CVG 2140 Solution

(1)

(2)

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 Stress–Strain Diagram 3.3 Hooke’s Law 3.4 Poisson’s 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-Venant’s 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

(3)

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

(4)

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 Mohr’s 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 Mohr’s Circle for Plane Strain

13.7 Strain Measurement and Strain Rosettes

13.8 Generalized Hooke’s 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

(5)

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 Formulas—Centric 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 Castigliano’s Second Theorem

17.12 Calculating Deflections of Trusses by Castigliano’s Theorem 17.13 Calculating Deflections of Beams by Castigliano’s 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 Mohr’s Circle for Principal Moments of Inertia

Appendix B: Geometric Properties of Structural Steel Shapes Appendix C: Table of Beam Slopes and Deflections

(6)

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 mm

4 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 2

max allow (200 N/mm )(863.938 mm ) 172,788 N 172.8 kN

(7)

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 kips 1.500 in. 18 ksi P P A 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 2

D 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 d t       Ans.

(8)

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) y F 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) y F 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

2 1 1,min 15 kips 0.375 in. 40 ksi F A    

The minimum rod diameter is therefore

2 2

1,min 1 0.375 in. 1 0.69099 i 0.691

4 n. in.

A d   d  Ans.

Similarly, the normal stress in rod (2) must be limited to 40 ksi, which requires a minimum area of 2 2 2,min 75 kips 1.875 in. 40 ksi F A    

The minimum diameter for rod (2) is therefore

2 2

1.875 in. 1.545097 in. 1.545 in.

(9)

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).

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) y F 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) y F F F           

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

2 2

1 (1.75 in.) 2.4053 in. 4

A  

and thus, the normal stress in rod (1) is 1 1 2 1 15 kips 6.23627 ksi 2.4053 in. 6.24 ksi (C) F A        Ans.

2 2



(10)

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.

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 8 kips 4 kips 4 kips 0 1 16 kips 16 kips (C)

y F 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 8 kips 4 kips 4 kips 15 kips 15 kips 0 2 14 kips 14 kips (T)

y

F F F

(11)

2 2

1 (2.00 in.) 3.1416 in. 4

A  

and thus, the normal stress in aluminum rod (1) is 1 1 2 1 16 kips 5.0930 ksi 3.1416 in. 5.09 ksi (C) F A        Ans.

2 2

2 (1.50 in.) 1.7671 in. 4

A  

Accordingly, the normal stress in brass rod (2) is 2 2 2 2 14 kips 7.9224 ksi 1.7671 in. 7.92 ksi (T) F A      Ans.

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

2 2

3 (3.00 in.) 7.0686 in. 4

A   

and the normal stress in the steel rod is 3 3 2 3 26 kips 3.6782 ksi 7.0686 in. 3.68 ksi (C) F A        Ans.

(12)

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 8 kips 4 kips 4 kips 0 1 16 kips 16 kips (C)

y F 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):

(13)

3 3

8 kips 4 kips 4 kips 15 kips 15 kips 20 kips 20 kips 0 26 kips 26 kips (C) y F 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

2 1 1,min 1 16 kips 0.8889 in. 18 ksi F A    

2 2

1,min 1 0.8889 in. 1 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 2 2 2,min 2 14 kips 0.5600 in. 25 ksi F A    

which requires a minimum diameter for rod (2) of

2 2

2,min 2 0.5600 in. 2 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:

2 3 3,min 3 26 kips 1.7333 in. 15 ksi F A    

2 2

3,min 3 1.7333 in. 3 1.4856 in. 1.486 in. 4

(14)

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.

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

4.0 m

tan 1.600 57.9946

2.5 m

    

and the angle  between rod (2) and the horizontal axis: 2.3 m

tan 0.7188 35.7067

3.2 m

    

Write equilibrium equations for the sum of forces in the horizontal and vertical directions. Note: Rods (1) and (2) are two-force members.

2cos(35.7067 ) 1cos(57.9946 ) 0 x F F F       (a) 2sin(35.7067 ) 1sin(57.9946 ) 0 y F 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 1

cos(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 P F                  

(15)

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

2 1 1,min 2 1 (40.6856 kN)(1,000 N/kN) 312.9664 mm 130 N/mm F A    

2 2

1,min 1 312.9664 mm 1 19.9620 19.

4 mm 96 mm

A  d   d  Ans.

The minimum area required for rod (2) is

2 2 2,min 2 2 (26.5553 kN)(1,000 N/kN) 204.2718 mm 130 N/mm F A    

2 2

2,min 2 204.2718 mm 2 16.1272 16.

4 mm 13 mm

(16)

P1.8 Two solid cylindrical rods support a load

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

FIGURE P1.7/8

Solution

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

4.0 m

tan 1.600 57.9946

2.5 m

    

and the angle  between rod (2) and the horizontal axis: 2.3 m

tan 0.7188 35.7067

3.2 m

    

Write equilibrium equations for the sum of forces in the horizontal and vertical directions. Note: Rods (1) and (2) are two-force members. 2cos(35.7067 ) 1cos(57.9946 ) 0 x F F F       (a) 2sin(35.7067 ) 1sin(57.9946 ) 0 y F 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 1

cos(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 P F                  

(17)

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 2

1 (16 mm) 201.0619 mm 4

A   

and the normal stress in rod (1) is:

2 1 1 2 1 (21.9702 kN)(1,000 N/kN) 109.2710 N/mm 201.0619 mm 109.3 MPa (T) F A      Ans.

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

2 2

2 (12 mm) 113.0973 mm 4

A  

and the normal stress in rod (2) is:

2 2 2 2 2 (14.3399 kN)(1,000 N/kN) 126.7924 N/mm 113.0973 mm 126.8 MPa (T) F A      Ans.

(18)

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 x F A A        (6 ft) (5 kips)(14 ft) (2 kips)(7 ft) 14 kips 0 y A By 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 ft

(19)

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 ) 0 x AC AB x F F F A       (a) sin(26.565 ) 0 y AC y F F A      (b)

Solve Eq. (b) for FAC:

9 kips

sin(26.565 ) sin(26.565 ) 20.125 kips

y AC

A

F      

 

and then compute FAB using Eq. (a):

cos(26.565 )

(20.125 kips) cos(26.565 ) ( 2 kips) 16.000 kips

AB AC x

F  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 ) 0 x AB BC F F F       (c) sin(41.186 ) 0 y BC y F F B      (d)

Solve Eq. (d) for FBC:

14 kips

sin(41.186 ) sin(41.186 ) 21.260 kips

y BC

B

F      

 

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 BC

F F F

    

       Checks!

Section properties:

For each of the three truss members:

2 2 2

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

(20)

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

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 kN x x x F A A        (1 m) (15 kN)(4.3 m) 0 64.5 kN y A By B M       15 kN 49.5 kN 0 y y y y F A A B         Method of joints:

Before beginning the process of determining the internal forces in the axial members, the geometry of the truss will be used to determine the magnitude of the inclination angles of members AB and BC. Use the definition of the tangent function to determine AB and BC:

1.5 m tan 1.50 56.310 1.0 m 1.5 m tan 0.454545 24.444 3.3 m AB AB BC BC              

(21)

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(56.310 ) 0 x AC AB x F F F A       (a) sin(56.310 ) 0 y y AB F A F      (b)

Solve Eq. (b) for FAB:

49.5 kN sin(56.310 ) sin(56.310 ) 59.492 kN y AB A F      

and then compute FAC using Eq. (a):

cos(56.310 ) ( 59.492 kN) cos(56.310 ) ( 12 kN) 45.000 kN AC AB x F  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 0 x AC BC F F F        (c) sin(24.444 ) 15 kN 0 y BC F F       (d)

Solve Eq. (d) for FBC:

15 kN

sin(24.444 ) 36.249 kN

BC

F   



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 BC F F F               Checks! Section properties:

2 2 2

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

4

(22)

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

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 0 y y y F 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 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:

(23)

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(19.573 ) 0 x x AC F A F      (a) sin(19.573 ) 0 y y AC AB F A F F       (b)

Solve Eq. (a) for FAC:

34.821 kN cos(19.573 ) cos(19.573 ) 36.957 kN x AC A F     

and then compute FAB using Eq. (b):

sin(19.573 ) (30.000 kN) (36.957 kN)sin(19.573 ) 17.619 kN AB y AC F  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 ) 0 x x BC F B F      (c) sin(41.634 ) 0 y BC AB F F F      (d)

Solve Eq. (c) for FBC:

( 19.821 kN) cos(41.634 ) cos(41.634 ) 26.520 kN x BC B F      

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 AB F F F          Checks! Section properties:

2 2 2

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

4

d    A _  _

Normal stress in each truss member:

(17.619 kN)(1,000 N/kN) F

(24)

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

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

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: 2 1 1 2 1 (8.100 kN)(1,000 N/kN) 46.286 N/mm 175 mm 46.3 MPa F A      Ans. 2 2 2 2 2 (18.900 kN)(1,000 N/kN) 63.000 N/mm 300 mm 63.0 MPa F A      Ans.

(25)

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.

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

2

1,max 1 1 (30 ksi)(0.75 in. ) 22.5 kips

F  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 A M 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: 1

6 ft 6 ft

(22.5 kips)

10 ft 10 ft 13.50 kips

(26)

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) x F 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 ) x F 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 mm)(40 mm) 15.17 MPa at x 0.5 m

(27)

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.

(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 x F 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 lb 4,379.944 psi 4,380 p 1.227185 in. si 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 x F x F F x            At x = 30 in., F = 2,125 lb.

(28)

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 lb 5,000 lb 2 x F 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 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 we’ve discussed only board (2), the same rationale applies to board (1). For both boards (1) and (2), the glue must be applied along a length of at least 6.9444 in. on both the top and bottom of the boards in order to resist the 10,000 lb applied load.

The glue applied to boards (1) and (2) must be matched by glue applied to the splice plates. Therefore, the splice plates must be at least 6.9444 in. + 6.9444 in. = 13.8889 in. long. However, we are told that a 0.5-in. gap is required between boards (1) and (2); therefore, the splice plates must be 0.5-in. longer. Altogether, the length of the splice plates must be at least

min 6.9444 in. 6.9444 in. 0.5 in. 14.39 in.

(29)

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.

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 2

x

F 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 2

pin pin (10 mm) 78.539816 mm

4 4

A  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 2

bolt (95 N/mm )(78.539816 mm ) 7, 461.283 N

V A  

Therefore, the maximum load P that may be applied to the connection is 2 2(7, 461.283 N) 14,922.565 N 14.92 kN

(30)

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.

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 lb 4 bolts

V  

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

2 2 2 2

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

4 4 4

A  d   

Therefore, the average shear stress in each bolt is 2 bolt 625 lb 5,658.8427 psi 0.110447 in. 5,660 psi V A      Ans.

(31)

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.

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 N 5 bolts

V   

Since the average shear stress must be limited to 120 MPa, each bolt must provide a shear area of at least: 2 2 53,000 N 441.6667 mm 120 N/mm V A  

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 mm

220.8333 mm per surface 2 surfaces per bolt 2 surfaces

V

A

A   

The minimum bolt diameter must be

2 2

bolt 220.8333 mm bolt 16.7682 mm 16.77 mm

4d d

 _ _ _ _

(32)

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.

To resist a shear force of 5,000 lb, the area of adhesive required on each pipe is 2 adhesive 5, 000 lb 12.5 in. 400 psi V V A    

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 in. 1.989 i n. V A L C     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 in. 2.65 in. V A L C     Ans.

(33)

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.

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

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

V

A    

Given that the plate shears at  = 30 ksi, the force required to remove the slug is therefore 2

min V (30 ksi)(4.17810 in. ) 125.343 kips 125.3 kips

(34)

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.

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 O M V V         _ _ _ _{ } _         

Since the average shear stress in the key must not exceed 150 MPa, the shear area required is 2 2 45,500 N 303.3333 mm 150 N/mm V V A    

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

2 303.3333 mm 12.1333 mm 25 mm 12.1 3 mm V A a L     Ans.

(35)

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.

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

2 2

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

P 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 N 135,937.5 mm 8 N/mm b P A    

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 mm 324 mm A a b a       Ans.

(36)

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 2

bolt bolt bolt (9,000 psi) (0.5 in.) (9,000 psi)(0.196350 in. ) 1,767.146 lb 4

F  A    

The contact area between the washer and the wood is

2 2 2

washer (2 in.) (0.625 in.) 2.834796 in. 4

A _  _

Thus, the bearing stress between the washer and the wood is 2

1,767.146 lb

2.834796 in. 623 psi

b

(37)

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

allowable bearing stress for the material under the supports at A and B is b = 800 psi.

Assume w = 2,100 lb/ft, P = 4,600 lb, a = 20 ft, and b = 8 ft. Determine the size of square bearing plates required to support the loading shown. Dimension the plates to the nearest ½ in.

FIGURE P1.25

Solution

Equilibrium: Using the FBD shown,

calculate the beam reaction forces.

20 ft (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

2 19,160 lb 23.950 in. 800 psi A A  

Since the plate is to be square, its dimensions must be 2

use 5 in.

23.950 in. 4.894 in. 5 in. bearing plate atA

width   Ans.

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

2 27, 440 lb 34.300 in. 800 psi B A  

Since the plate is to be square, its dimensions must be 2

use 6 in.

34.300 in. 5.857 in. 6 in. bearing plate atB

(38)

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.

The cross-sectional area of the rod is:

2 2

rod (15 mm) 176.715 mm 4

A  

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

2 2

rod rod rod (225 N/mm )(176.715 mm ) 39, 760.782 N

F  A  

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

2 2 2

contact (30 mm) (20 mm) 392.699 mm 4

A  _  _

Thus, the bearing stress between the support plate and the rod head is 2

39,760.782 N

392.699 mm 101.3 MPa

b

   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 mm

V

A  

and therefore, the average shear stress in the rod head is 2

39,760.782 N

471.239 mm 84.4 MPa

   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 mm

(39)

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.

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 in.) 23,900 psi b     Ans.

(40)

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 2

pipe (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 kips 12.1623 ksi 6.5777 in. 12.16 ksi b 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

2 80 kips 160 in. 0.5 ksi b b P A    

If the width of the timber beam is 10.75 in., then the length L of the upper bearing plate must be 2

160 in.

14.8837 in.

beam width 10.75in. 14.88 in.

b

A

L    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

2 80 kips 88.8889 in. 0.9 ksi b b P A    

(41)

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

(a) The normal stress in the hanger rod is

2 2 rod rod 2 (0.5 in.) 0.196350 in. 4 2,000 lb 10,185.917 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 2

2,000 lb

3, 259.493 psi

2(0.306796 in. ) 3, 260 psi

    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 psi

2(0.625 in.)(3/16 in.) 8,530 psi

b

(42)

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

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 A M 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 N

A    

(a) Normal stress in rod (1).

2 2 rod rod 2 (5 mm) 19.635 mm 4 1,843.092 N 19.635 mm 93.9 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 mm 4

A  

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

pin 2

1,843.092 N

2(38.485 mm ) 23.9 M aP

B

(43)

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 B M 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 N

B    

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

2 2

pin (10 mm) 78.540 mm 4

A  

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

pin 2

909.376 N

78.540 mm 11.58 MPa

B

(44)

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°.

Equilibrium: Using the FBD shown,

calculate the reaction forces that act on the beam. 1 1 1.8 m sin(35 )(1.8 m) (30 kN/m)(1.8 m) 0 2 47.0731 kN C M 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 kN

C   

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

2 2

pin (25 mm) 490.8739 mm 4

A  

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

pin 2

47,073.1 N

490.8739 mm 95.9 MPa

A

   Ans.

Shear stress in pin C.

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

pin 2

47,073.1 N

2(490.8739 mm ) 47.9 M aP

C

(45)

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.

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 B M 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 N

B    

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

at B is 2 2 13,061.423 N 326.536 mm 40 N/mm V A  

(46)

(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 2 2 13,061.423 N 130.614 mm 100 N/mm b A  

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: 2 130.614 mm

8mm 16.33 mm

d   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

2 2 13,061.423 N 79.160 mm 165 N/mm b A   2 79.160 mm 2(6 mm) 6.60 mm d   Ans.

(47)

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 max

max

2 40 MPa



(48)

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 kips 3.0667 in. 30 ksi F A    

For the 12-ksi shear stress limit, the minimum cross-sectional area of the rod must be 2 min max 92 kips 3.8333 in. 2 2(12 ksi) F A    

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 2

min 3.8333 in. min 2.2092 in. 2.21 in.

4d d



(49)

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.

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 kN

NP   

and the shear force V parallel to plane AB is sin (70 kN)sin 35 40.1504 kN

V P   

The cross-sectional area of the bar is 400 mm2, but the area along inclined plane AB is 2 2 400 mm 488.3098 mm cos cos35 n A A     

The normal stress n perpendicular to plane AB is

2 (57.3406 kN)(1,000 N/kN) 117.4268 MPa 488.3098 mm 117.4 MPa n n N A      Ans.

The shear stress nt parallel to plane AB is

2 (40.1504 kN)(1,000 N/kN) 82.2231 MPa 488.309 8 mm 82.2 MPa nt n V A      Ans. 

(50)

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.

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 kips

NP   

and the shear force V parallel to plane AB is sin (18 kips)sin 60 15.5885 kips

V 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 2 1.3125 in. / cos 2.6250 in. cos 60 n A  A    

2 9.0 kips 3.4286 ksi 2.6250 in. 3.43 ksi n n N A      Ans.

2 15.5885 kips 5.9385 ksi 2.6250 in. 5.94 ksi nt n V A      Ans. 

(51)

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.

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 kips

N P   

and the shear force V parallel to plane AB is sin (80 kips)sin 55 65.5322 kips

V 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 2 16 in. / cos 27.8951 in. cos55 n A  A    

2 45.8861 kips 1.6449 ksi 27.8951 in. 1.645 ksi n n N A      Ans.

2 65.5322 kips 2.3492 ksi 27.8951 in. 2.35 ksi nt n V A      Ans.

(52)

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.

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 2 2 nt P A    (b)

The cross-sectional area of the square bar is A = (50 mm)2 = 2,500 mm2, and the angle  for plane AB is 55°.

The normal stress on plane AB is limited to 120 MPa; therefore, the maximum load P that can be supported by the square bar is found from Eq. (a):

2 2 2 2(2,500 mm )(120 N/mm ) 911,882 N 1 cos 2 1 cos 2(55 ) n A P        

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 2 2 2(2,500 mm )(90 N/mm ) 478,880 N sin 2 sin 2(55 ) nt A P      

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

max 479 kN

(53)

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.

The cross-sectional area of the square post is A = (6 in.)2 = 36 in.2, and the angle  for plane AB is 40°. The normal stress on plane AB is limited to 800 psi; therefore, the maximum load P that can be

supported by the square post is found from Eq. (a): 2 2 2(36 in. )(800 psi) 49,078 lb 1 cos 2 1 cos 2(40 ) n A P        

The shear stress on plane AB is limited to 400 psi. From Eq. (b), the maximum load P based the shear stress limit is 2 2 2(36 in. )(400 psi) 29, 244 lb sin 2 sin 2(40 ) nt A P      

Thus, the maximum load that can be supported by the post is max 29, 200 lb 29.2 ipk s

(54)

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.

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 n 2(150 N/mm ) P A        

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 _nt 2(100 N/mm ) P A      

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 mm 15.3192 mm 90 mm 15.3 2 mm t    Ans.