)

sin(19.573 ) 0y y AC ABF A F F (b)

Solve Eq. (a) for FAC:

34.821 kN

cos(19.573 ) cos(19.573 )36.957 kNxAC

AF

and then compute FAB using Eq. (b):

sin(19.573 )

(30.000 kN) (36.957 kN)sin(19. 17.619 573 ) kN

AB y ACF A F

Joint B:




each truss member.

cos(41.634 ) 0x x BCF B F (c)

sin(41.634 ) 0y BC ABF F F (d)

Solve Eq. (c) for FBC:

( 19.821 kN)

cos(41.634 ) cos(41.634 )26.520 kNxBC

BF

Eq. (d) can be used as a check on our calculations:

sin(41.634 )

( 26.520 kN)sin(41.634 ) (17.619 kN) 0

y BC ABF F F

Checks!

Section properties:


2 2 242 mm 2(3.5 mm) 35 mm (42 mm) (35 mm) 423.3296 mm4

d A


2

(17.619 kN)(1,000 N/kN)41.620 MPa

4241.6 MPa

3.3296)

mm(TABAB

AB

F

A Ans.

2

(36.957 kN)(1,000 N/kN)87.301 MPa

4287.3 MPa

3.3296)

mm(TACAC

AC

F

A Ans.

2

( 26.520 kN)(1,000 N/kN)62.647 MPa

4262.6 MPa (C)

3.3296 mm

BCBC

BC

F

A

Ans.




1.14 The members of the truss shown in Fig.

P1.14 are aluminum pipes that have an outside

diameter of 4.50 in. and a wall thickness of

0.237 in. Determine the normal stress in each

truss member.

Fig. P1.14

Solution



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:

(15 kips)cos5

9.642 k

0

ip

0

sx

x xF A

A

(4 ft) (15 kips)(4 ft)cos50 (15 kips)(18 ft)sin50 0

61.350 kips

(15 kips)sin50

49.859 kips

0

A y

y y y

y

y

M B

F A B

B

A

Method of joints:


the truss will be used to determine the magnitude of the inclination angles of members AB, AC, and BC.

Use the definition of the tangent function to determine AB, AC, and BC:

6 fttan 1.5 56.3099

4 ft

4 fttan 0.222222 12.5288

18 ft

10 fttan 0.714286 35.5377

14 ft

AB AB

AC AC

BC BC




Joint A:






cos(12.5288 ) cos(56.3099 ) 0x AC AB xF F F A (a)

sin(12.5288 ) sin(56.3099 ) 0y AC AB yF F F A (b)

Solve Eqs. (a) and (b) simultaneously to obtain:

49.948 kips

38.259 kips

AB

AC

F

F

Joint B:




each truss member.

cos(35.5377 ) cos(56.3099 ) 0x BC ABF F F (c)

sin(35.5377 ) sin(56.3099 ) 0y BC AB yF F F B (d)

Solve Eq. (c) for FBC:

cos(56.3099 ) cos(56.3099 )

( 49.9484)cos(35.5377 ) cos(35.5377

34.048 k s)

ipBC ABF F

Eq. (d) can be used as a check on our calculations:

sin(35.5377 ) sin(56.3099 )

( 34.0485 kips)sin(35.5377 ) ( 49.9484 kips)sin(56.3099 ) 61.350 kips 0

y BC AB yF F F B

Checks!

Section properties:


2 2 24.50 in. 2(0.237 in.) 4.026 in. (4.50 in.) (4.026 in.) 3.17405 in.4

d A


2

49.948 kips15.736 ksi

3.17405 15.74 ksi (C

in)

.

ABAB

AB

F

A

Ans.

2

38.259 kips12.054 ksi

3.17405 12.05 ksi (T

n)

i .

ACAC

AC

F

A Ans.

2

34.048 kips10.727 ksi

3.17405 10.73 ksi (C

in)

.

BCBC

BC

F

A

Ans.




1.15 Bar (1) in Fig. P1.15 has a cross-

sectional 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.

Fig. P1.15

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

21,max 1 1 (30 ksi)(0.75 in. ) 22.5 kipsF A

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) 0

6 ft

10 ft

AM F P

P F

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)10 ft 10 ft

13.50 kipsP F Ans.




1.16 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).

Fig. 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

0

10,000 lb5,000 lb

2

xF P V V

V

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

2

min

5,000 lb41.6667 in.

120 psiA

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

2

glue joint

41.6667 in.6.9444 in.

6 in.L

Although weve 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.




1.17 For the clevis connection shown in Fig. P1.17,

determine the shear stress in the 22-mm diameter bolt

for an applied load of P = 90 kN.

Fig. P1.17

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

90 kN

0 45 kN2

xF P V V V

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

2 2 2bolt bolt (22 mm) 380.1327 mm4 4

A d

Therefore, the shear stress in the bolt is

2

2

bolt

(45 kN)(1,000 N/kN)118.3797 N/mm

380.1118.4 MPa

327 mm

V

A Ans.




1.18 For the clevis connection shown in Fig.

P1.18, the shear stress in the 3/8 in. diameter

bolt must be limited to 36 ksi. Determine the

maximum load P that may be applied to the

connection.

Fig. P1.18

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 V

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

2 2 2 2bolt bolt (3 / 8 in.) (0.3750 in.) 0.1104466 in.4 4 4

A d

If the shear stress in the bolt must be limited to 36 ksi, the maximum shear force V on a single cross-

sectional surface must be limited to

2

bolt (36 ksi)(0.1104466 in. ) 3.976078 kipsV A

Therefore, the maximum load P that may be applied to the connection is

2 2(3.976078 kips) 7.952156 k 7ips .95 kipsP V Ans.




1.19 For the connection shown in Fig. P1.19,

determine the average shear stress in the 7/8-in.

diameter bolts if the load is P = 45 kips.

Fig. P1.19

Solution

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

2 2 2 2bolt bolt (7 / 8 in.) (0.875 in.) 0.6013205 in.4 4 4

A d

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

2 2

bolt5 5(0.6013205 in. ) 3.006602 in.VA A

Therefore, the shear stress in each bolt is

2

45 kips14.96706 ksi

3.006602 in.14.97 ksi

V

P

A Ans.




1.20 The five-bolt connection shown in Fig.

P1.20 must support an applied load of P = 300

kN. If the average shear stress in the bolts must

be limited to 225 MPa, determine the minimum

bolt diameter that may be used in the connection.

Fig P1.20

Solution

To support a load of 300 kN while not exceeding an average shear stress of 225 MPa, the total shear

area provided by the bolts must be at least

22

(300 kN)(1,000 N/kN)1,333.3333 mm

225 N/mmV

PA

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

2

bolt

1,333.3333 mm266.6667 mm

5 5

VAA

The minimum bolt diameter is therefore

2 2bolt bolt bolt 18.43 266.6667 mm 18.42 mm64 mm4

A d d

Ans.




1.21 The three-bolt connection shown in Fig. P1.21

must support an applied load of P = 40 kips. If the

average shear stress in the bolts must be limited to

24 ksi, determine the minimum bolt diameter that

may be used in the connection.

Fig. P1.21

Solution

The shear force V that must be provided by the bolts equals the applied load of P = 40 kips. The total

shear area required is thus

240 kips

1.66667 in.24 ksi

V

VA

The three bolts in this connection act in double shear; therefore, six cross-sectional bolt surfaces are

available to transmit shear stress.

2

2

bolt

1.66667 in.0.27778 in. per surface

(2 surfaces per bolt)(3 bolts) 6 surfaces

VAA

The minimum bolt diameter must be

2 2bolt bolt0.27778 in. 0.59471 in. 0.595 i .4

nd d

Ans.




1.22 For the connection shown in Fig. P1.22,

the average shear stress in the 12-mm-diameter

bolts must be limited to 160 MPa. Determine

the maximum load P that may be applied to the

connection.

Fig. P1.22

Solution

The cross-sectional area of a 12-mm-diameter bolt is

2 2 2bolt bolt (12 mm) 113.097355 mm4 4

A d

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

2 2

bolt2(3 bolts) 2(3 bolts)(113.097355 mm ) 678.58401 mmVA A

Since the shear stress must be limited to 160 MPa, the total shear force that can be resisted by the three

bolts is

2 2

max (160 N/mm )(678.58401 mm ) 108,573.442 NVV A

In this connection, the shear force in the bolts is equal to the applied load P; therefore,

max 108.6 kNP Ans.




1.23 A hydraulic punch press is used to

punch a slot in a 0.50-in. thick plate, as

illustrated in Fig. P1.23. If the plate shears

at a stress of 30 ksi, determine the

minimum force P required to punch the

slot.

Fig. P1.23

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

2perimeter plate thickness (8.35619 in.)(0.50 in.) 4.17810 in.VA

Given that the plate shears at = 30 ksi, the force required to remove the slug is therefore

2min (30 ksi)(4.17810 in. ) 125.343 kips 125.3 kipsVP A Ans.




1.24 A coupling is used to connect a 2 in. diameter

plastic pipe (1) to a 1.5 in. diameter pipe (2), as

shown in Fig. P1.24. 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.

Fig. P1.24

Solution

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

2

adhesive

5,000 lb12.5 in.

400 psiV

VA

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 D

The minimum length L1 is therefore

2

1

1

12.5 in.1.9894 in.

6.2832 i1.989 i

n n.

.

VALC

Ans.

Consider the coupling on pipe (2). The circumference C2 of pipe (2) is

2 2 (1.5 in.) 4.7124 in.C D

The minimum length L2 is therefore

2

2

2

12.5 in.2.6526 in.

4.7124 2.65 in.

in.

VALC

Ans.




1.25 A lever is attached to a shaft with a

square shear key, as shown in Fig. P1.25.

The force applied to the lever is P = 400 N.

If the shear stress in the key must not exceed

90 MPa, determine the minimum dimension

a that must be used if the key is 15 mm long.

Fig. P1.25

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):

50 mm

(400 N)(750 mm) 0 12,000 N2

OM V V

Since the shear stress in the key must not exceed 90 MPa, the shear area required is

22

12,000 N133.3333 mm

90 N/mmV

VA

The shear area in the key is given by the product of its length L (i.e., 15 mm) and its width a. Therefore,

the minimum key width a is

2133.3333 mm

8.8889 mm15

8.89 mmmm

VAaL

Ans.




1.26 A common trailer hitch connection is shown in

Fig. P1.26. 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 used

for the pin.

Fig. P1.26

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

VA

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

2

pin pin

0.1333 in.2 0.0667 in.

2 2

V

V

AA A A

and the minimum pin diameter is

2 2pin pin0.0667 in. 0.2913 in. 0.291 n.4

id d

Ans.




1.27 An axial load P is supported by a short steel

column, which has a cross-sectional area of

11,400 mm2. If the average normal stress in the steel

column must not exceed 110 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.

Fig. P1.27

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 A

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

2

min 2

1,254,000 N156,750 mm

8 N/mmb

PA

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

2

min

396

156,

mm

750 mm

395.9167 mm

A a a

a

Ans.




1.28 The steel pipe column shown in Fig. 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.

Fig. P1.28

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 4

A D d

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

2

80 kips12.1623 ksi

6.5777 in.12.16 ksib

b

P

A 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

PA

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.beam width 10.75

14.88 in

.in.

bAL Ans.

(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

PA

Since the lower bearing plate is square, its dimension a must be

288.8889 in. 9.4 9.43 in281 n. .ibA a a a Ans.




1.29 A vertical shaft is supported by a thrust

collar and bearing plate, as shown in Fig.

P1.29. The average shear stress in the collar

must be limited to 18 ksi. The average bearing

stress between the collar and the plate must be

limited to 24 ksi. Based on these limits,

determine the maximum axial load P that can

be applied to the shaft.

Fig. P1.29

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

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 A

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.4

A

is reduced by the area taken up by the shaft

2 2shaft (1.0 in.) 0.7854 in.4

A

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 A

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.




1.30 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)

(25 mm)(75 mm)80 MPa

F

A Ans.

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

maxmax2

40 MPa

Ans.




1.31 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

2minmax

92 kips3.0667 in.

30 ksi

FA

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

2

min

max

92 kips3.8333 in.

2 2(12 ksi)

FA

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.4

d d

Ans.




1.32 An axial load P is applied to the

rectangular bar shown in Fig. P1.32. 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.

Fig. P1.32

Solution

The angle for the inclined plane is 35. The normal force N perpendicular to plane AB is

found from

cos (40 kN)cos35 57.3406 kNN P

and the shear force V parallel to plane AB is

sin (70 kN)sin35 40.1504 kNV P

The cross-sectional area of the bar is 400 mm2, but the area along inclined plane AB is

2

2400 mm 488.3098 mmcos cos35

n

AA

The normal stress n perpendicular to plane AB is

2

(57.3406 kN)(1,000 N/kN)117.4268 MPa

488117.4 MP

.3098 mman

n

N

A Ans.

The shear stress nt parallel to plane AB is

2

(40.1504 kN)(1,000 N/kN)82.2231 MPa

482.2 MP

88.309 a

8 mmnt

n

V

A Ans.




1.33 An axial load P is applied to the 1.75 in.

by 0.75 in. rectangular bar shown in Fig.

P1.33. 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.

Fig. P1.33

Solution

The angle for the inclined plane is 60. The normal force N perpendicular to plane AB is

found from

cos (18 kips)cos60 9.0 kipsN P


sin (18 kips)sin60 15.5885 kipsV P

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

2

21.3125 in./ cos 2.6250 in.cos60

nA A


2

9.0 kips3.4286 ksi

2.6250 in3.43 ks

.in

n

N

A Ans.


2

15.5885 kips5.9385 ksi

2.62505.94 ks

ii

n.nt

n

V

A Ans.




1.34 A compression load of P = 80 kips is applied to a 4 in.

by 4 in. square post, as shown in Fig. P1.34. Determine the

normal stress perpendicular to plane AB and the shear stress

parallel to plane AB.

Fig. P1.34

Solution

The angle 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


sin (80 kips)sin55 65.5322 kipsV P

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.cos55

nA A


2

45.8861 kips1.6449 ksi

27.8951 1.645 ksi

in.n

n

N

A Ans.


2

65.5322 kips2.3492 ksi

27.89512.35 ks

ii

n.nt

n

V

A Ans.




1.35 Specifications for the 50 mm 50 mm square bar

shown in Fig. P1.35 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.

Fig. P1.35

Solution

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

(1 cos 2 )2

n

P

A (a)

and

sin 22

nt

P

A (b)

The cross-sectional area of the square bar is A = (50 mm)2 = 2,500 mm

2, 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 22 2(2,500 mm )(120 N/mm )

911,882 N1 cos2 1 cos2(55 )

nAP

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 22 2(2,500 mm )(90 N/mm )

478,880 Nsin 2 sin 2(55 )

ntAP

Thus, the maximum load that can be supported by the bar is

max 479 kNP Ans.




1.36 Specifications for the 6 in. 6 in. square post shown in

Fig. P1.36 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.

Fig. P1.36

Solution


(1 cos 2 )2

n

P

A (a)

and

sin 22

nt

P

A (b)

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):

22 2(36 in. )(800 psi)

49,078 lb1 cos2 1 cos2(40 )

nAP

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 lbsin 2 sin 2(40 )

ntAP

Thus, the maximum load that can be supported by the post is

max 29,200 l 29.2 ipb k sP Ans.




1.37 A 90 mm wide bar will be used to carry an axial

tension load of 280 kN. 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.

Fig. P1.37

Solution


(1 cos 2 )2

n

P

A (a)

and

sin 22

nt

P

A (b)

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):

2

2

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

2 2(150 N/mm )n

PA

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):

2

2

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

2 2(100 N/mm )nt

PA

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

min

1,378.7309 mm15.3192 mm

90 m15.3

m2 mmt Ans.




1.38 A rectangular bar having width w = 6.00

in. and thickness t = 1.50 in. is subjected to a

tension load P. 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. Fig. P1.38

Solution


(1 cos 2 )2

n

P

A (a)

and

sin 22

nt

P

A (b)

The angle for inclined plane AB is calculated from

3

tan 3 71.56511

The cross-sectional area of the bar is A = wt = (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 ksi1 cos2 1 cos2(71.5651 )

nAP

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 kipssin 2 sin 2(71.5651 )

ntAP

Thus, the maximum load that can be supported by the bar is

max 240 kipsP Ans.




1.39 In Fig. P1.39, 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.

Fig. P1.39

Solution


(1 cos 2 )2

n

P

A (a)

and

sin 22

nt

P

A (b)

The angle for inclined plane AB is calculated from

3

tan 3 71.56511

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):

230 kips(1 cos2 ) (1 cos2(71.5651 )) 0.2500 in.

2 2(12 ksi)n

PA

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):

230 kipssin 2 sin 2(71.5651 ) 1.1250 in.

2 2(8 ksi)nt

PA

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

2

min

1.1250 in.0.900 in.

1.25 in0.900 .

. int Ans.




1.40 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 Fig. P1.40 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.

Fig. P1.40

Solution

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

(1 cos 2 )2

n

P

A (a)

and the angle for inclined plane AB is

3

tan 0.75 36.86994

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 kips1 cos2 1 co

56.3 kips2(36.8699 )

snA

P

Ans.

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

sin 22

nt

P

A

therefore, the shear stress on plane AB is

2

56.25 kipssin 2(36.8699 )

2(6.00 in. )4.50 ksint Ans.

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

max 256.25 kips

9.3750 ksi6.00 i

9.38 kn

si.

P

A Ans.

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

max 256.25 kips

4.6875 ksi2 2(6.00

4.69 ksiin. )

P

A Ans.




1.41 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

Fig. P1.41 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. P1.41

Solution

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

sin 22

nt

P

A (a)

and the angle for inclined plane AB is

3

tan 0.75 36.86994

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 Nsin 2 sin 2(36.869

187.5 k9

N)

ntAP

Ans.

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

(1 cos 2 )2

n

P

A

therefore, the normal stress on plane AB is

2

187,500 N(1 cos2(36.8699 ))

2(7,500 mm16.00 MPa

)n Ans.

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

max 2187,500 N

7,500 mm25.0 MPa

P

A Ans.

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

max 2187,500 N

2 2(7,500 mm12.50 MPa

)

P

A Ans.




2.1 When an axial load is applied to the

ends of the bar shown in Fig. P2.1, the total

elongation of the bar between joints A and C

is 0.15 in. In segment (2), the normal strain

is measured as 1,300 in./in. Determine:

(a) the elongation of segment (2). (b) the normal strain in segment (1) of the bar.

Fig. P2.1

Solution

(a) From the definition of normal strain, the elongation in segment (2) can be computed as

62 2 2 (1,300 10 )(90 in. 0.1170 i) n.L Ans.

(b) The combined elongations of segments (1) and (2) is given as 0.15 in. Therefore, the elongation that

mechanics of material solution maunl

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