cvg 2140 solution

Typewritten TextINSTRUCTORSOLUTIONSMANUAL

CONTENTS Philpot MoM 3rd

1. Stress 1.1 Introduction 1.2 Normal Stress Under Axial Loading 1.3 Direct Shear Stress 1.4 Bearing Stress 1.5 Stresses on Inclined Sections 1.6 Equality of Shear Stresses on Perpendicular Planes

2. Strain 2.1 Displacement, Deformation, and the Concept of Strain 2.2 Normal Strain 2.3 Shear Strain 2.4 Thermal Strain

3. Mechanical Properties of Materials 3.1 The Tension Test 3.2 The StressStrain Diagram 3.3 Hookes Law 3.4 Poissons Ratio

4. Design Concepts 4.1 Introduction 4.2 Types of Loads 4.3 Safety 4.4 Allowable Stress Design 4.5 Load and Resistance Factor Design

5. Axial Deformation 5.1 Introduction 5.2 Saint-Venants Principle 5.3 Deformations in Axially Loaded Bars 5.4 Deformations in a System of Axially Loaded Bars 5.5 Statically Indeterminate Axially Loaded Members 5.6 Thermal Effects on Axial Deformation 5.7 Stress Concentrations

6. Torsion 6.1 Introduction 6.2 Torsional Shear Strain 6.3 Torsional Shear Stress 6.4 Stresses on Oblique Planes 6.5 Torsional Deformations 6.6 Torsion Sign Conventions 6.7 Gears in Torsion Assemblies

6.8 Power Transmission 6.9 Statically Indeterminate Torsion Members 6.10 Stress Concentrations in Circular Shafts Under Torsional Loadings 6.11 Torsion of Noncircular Sections 6.12 Torsion of Thin-Walled Tubes: Shear Flow

7. Equilibrium of Beams 7.1 Introduction 7.2 Shear and Moment in Beams 7.3 Graphical Method for Constructing Shear and Moment Diagrams 7.4 Discontinuity Functions to Represent Load, Shear, and Moment

8. Bending 8.1 Introduction 8.2 Flexural Strains 8.3 Normal Stresses in Beams 8.4 Analysis of Bending Stresses in Beams 8.5 Introductory Beam Design for Strength 8.6 Flexural Stresses in Beams of Two Materials 8.7 Bending Due to Eccentric Axial Load 8.8 Unsymmetric Bending 8.9 Stress Concentrations Under Flexural Loadings

9. Shear Stress in Beams 9.1 Introduction 9.2 Resultant Forces Produced by Bending Stresses 9.3 The Shear Stress Formula 9.4 The First Moment of Area Q 9.5 Shear Stresses in Beams of Rectangular Cross Section 9.6 Shear Stresses in Beams of Circular Cross Section 9.7 Shear Stresses in Webs of Flanged Beams 9.8 Shear Flow in Built-Up Members 9.9 Shear Stress and Shear Flow in Thin-Walled Members 9.10 Shear Centers of Thin-Walled Open Sections

10. Beam Deflections 10.1 Introduction 10.2 Moment-Curvature Relationship 10.3 The Differential Equation of the Elastic Curve 10.4 Deflections by Integration of a Moment Equation 10.5 Deflections by Integration of Shear-Force or Load Equations 10.6 Deflections Using Discontinuity Functions 10.7 Method of Superposition

11. Statically Indeterminate Beams 11.1 Introduction 11.2 Types of Statically Indeterminate Beams 11.3 The Integration Method 11.4 Use of Discontinuity Functions for Statically Indeterminate Beams 11.5 The Superposition Method

12. Stress Transformations 12.1 Introduction 12.2 Stress at a General Point in an Arbitrarily Loaded Body 12.3 Equilibrium of the Stress Element 12.4 Plane Stress 12.5 Generating the Stress Element 12.6 Equilibrium Method for Plane Stress Transformations 12.7 General Equations of Plane Stress Transformation 12.8 Principal Stresses and Maximum Shear Stress 12.9 Presentation of Stress Transformation Results 12.10 Mohrs Circle for Plane Stress 12.11 General State of Stress at a Point

13. Strain Transformations 13.1 Introduction 13.2 Two-Dimensional or Plane Strain 13.3 Transformation Equations for Plane Strain 13.4 Principal Strains and Maximum Shearing Strain 13.5 Presentation of Strain Transformation Results 13.6 Mohrs Circle for Plane Strain 13.7 Strain Measurement and Strain Rosettes 13.8 Generalized Hookes Law for Isotropic Materials

14. Thin-Walled Pressure Vessels 14.1 Introduction 14.2 Spherical Pressure Vessels 14.3 Cylindrical Pressure Vessels 14.4 Strains in Pressure Vessels

15. Combined Loads 15.1 Introduction 15.2 Combined Axial and Torsional Loads 15.3 Principal Stresses in a Flexural Member 15.4 General Combined Loadings 15.5 Theories of Failure

16. Columns 16.1 Introduction 16.2 Buckling of Pin-Ended Columns 16.3 The Effect of End Conditions on Column Buckling 16.4 The Secant Formula 16.5 Empirical Column FormulasCentric Loading 16.6 Eccentrically Loaded Columns

17. Energy Methods 17.1 Introduction 17.2 Work and Strain Energy 17.3 Elastic Strain Energy for Axial Deformation 17.4 Elastic Strain Energy for Torsional Deformation 17.5 Elastic Strain Energy for Flexural Deformation 17.6 Impact Loading 17.7 Work-Energy Method for Single Loads 17.8 Method of Virtual Work 17.9 Deflections of Trusses by the Virtual-Work Method 17.10 Deflections of Beams by the Virtual-Work Method 17.11 Castiglianos Second Theorem 17.12 Calculating Deflections of Trusses by Castiglianos Theorem 17.13 Calculating Deflections of Beams by Castiglianos Theorem

Appendix A: Geometric Properties of an Area A.1 Centroid of an Area A.2 Moment of Inertia for an Area A.3 Product of Inertia for an Area A.4 Principal Moments of Inertia A.5 Mohrs Circle for Principal Moments of Inertia

Appendix B: Geometric Properties of Structural Steel Shapes

Appendix C: Table of Beam Slopes and Deflections

Appendix D: Average Properties of Selected Materials

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P1.1 A stainless steel tube with an outside diameter of 60 mm and a wall thickness of 5 mm is used as

a compression member. If the axial normal stress in the member must be limited to 200 MPa,

determine the maximum load P that the member can support.

Solution

The cross-sectional area of the stainless steel tube is

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.




P1.2 A 2024-T4 aluminum tube with an outside diameter of 2.50 in. will be used to support a 27-kip

load. If the axial normal stress in the member must be limited to 18 ksi, determine the wall thickness

required for the tube.

Solution

From the definition of normal stress, solve for the minimum area required to support a 27-kip load

without exceeding a stress of 18 ksi

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

min

2.50 in. 2.08330 in.0.20835 in. 0.208 in.

2 2

D dt

Ans.




P1.3 Two solid cylindrical rods (1) and (2)

are joined together at flange B and loaded, as

shown in Figure P1.3/4. If the normal stress

in each rod must be limited to 40 ksi,

determine the minimum diameter required

for each rod.

FIGURE P1.3/4

Solution

Cut a FBD through rod (1). The FBD should include the free end of the rod at A.

As a matter of course, we will assume that the internal force in rod (1) is tension

(even though it obviously will be in compression). From equilibrium,

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,min

15 kips0.375 in.

40 ksi

FA

The minimum rod diameter is therefore

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.




P1.4 Two solid cylindrical rods (1) and (2) are

joined together at flange B and loaded, as shown in

Figure P1.3/4. The diameter of rod (1) is 1.75 in.

and the diameter of rod (2) is 2.50 in. Determine the

normal stresses in rods (1) and (2).

FIGURE P1.3/4

Solution

Cut a FBD through rod (1). The FBD should include the free end of the rod at A. We

will assume that the internal force in rod (1) is tension (even though it obviously will

be in compression). From equilibrium,

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

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

2 21 (1.75 in.) 2.4053 in.

4A

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.

4A

Accordingly, the normal stress in rod (2) is

22 22

75 kips15.2789 ksi

2.4053 in.15.28 ksi (C)

F

A

Ans.




P1.5 Axial loads are applied with rigid bearing plates to the

solid cylindrical rods shown in Figure P1.5/6. The diameter

of aluminum rod (1) is 2.00 in., the diameter of brass rod (2)

is 1.50 in., and the diameter of steel rod (3) is 3.00 in.

Determine the axial normal stress in each of the three rods.

FIGURE P1.5/6

Solution

Cut a FBD through rod (1). The FBD should include the free end A. We will assume that the internal

force in rod (1) is tension (even though it obviously will be in compression). From equilibrium,

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.

4A

and the normal stress in the steel rod is

33 23

26 kips3.6782 ksi

7.0686 in3.68 ksi (C)

.

F

A

Ans.




P1.6 Axial loads are applied with rigid bearing plates to the

solid cylindrical rods shown in Figure P1.5/6. The normal

stress in aluminum rod (1) must be limited to 18 ksi, the

normal stress in brass rod (2) must be limited to 25 ksi, and

the normal stress in steel rod (3) must be limited to 15 ksi.

Determine the minimum diameter required for each of the

three rods.

FIGURE P1.5/6

Solution

The internal forces in the three rods must be determined. Begin with a FBD cut through rod (1) that

includes the free end A. We will assume that the internal force in rod (1) is tension (even though it

obviously will be in compression). From equilibrium,

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

21

1,min

1

16 kips0.8889 in.

18 ksi

FA


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

4A 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..

4A 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.




P1.7 Two solid cylindrical rods support a load of

P = 50 kN, as shown in Figure P1.7/8. If the

normal stress in each rod must be limited to 130

MPa, determine the minimum diameter required

for each rod.

FIGURE P1.7/8

Solution

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

4.0 m

tan 1.600 57.99462.5 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.

496 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.

413 mmmmA d d

Ans.




P1.8 Two solid cylindrical rods support a load

of P = 27 kN, as shown in Figure P1.7/8. Rod

(1) has a diameter of 16 mm and the diameter

of rod (2) is 12 mm. Determine the axial

normal stress in each rod.

FIGURE P1.7/8

Solution

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

4.0 m

tan 1.600 57.99462.5 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 mm

4A

and the normal stress in rod (1) is:

21

1 2

1

(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 mm

4A

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.




P1.9 A simple pin-connected truss is loaded

and supported as shown in Figure P1.9. All

members of the truss are aluminum pipes that

have an outside diameter of 4.00 in. and a wall

thickness of 0.226 in. Determine the normal

stress in each truss member.

FIGURE P1.9

Solution

Overall equilibrium:

Begin the solution by determining the

external reaction forces acting on the

truss at supports A and B. Write

equilibrium equations that include all

external forces. Note that only the

external forces (i.e., loads and

reaction forces) are considered at this

time. The internal forces acting in the

truss members will be considered

after the external reactions have been

computed. The free-body diagram

(FBD) of the entire truss is shown.

The following equilibrium equations

can be written for this structure:

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


FIGURE P1.10

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:

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


FIGURE P1.11

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:




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

assumed in each truss member.

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:

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(41.634 ) 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.




P1.12 The rigid beam BC shown in Figure P1.12 is

supported by rods (1) and (2) that have cross-sectional

areas of 175 mm2 and 300 mm

2, respectively. For a

uniformly distributed load of w = 15 kN/m, determine

the normal stress in each rod. Assume L = 3 m and a =

1.8 m.

FIGURE P1.12

Solution

Equilibrium: Calculate the internal forces in members (1) and (2).

1

2

1

2

1.8 m(3 m) (15 kN/m)(1.8 m) 0

2

1.8 m(3 m) (15 kN/m)(1.8 m) 3 m 0

8.100 kN

18.900

2

kN

C

B

M F

M

F

F

F

Stresses:

21

1 2

1

(8.100 kN)(1,000 N/kN)46.286 N/mm

175 m46.3 MPa

m

F

A Ans.

22

2 2

2

(18.900 kN)(1,000 N/kN)63.000 N/mm

300 m63.0 MPa

m

F

A Ans.




P1.13 Bar (1) in Figure 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.

FIGURE P1.13

Solution

Given that the cross-sectional area of bar (1) is 0.75 in.2 and its normal stress must be limited to 30 ksi,

the maximum force that may be carried by bar (1) is

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.




P1.14 The rectangular bar shown in Figure

P1.14 is subjected to a uniformly distributed

axial loading of w = 13 kN/m and a

concentrated force of P = 9 kN at B.

Determine the magnitude of the maximum

normal stress in the bar and its location x.

Assume a = 0.5 m, b = 0.7 m, c = 15 mm, and

d = 40 mm.

FIGURE P1.14

Solution

Equilibrium: Draw a FBD for the interval between A and B where

0 x a . Write the following equilibrium equation:

(13 kN/m)(1.2 m ) (9 kN) 0

(13 kN/m)(1.2 m ) (9 kN)

xF x F

F x

The largest force in this interval occurs at x = 0 where F = 6.6

kN.

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

(13 kN/m)(1.2 m ) 0

(13 kN/m)(1.2 m )

xF x F

F x

The largest force in this interval occurs at x = a where F = 9.1

kN.

Maximum Normal Stress:

max(9.1 kN)(1,000 N/kN)

(15 mm15.17 MPa

)(40 mm)at 0.5 mx Ans.




P1.15 The solid 1.25-in.-diameter rod shown in

Figure P1.15 is subjected to a uniform axial

distributed loading along its length of w = 750 lb/ft.

Two concentrated loads also act on the rod: P =

2,000 lb and Q = 1,000 lb. Assume a = 16 in. and b

= 32 in. Determine the normal stress in the rod at

the following locations:

(a) x = 10 in.

(b) x = 30 in.

FIGURE P1.15

Solution

(a) x = 10 in.

Equilibrium: Draw a FBD for the interval between A and B

where 0 x a , and write the following equilibrium equation:

(750 lb/ft)(1 ft/12 in.)(48 in. )

(2,000 lb) (1,000 lb) 0

(62.5 lb/in.)(48 in. ) 3,000 lb

xF x

F

F x

At x = 10 in., F = 5,375 lb.

Stress: The normal stress at this location can be calculated as follows.

2 2

2

(1.25 in.) 1.227185 in.4

5,375 lb4,379 4,380 p.944 psi

1si

.227185 in.

A

Ans.

(b) x = 30 in.

Equilibrium: Draw a FBD for the interval between B and C

where a x a b , and write the following equilibrium equation:

(750 lb/ft)(1 ft/12 in.)(48 in. )

(1,000 lb) 0

(62.5 lb/in.)(48 in. ) 1,000 lb

xF x

F

F x

At x = 30 in., F = 2,125 lb.

Stress: The normal stress at this location can be calculated as follows.

2

1,730 ps2,125 lb

1,731.606 psi1.227185 i

in.

Ans.




P1.16 Two 6 in. wide wooden boards are

to be joined by splice plates that will be

fully glued on the contact surfaces. The

glue to be used can safely provide a shear

strength of 120 psi. Determine the smallest

allowable length L that can be used for the

splice plates for an applied load of P =

10,000 lb. Note that a gap of 0.5 in. is

required between boards (1) and (2).

FIGURE P1.16

Solution

Consider a FBD of board (2). The glue on the splice plates provides resistance to the 10,000 lb applied

load on both the top and bottom surfaces of board (2). Denoting the shear resistance on a glue surface as

V, equilibrium in the horizontal direction requires

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

2min5,000 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 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.




P1.17 For the clevis connection shown in Figure

P1.17, determine the maximum applied load P that

can be supported by the 10-mm-diameter pin if the

average shear stress in the pin must not exceed 95

MPa.

FIGURE P1.17

Solution

Consider a FBD of the bar that is connected by the clevis,

including a portion of the pin. If the shear force acting on each

exposed surface of the pin is denoted by V, then the shear force

on each pin surface is related to the load P by:

0 2xF P V V P V

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

2 2 2pin pin (10 mm) 78.539816 mm

4 4A d

If the average shear stress in the pin must be limited to 95 MPa, the maximum shear force V on a single

cross-sectional surface must be limited to

2 2bolt (95 N/mm )(78.539816 mm ) 7,461.283 NV A

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

2 2(7,461.283 N) 14,922.565 14.92 kNNP V Ans.




P1.18 For the connection shown in Figure P1.18,

determine the average shear stress produced in the 3/8-

in. diameter bolts if the applied load is P = 2,500 lb.

FIGURE P1.18

Solution

There are four bolts, and it is assumed that each bolt supports an equal portion of the external load P.

Therefore, the shear force carried by each bolt is

2,500 lb

625 lb4 bolts

V

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

2 2 2 2bolt bolt (3 / 8 in.) (0.375 in.) 0.110447 in.

4 4 4A d

Therefore, the average shear stress in each bolt is

2

bolt

625 lb5,658.8427 psi

0.110447 in.5,660 psi

V

A Ans.




P1.19 The five-bolt connection shown in Figure P1.19 must

support an applied load of P = 265 kN. If the average shear stress

in the bolts must be limited to 120 MPa, determine the minimum

bolt diameter that may be used for this connection.

FIGURE P1.19

Solution

There are five bolts, and it is assumed that each bolt supports an equal portion of the external load P.

Therefore, the shear force carried by each bolt is

265 kN

53 kN 53,000 N5 bolts

V

Since the average shear stress must be limited to 120 MPa, each bolt must provide a shear area of at

least:

22

53,000 N441.6667 mm

120 N/mmVA

Each bolt in this connection acts in double shear; therefore, two cross-sectional bolt surfaces are

available to transmit shear stress in each bolt.

2

2

bolt

441.6667 mm220.8333 mm per surface

2 surfaces per bolt 2 surfaces

VAA

The minimum bolt diameter must be

2 2bolt bolt 16.77 m220.8333 mm 16.7682 mm m

4d d

Ans.




P1.20 A coupling is used to connect a 2 in. diameter

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

shown in Figure P1.20. If the average shear stress in

the adhesive must be limited to 400 psi, determine

the minimum lengths L1 and L2 required for the joint

if the applied load P is 5,000 lb.

FIGURE P1.24

Solution

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

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.




1.21 A hydraulic punch press is used to

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

illustrated in Fig. P1.21. If the plate shears

at a stress of 30 ksi, determine the

minimum force P required to punch the

slot.

FIGURE P1.21

Solution

The shear stress associated with removal of the slug exists on its perimeter. The perimeter of the slug is

given by

perimeter 2(3.00 in.) + (0.75 in.) 8.35619 in.

Thus, the area subjected to shear stress is

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.




P1.22 The handle shown in Figure P1.22 is

attached to a 40-mm-diameter shaft with a

square shear key. The forces applied to the

lever are P = 1,300 N. If the average shear

stress in the key must not exceed 150 MPa,

determine the minimum dimension a that must

be used if the key is 25 mm long. The overall

length of the handle is L = 0.70 m. FIGURE P1.22

Solution

To determine the shear force V that must be resisted by the shear key, sum moments about the center of

the shaft (which will be denoted O):

700 mm 700 mm 40 mm(1,300 N) (1,300 N) 0

2 2 2

45,500 N

OM V

V

Since the average shear stress in the key must not exceed 150 MPa, the shear area required is

22

45,500 N303.3333 mm

150 N/mmV

VA

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

2303.3333 mm

12.1333 mm25 m

12.1 m

3 mmVA

aL

Ans.




P1.23 An axial load P is supported by the short steel

column shown in Figure P1.23. The column has a cross-

sectional area of 14,500 mm2. If the average normal stress

in the steel column must not exceed 75 MPa, determine the

minimum required dimension a so that the bearing stress

between the base plate and the concrete slab does not

exceed 8 MPa. Assume b = 420 mm.

FIGURE P1.23

Solution

Since the normal stress in the steel column must not exceed 75 MPa, the maximum column load is

2 2max (75 N/mm )(14,500 mm ) 1,087,500 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

1,087,500 N135,937.5 mm

8 N/mmb

PA

The area of the plate is a b. Since b = 420, the minimum length of a must be

2

min

2

135,937.5 mm

135,937.5 mm

420 mm324 mm

A a b

a

Ans.




P1.24 The two wooden boards shown in Figure P1.24 are

connected by a 0.5-in.-diameter bolt. Washers are installed

under the head of the bolt and under the nut. The washer

dimensions are D = 2 in. and d = 5/8 in. The nut is tightened

to cause a tensile stress of 9,000 psi in the bolt. Determine

the bearing stress between the washer and the wood.

FIGURE P1.24

Solution

The tensile stress in the bolt is 9,000 psi; therefore, the tension force that acts in the bolt is

2 2bolt bolt bolt (9,000 psi) (0.5 in.) (9,000 psi)(0.196350 in. ) 1,767.146 lb

4F A

The contact area between the washer and the wood is

2 2 2washer (2 in.) (0.625 in.) 2.834796 in.

4A

Thus, the bearing stress between the washer and the wood is

2

1,767.146 lb

2.834796 in.623 psib Ans.




P1.25 For the beam shown in Figure P1.25, the

allowable bearing stress for the material under

the supports at A and B is b = 800 psi. Assume w = 2,100 lb/ft, P = 4,600 lb, a = 20 ft,

and b = 8 ft. Determine the size of square

bearing plates required to support the loading

shown. Dimension the plates to the nearest

in.

FIGURE P1.25

Solution

Equilibrium: Using the FBD shown,

calculate the beam reaction forces.

20 ft(20 ft) (2,100 lb/ft)(20 ft) (4,600 lb)(28 ft) 0

2

27,440 lb

A y

y

M B

B

20 ft(20 ft) (2,100 lb/ft)(20 ft) (4,600 lb)(8 ft) 0

2

19,160 lb

B y

y

M A

A

Bearing plate at A: The area of the bearing plate required for support A is

219,160 lb

23.950 in.800 psi

AA

Since the plate is to be square, its dimensions must be

2 use 5 in. 23.95 5 in. bearing plate at0 in. 4.894 in . Awidth Ans.

Bearing plate at B: The area of the bearing plate required for support B is

227,440 lb

34.300 in.800 psi

BA

Since the plate is to be square, its dimensions must be

2 use 6 in. 34.30 6 in. bearing plate at0 in. 5.857 in . Bwidth Ans.




P1.26 The d = 15-mm-diameter solid rod shown in

Figure P1.26 passes through a D = 20-mm-diameter

hole in the support plate. When a load P is applied

to the rod, the rod head rests on the support plate.

The support plate has a thickness of b = 12 mm.

The rod head has a diameter of a = 30 mm and the

head has a thickness of t = 10 mm. If the normal

stress produced in the rod by load P is 225 MPa,

determine:

(a) the bearing stress acting between the support

plate and the rod head.

(b) the average shear stress produced in the rod

head.

(c) the punching shear stress produced in the

support plate by the rod head.

FIGURE P1.26

Solution

The cross-sectional area of the rod is:

2 2

rod (15 mm) 176.715 mm4

A

The tensile stress in the rod is 225 MPa; therefore, the tension force in the rod is

2 2rod rod rod (225 N/mm )(176.715 mm ) 39,760.782 NF A

(a) The contact area between the support plate and the rod head is

2 2 2contact (30 mm) (20 mm) 392.699 mm

4A

Thus, the bearing stress between the support plate and the rod head is

2

39,760.782 N

392.699 mm101.3 MPab Ans.

(b) In the rod head, the area subjected to shear stress is equal to the perimeter of the rod times the

thickness of the head.

2(15 mm)(10 mm) 471.239 mmVA

and therefore, the average shear stress in the rod head is

2

39,760.782 N

471.84.4 MP

ma

239 m Ans.

(c) In the support plate, the area subjected to shear stress is equal to the product of the rod head

perimeter and the thickness of the plate.

2(30 mm)(12 mm) 1,130.973 mmVA

and therefore, the average punching shear stress in the support plate is

2

39,760.782 N

1,13035.2 MPa

.973 mm Ans.




P1.27 The rectangular bar is connected to the support bracket with

a circular pin, as shown in Figure P1.27. The bar width is w = 1.75

in. and the bar thickness is 0.375 in. For an applied load of P =

5,600 lb, determine the average bearing stress produced in the bar

by the 0.625-in.-diameter pin.

FIGURE P1.27

Solution

The average bearing stress produced in the bar by the pin is based on the projected area of the pin. The

projected area is equal to the pin diameter times the bar thickness. Therefore, the average bearing stress

in the bar is

5,600 lb

23,893.33 psi(0.625 in.)(0.375 i

23,n.

900)

psib Ans.




P1.28 The steel pipe column shown in Figure P1.28 has an

outside diameter of 8.625 in. and a wall thickness of 0.25 in.

The timber beam is 10.75 in wide, and the upper plate has the

same width. The load imposed on the column by the timber

beam is 80 kips. Determine

(a) The average bearing stress at the surfaces between the pipe

column and the upper and lower steel bearing plates.

(b) The length L of the rectangular upper bearing plate if its

width is 10.75 in. and the average bearing stress between the

steel plate and the wood beam is not to exceed 500 psi.

(c) The dimension a of the square lower bearing plate if the average bearing stress between the lower bearing plate and

the concrete slab is not to exceed 900 psi.

Figure P1.28

Solution

(a) The area of contact between the pipe column and one of the bearing plates is simply the 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 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.




P1.29 A clevis-type pipe hanger supports an 8-in-

diameter pipe as shown in Figure P1.29. The hanger

rod has a diameter of 1/2 in. The bolt connecting the

top yoke and the bottom strap has a diameter of 5/8 in.

The bottom strap is 3/16-in.-thick by 1.75-in.-wide by

36-in.-long. The weight of the pipe is 2,000 lb.

Determine the following:

(a) the normal stress in the hanger rod

(b) the shear stress in the bolt

(c) the bearing stress in the bottom strap

FIGURE P1.29

Solution

(a) The normal stress in the hanger rod is

2 2

rod

rod 2

(0.5 in.) 0.196350 in.4

2,000 lb10,185.917 psi

0.196350 in.10,190 psi

A

Ans.

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

2 2

bolt (0.625 in.) 0.306796 in.4

A

The bolt acts in double shear; therefore, its average shear stress is

bolt 22,000 lb

3,259.493 psi2(0.306796 in.

3,260 i)

ps Ans.

(c) The bearing stress in the bottom strap is based on the projected area of the bolt in contact with the

strap. Also, keep in mind that there are two ends of the strap that contact the bolt. The bearing stress is

thus

2,000 lb

8,533.334 psi2(0.625 in.)(3/16

8,530 psiin.)

b Ans.




P1.30 Rigid bar ABC shown in Figure P1.30 is

supported by a pin at bracket A and by tie rod (1).

Tie rod (1) has a diameter of 5 mm, and it is

supported by double-shear pin connections at B and

D. The pin at bracket A is a single-shear

connection. All pins are 7 mm in diameter.

Assume a = 600 mm, b = 300 mm, h = 450 mm, P

= 900 N, and = 55. Determine the following: (a) the normal stress in rod (1)

(b) the shear stress in pin B

(c) the shear stress in pin A

FIGURE P1.30

Solution

Equilibrium: Using the FBD shown, calculate

the reaction forces that act on rigid bar ABC.

1

1

sin(36.87 )(600 mm)

(900 N)sin (55 )(900 mm) 0

1,843.092 N

AM F

F

(1,843.092 N)cos(36.87 ) (900 N)cos(55 ) 0

958.255 N

x x

x

F A

A

(1,843.092 N)sin (36.87 ) (900 N)sin (55 ) 0

368.618 N

y y

y

F A

A

The resultant force at A is

2 2(958.255 N) ( 368.618 N) 1,026.709 NA

(a) Normal stress in rod (1).

2 2

rod

rod 2

(5 mm) 19.635 mm4

1,843.092 N

19.635 mm93.9 MPa

A

Ans.

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

2 2

pin (7 mm) 38.485 mm4

A

Pin B is a double shear connection; therefore, its average shear stress is

pin 21,843.092 N

2(38.485 m23.9 M a

)P

mB Ans.

(c) Shear stress in pin A.

Pin A is a single shear connection; therefore, its average shear stress is

pin 21,026.709 N

38.485 m26.7 MPa

mA Ans.




P1.31 The bell crank shown in Figure P1.31 is in

equilibrium for the forces acting in rods (1) and (2).

The bell crank is supported by a 10-mm-diameter

pin at B that acts in single shear. The thickness of

the bell crank is 5 mm. Assume a = 65 mm, b =

150 mm, F1 = 1,100 N, and = 50. Determine the following:

(a) the shear stress in pin B

(b) the bearing stress in the bell crank at B

FIGURE P1.31

Solution

Equilibrium: Using the FBD shown, calculate the

reaction forces that act on the bell crank.

2

2

(1,100 N)sin(50 )(65 mm)

(150 mm) 0

365.148 N

BM

F

F

(1,100 N)cos(50 )

365.148 N 0

341.919 N

x x

x

F B

B

(1,100 N)sin(50 ) 0

842.649 N

y y

y

F B

B

The resultant force at B is

2 2(341.919 N) ( 842.649 N) 909.376 NB

(a) Shear stress in pin B. The cross-sectional area of the 10-mm-diameter pin is:

2 2

pin (10 mm) 78.540 mm4

A

Pin B is a single shear connection; therefore, its average shear stress is

pin 2909.376 N

78.540 mm11.58 MPaB Ans.

(b) Bearing stress in the bell crank at B. The average bearing stress produced in the bell crank by the

pin is based on the projected area of the pin. The projected area is equal to the pin diameter times the

bell crank thickness. Therefore, the average bearing stress in the bell crank is

909.376 N

(10 mm)(5 mm)18.19 MPab Ans.




P1.32 The beam shown in Figure P1.32 is

supported by a pin at C and by a short link

AB. If w = 30 kN/m, determine the average

shear stress in the pins at A and C. Each

pin has a diameter of 25 mm. Assume L =

1.8 m and = 35.

FIGURE P1.32

Solution

Equilibrium: Using the FBD shown,

calculate the reaction forces that act on the

beam.

1

1

1.8 msin(35 )(1.8 m) (30 kN/m)(1.8 m) 0

2

47.0731 kN

CM F

F

(47.0731 kN)cos(35 ) 0

38.5600 kN

x x

x

F C

C

1.8 m(1.8 m) (30 kN/m)(1.8 m) 0

2

27.0000 kN

B y

y

M C

C

The resultant force at C is

2 2(38.5600 kN) (27.0000 kN) 47.0731 kNC

Shear stress in pin A. The cross-sectional area of a 25-mm-diameter pin is:

2 2

pin (25 mm) 490.8739 mm4

A

Pin A is a single shear connection; therefore, its average shear stress is

pin 247,073.1 N

490.8739 m95.9 MPa

mA Ans.

Shear stress in pin C.

Pin C is a double shear connection; therefore, its average shear stress is

pin 247,073.1 N

2(490.8739 m47.9 M a

)P

mC Ans.




P1.33 The bell-crank mechanism shown

in Figure P1.33 is in equilibrium for an

applied load of P = 7 kN applied at A.

Assume a = 200 mm, b = 150 mm, and = 65. Determine the minimum diameter d

required for pin B for each of the following

conditions:

(a) The average shear stress in the pin may

not exceed 40 MPa.

(b) The bearing stress in the bell crank

may not exceed 100 MPa.

(c) The bearing stress in the support

bracket may not exceed 165 MPa.

FIGURE P1.33

Solution

Equilibrium: Using the FBD shown, calculate

the reaction forces that act on the bell crank.

2

2

(7,000 N)sin(65 )(200 mm)

(150 mm) 0

8,458.873 N

BM

F

F

(7,000 N)cos(65 )

8,458.873 N 0

11,417.201 N

x x

x

F B

B

(7,000 N)sin(65 ) 0

6,344.155 N

y y

y

F B

B

The resultant force at B is

2 2( 11,417.201 N) (6,344.155 N) 13,061.423 NB

(a) The average shear stress in the pin may not exceed 40 MPa. The shear area required for the pin

at B is

22

13,061.423 N326.536 mm

40 N/mmVA

Since the pin at B is supported in a double shear connection, the required cross-sectional area for the pin

is

2pin 163.268 mm2

VAA

and therefore, the pin must have a diameter of

24 (163.268 m 14.m 4) 2 mmd

Ans.




(b) The bearing stress in the bell crank may not exceed 100 MPa. The projected area of pin B on the

bell crank must equal or exceed

22

13,061.423 N130.614 mm

100 N/mmbA

The bell crank thickness is 8 mm; therefore, the projected area of the pin is Ab = (8 mm)d. Calculate the

required pin diameter d:

2130.614 mm

816.

3

mm3 mmd Ans.

(c) The bearing stress in the support bracket may not exceed 165 MPa. The pin at B bears on two 6-

mm-thick support brackets. Thus, the minimum pin diameter required to satisfy the bearing stress limit

on the support bracket is

22

13,061.423 N79.160 mm

165 N/mmbA

279.160 mm

2(6 mm)6.60 mmd Ans.




P1.34 A structural steel bar with a 25 mm 75 mm rectangular cross section is subjected to an axial

load of 150 kN. Determine the maximum normal and shear stresses in the bar.

Solution

The maximum normal stress in the steel bar is

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

maxmax

240 MPa

Ans.




P1.35 A steel rod of circular cross section will be used to carry an axial load of 92 kips. The

maximum stresses in the rod must be limited to 30 ksi in tension and 12 ksi in shear. Determine the

required diameter for the rod.

Solution

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

2

min

max

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.

4d d

Ans.




P1.36 An axial load P is applied to the

rectangular bar shown in Figure P1.36. The

cross-sectional area of the bar is 400 mm2.

Determine the normal stress perpendicular to

plane AB and the shear stress parallel to

plane AB if the bar is subjected to an axial

load of P = 70 kN.

FIGURE P1.36

Solution

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

found from

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.




P1.37 An axial load P is applied to the 1.75 in.

by 0.75 in. rectangular bar shown in Figure

P1.37. Determine the normal stress

perpendicular to plane AB and the shear stress

parallel to plane AB if the bar is subjected to

an axial load of P = 18 kips.

FIGURE P1.37

Solution

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

found from

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.




P1.38 A compression load of P = 80 kips is applied to a 4 in. by 4

in. square post, as shown in Figure P1.38/39. Determine the normal

stress perpendicular to plane AB and the shear stress parallel to

plane AB.

FIGURE P1.38/39

Solution

The angle for the inclined plane is 55. The normal force N perpendicular to plane AB is found from

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.




P1.39 Specifications for the 50 mm 50 mm square bar

shown in Figure P1.38/39 require that the normal and shear

stresses on plane AB not exceed 120 MPa and 90 MPa,

respectively. Determine the maximum load P that can be

applied without exceeding the specifications.

FIGURE P1.38/39

Solution

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

(1 cos2 )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.




P1.40 Specifications for the 6 in. 6 in. square post shown in

Figure P1.40 require that the normal and shear stresses on

plane AB not exceed 800 psi and 400 psi, respectively.

Determine the maximum load P that can be applied without

exceeding the specifications.

FIGURE P1.40

Solution


(1 cos2 )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.




P1.41 A 90 mm wide bar will be used to carry an axial

tension load of 280 kN as shown in Figure P1.41. The

normal and shear stresses on plane AB must be limited

to 150 MPa and 100 MPa, respectively. Determine the

minimum thickness t required for the bar.

FIGURE P1.41

Solution


(1 cos2 )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 cos2 ) (1 cos2(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.

Excerpts from this work may be reproduced by inst

cvg 2140 solution

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