the bolts used for the connections of this steel framework...

The bolts used for the connections of this steel framework are subjected tostress. In this chapter we will discuss how engineers design these connectionsand their fasteners.

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3

CHAPTER OBJECTIVES

Stress

In this chapter we will review some of the important principles of statics andshow how they are used to determine the internal resultant loadings in a body.Afterwards the concepts of normal and shear stress will be introduced, andspecific applications of the analysis and design of members subjected to an axialload or direct shear will be discussed.

Mechanics of materials is a branch of mechanics that studies therelationships between the external loads applied to a deformable bodyand the intensity of internal forces acting within the body. This subjectalso involves computing the deformations of the body, and it provides astudy of the body’s stability when the body is subjected to external forces.

In the design of any structure or machine, it is first necessary to usethe principles of statics to determine the forces acting both on and withinits various members. The size of the members, their deflection, and theirstability depend not only on the internal loadings, but also on the typeof material from which the members are made. Consequently, an accuratedetermination and fundamental understanding of material behavior willbe of vital importance for developing the necessary equations used inmechanics of materials. Realize that many formulas and rules for design,as defined in engineering codes and used in practice, are based on thefundamentals of mechanics of materials, and for this reason anunderstanding of the principles of this subject is very important.

C H A P T E R

1

1.1 Introduction


4 • CHAPTER 1 Stress

Historical Development. The origin of mechanics of materials dates backto the beginning of the seventeenth century,at which time,Galileo performedexperiments to study the effects of loads on rods and beams made of variousmaterials. For a proper understanding, however, it was necessary to establishaccurate experimental descriptions of a material’s mechanical properties.Methods for doing this were remarkably improved at the beginning of theeighteenth century.At that time both experimental and theoretical studies inthis subject were undertaken primarily in France by such notables as Saint-Venant, Poisson, Lamé, and Navier. Because their efforts were based onmaterial-body applications of mechanics, they called this study “strength ofmaterials.” Currently, however, it is usually referred to as “mechanics ofdeformable bodies” or simply “mechanics of materials.”

Over the years, after many of the fundamental problems of mechanics ofmaterials had been solved, it became necessary to use advancedmathematical and computer techniques to solve more complex problems.As a result, this subject expanded into other subjects of advanced mechanicssuch as the theory of elasticity and the theory of plasticity. Research in thesefields is ongoing, not only to meet the demands for solving advancedengineering problems, but to justify further use and the limitations uponwhich the fundamental theory of mechanics of materials is based.

1.2 Equilibrium of a Deformable Body

Since statics plays an important role in both the development andapplication of mechanics of materials, it is very important to have a goodgrasp of its fundamentals. For this reason we will review some of themain principles of statics that will be used throughout the text.

External Loads. A body can be subjected to several different types ofexternal loads; however, any one of these can be classified as either asurface force or a body force, Fig. 1–1.

Surface Forces. As the name implies, surface forces are caused by thedirect contact of one body with the surface of another. In all cases theseforces are distributed over the area of contact between the bodies. If thisarea is small in comparison with the total surface area of the body, thenthe surface force can be idealized as a single concentrated force, which isapplied to a point on the body. For example, the force of the ground on thewheels of a bicycle can be considered as a concentrated force when studyingthe loading on the bicycle. If the surface loading is applied along a narrowarea, the loading can be idealized as a linear distributed load, w(s). Herethe loading is measured as having an intensity of force/length along thearea and is represented graphically by a series of arrows along the line s.The resultant force of w(s) is equivalent to the area under thedistributed loading curve, and this resultant acts through the centroid Cor geometric center of this area. The loading along the length of a beam isa typical example of where this idealization is often applied.

FR

w(s)

Concentrated forceidealization

Linear distributedload idealization

Surfaceforce

Bodyforce

s

C

G

FR W

Fig. 1–1


TA B L E 1 – 1 .

SECTION 1.2 Equilibrium of a Deformable Body • 5

Body Force. A body force is developed when one body exerts a force onanother body without direct physical contact between the bodies. Examplesinclude the effects caused by the earth’s gravitation or its electromagneticfield.Although body forces affect each of the particles composing the body,these forces are normally represented by a single concentrated force actingon the body. In the case of gravitation, this force is called the weight of thebody and acts through the body’s center of gravity.

Support Reactions. The surface forces that develop at the supportsor points of contact between bodies are called reactions. For two-dimensional problems, i.e., bodies subjected to coplanar force systems, thesupports most commonly encountered are shown in Table 1–1. Notecarefully the symbol used to represent each support and the type ofreactions it exerts on its contacting member. In general, one can alwaysdetermine the type of support reaction by imagining the attached memberas being translated or rotated in a particular direction. If the supportprevents translation in a given direction, then a force must be developedon the member in that direction. Likewise, if rotation is prevented, acouple moment must be exerted on the member. For example, a rollersupport can only prevent translation in the contact direction, perpendicularor normal to the surface. Hence, the roller exerts a normal force F on themember at the point of contact. Since the member can freely rotate aboutthe roller, a couple moment cannot be developed on the member.

Many machine elements are pin connectedin order to enable free rotation at theirconnections. These supports exert a force ona member, but no moment.

F

F

Type of connection Reaction

Cable

Roller

One unknown: F

One unknown: F

FSmooth support One unknown: F

External pin

Internal pin

Fx

Fy

Fx

Fy

Two unknowns: Fx, Fy

Fx

FyM

Fixed support Three unknowns: Fx, Fy, M

Two unknowns: Fx, Fy

Type of connection Reaction



Equations of Equilibrium. Equilibrium of a body requires both abalance of forces, to prevent the body from translating or havingaccelerated motion along a straight or curved path, and a balance ofmoments, to prevent the body from rotating. These conditions can beexpressed mathematically by the two vector equations

(1–1)

Here, represents the sum of all the forces acting on the body, andis the sum of the moments of all the forces about any point O

either on or off the body. If an x, y, z coordinate system is establishedwith the origin at point O, the force and moment vectors can be resolvedinto components along the coordinate axes and the above two equationscan be written in scalar form as six equations, namely,

(1–2)

Often in engineering practice the loading on a body can be representedas a system of coplanar forces. If this is the case, and the forces lie in thex–y plane, then the conditions for equilibrium of the body can bespecified by only three scalar equilibrium equations; that is,

(1–3)

In this case, if point O is the origin of coordinates, then moments willalways be directed along the z axis, which is perpendicular to the planethat contains the forces.

Successful application of the equations of equilibrium requirescomplete specification of all the known and unknown forces that act onthe body. The best way to account for these forces is to draw the body’sfree-body diagram. Obviously, if the free-body diagram is drawncorrectly, the effects of all the applied forces and couple moments canbe accounted for when the equations of equilibrium are written.

©Fx = 0©Fy = 0

©MO = 0

©Fx = 0 ©Fy = 0 ©Fz = 0©Mx = 0 ©My = 0 ©Mz = 0

© MO

© F

©F = 0©MO = 0



section

F4

F2

(a)

F1

F3

Fig. 1–2

*The body’s weight is not shown, since it is assumed to be quite small, and therefore negligiblecompared with the other loads.

Internal Resultant Loadings. One of the most important applicationsof statics in the analysis of mechanics of materials problems is to be able todetermine the resultant force and moment acting within a body, which arenecessary to hold the body together when the body is subjected to externalloads. For example, consider the body shown in Fig. 1–2a, which is held inequilibrium by the four external forces. *In order to obtain the internalloadings acting on a specific region within the body, it is necessary to usethe method of sections. This requires that an imaginary section or “cut” bemade through the region where the internal loadings are to be determined.The two parts of the body are then separated, and a free-body diagram ofone of the parts is drawn, Fig. 1–2b. Here it can be seen that there is actuallya distribution of internal force acting on the “exposed” area of the section.These forces represent the effects of the material of the top part of the bodyacting on the adjacent material of the bottom part.

Although the exact distribution of internal loading may be unknown,we can use the equations of equilibrium to relate the external forces onthe body to the distribution’s resultant force and moment, and at any specific point O on the sectioned area, Fig. 1–2c. When doing so,note that acts through point O, although its computed value will notdepend on the location of this point. On the other hand, doesdepend on this location, since the moment arms must extend from O tothe line of action of each external force on the free-body diagram. It willbe shown in later portions of the text that point O is most often chosenat the centroid of the sectioned area, and so we will always choose thislocation for O, unless otherwise stated. Also, if a member is long andslender, as in the case of a rod or beam, the section to be considered isgenerally taken perpendicular to the longitudinal axis of the member.This section is referred to as the cross section.

MRO,

FR

MRO,FR

F1F2

(b)

Fig. 1–2b

F1 F2

O

MRO FR

(c)

Fig. 1–2c


Fig. 1–2d


(d)

O

F1 F2

N

T

M

V

TorsionalMoment

BendingMoment

ShearForce

MROFR

NormalForce

Three Dimensions. Later in this text we will show how to relate theresultant loadings, and to the distribution of force on the sectionedarea, and thereby develop equations that can be used for analysis anddesign. To do this, however, the components of and acting bothnormal or perpendicular to the sectioned area and within the plane of thearea, must be considered, Fig. 1–2d. Four different types of resultantloadings can then be defined as follows:

Normal force, N. This force acts perpendicular to the area. It is developedwhenever the external loads tend to push or pull on the two segments ofthe body.

Shear force, V. The shear force lies in the plane of the area and isdeveloped when the external loads tend to cause the two segments of thebody to slide over one another.

Torsional moment or torque, T. This effect is developed when theexternal loads tend to twist one segment of the body with respect to theother.

Bending moment, M. The bending moment is caused by the externalloads that tend to bend the body about an axis lying within the plane ofthe area.

In this text, note that graphical representation of a moment or torqueis shown in three dimensions as a vector with an associated curl. By theright-hand rule, the thumb gives the arrowhead sense of the vector andthe fingers or curl indicate the tendency for rotation (twist or bending).Using an x, y, z coordinate system, each of the above loadings can bedetermined directly from the six equations of equilibrium applied toeither segment of the body.

MRO,FR

MRO,FR

F1 F2

O

MRO FR

(c)

Fig. 1–2 (cont.)


Fig. 1–3b


In order to design the members of thisbuilding frame, it is first necessary to findthe internal loadings at various pointsalong their length.

section

F4

F3F2

F1

(a)

Fig. 1–3

O

VMO

Nx

y

BendingMoment

ShearForce

NormalForce

(b)

F2

F1

Coplanar Loadings. If the body is subjected to a coplanar system of forces,Fig. 1–3a, then only normal-force, shear, and bending-moment componentswill exist at the section, Fig. 1–3b. If we use the x, y, z coordinate axes, withorigin established at point O, as shown on the left segment, then a directsolution for N can be obtained by applying and V can be obtaineddirectly from Finally, the bending moment can be determineddirectly by summing moments about point O (the z axis), in orderto eliminate the moments caused by the unknowns N and V.

© MO = 0,MO© Fy = 0.

© Fx = 0,

IMPORTANT POINTS• Mechanics of materials is a study of the relationship between the

external loads on a body and the intensity of the internal loadswithin the body.

• External forces can be applied to a body as distributed orconcentrated surface loadings, or as body forces which actthroughout the volume of the body.

• Linear distributed loadings produce a resultant force having amagnitude equal to the area under the load diagram, and havinga location that passes through the centroid of this area.

• A support produces a force in a particular direction on its attachedmember if it prevents translation of the member in that direction,and it produces a couple moment on the member if it preventsrotation.

• The equations of equilibrium and must besatisfied in order to prevent a body from translating withaccelerated motion and from rotating.

• When applying the equations of equilibrium, it is important tofirst draw the body’s free-body diagram in order to account forall the terms in the equations.

• The method of sections is used to determine the internal resultantloadings acting on the surface of the sectioned body. In general,these resultants consist of a normal force, shear force, torsionalmoment, and bending moment.

© M = 0© F = 0



PROCEDURE FOR ANALYSISThe method of sections is used to determine the resultant internalloadings at a point located on the section of a body. To obtain theseresultants, application of the method of sections requires thefollowing steps.

Support Reactions.

• First decide which segment of the body is to be considered. If thesegment has a support or connection to another body, then beforethe body is sectioned, it will be necessary to determine thereactions acting on the chosen segment of the body. Draw thefree-body diagram for the entire body and then apply thenecessary equations of equilibrium to obtain these reactions.

Free-Body Diagram.

• Keep all external distributed loadings, couple moments, torques,and forces acting on the body in their exact locations, then pass animaginary section through the body at the point where the resultantinternal loadings are to be determined.

• If the body represents a member of a structure or mechanicaldevice, the section is often taken perpendicular to the longitudinalaxis of the member.

• Draw a free-body diagram of one of the “cut” segments and indi-cate the unknown resultants N, V, M, and T at the section. Theseresultants are normally placed at the point representing the geo-metric center or centroid of the sectioned area.

• If the member is subjected to a coplanar system of forces, only N,V, and M act at the centroid.

• Establish the x, y, z coordinate axes with origin at the centroid andshow the resultant components acting along the axes.

Equations of Equilibrium.

• Moments should be summed at the section, about each of thecoordinate axes where the resultants act. Doing this eliminatesthe unknown forces N and V and allows a direct solution for M(and T).

• If the solution of the equilibrium equations yields a negative valuefor a resultant, the assumed directional sense of the resultant isopposite to that shown on the free-body diagram.

The following examples illustrate this procedure numerically and alsoprovide a review of some of the important principles of statics.


Fig. 1–4b

Fig. 1–4c


E X A M P L E 1.1

Determine the resultant internal loadings acting on the cross sectionat C of the beam shown in Fig. 1–4a.

180 N/m

540 N

2 m 4 mVC

MC

NC

(b)

BC

(a)

A B

C3 m 6 m

270 N/m

Fig. 1–4

Solution

Support Reactions. This problem can be solved in the most directmanner by considering segment CB of the beam, since then thesupport reactions at A do not have to be computed.

Free-Body Diagram. Passing an imaginary section perpendicular to thelongitudinal axis of the beam yields the free-body diagram of segmentCB shown in Fig. 1–4b. It is important to keep the distributed loadingexactly where it is on the segment until after the section is made. Onlythen should this loading be replaced by a single resultant force. Noticethat the intensity of the distributed loading at C is found by proportion,i.e., from Fig.1–4a, Themagnitude of the resultant of the distributed load is equal to the areaunder the loading curve (triangle) and acts throughthe centroid of thisarea.Thus, which acts from C as shown in Fig. 1–4b.

Equations of Equilibrium. Applying the equations of equilibriumwe have

Ans.

Ans.

Ans.

The negative sign indicates that acts in the opposite direction tothat shown on the free-body diagram. Try solving this problem usingsegment AC, by first obtaining the support reactions at A, which aregiven in Fig. 1–4c.

MC

MC = -1080 N # m

-MC - 540 N12 m2 = 0d+ © MC = 0;

VC = 540 N

VC - 540 N = 0+q © Fy = 0;

NC = 0

-NC = 0:+ © Fx = 0;

1>316 m2 = 2 mF = 121180 N>m216 m2 = 540 N,

w = 180 N>m.w>6 m = 1270 N>m2>9m,

1.5 m0.5 m

1 m

180 N/m90 N/m

540 N135 N

VC

MC

NC

(c)

1215 N

3645 NmCA


Fig. 1–5c

Fig. 1–5b


E X A M P L E 1.2

Determine the resultant internal loadings acting on the cross sectionat C of the machine shaft shown in Fig. 1–5a. The shaft is supportedby bearings at A and B, which exert only vertical forces on the shaft.

225 N

CD

200 mm100 mm 100 mm

50 mm50 mm

800 N/m

B

(a)

A

Fig. 1–5

0.275 m0.125 m 0.100 m

225 N

Ay By

(b)

B

SolutionWe will solve this problem using segment AC of the shaft.

Support Reactions. A free-body diagram of the entire shaft is shownin Fig. 1–5b. Since segment AC is to be considered, only the reactionat A has to be determined. Why?

The negative sign for indicates that acts in the opposite sense tothat shown on the free-body diagram.

Free-Body Diagram. Passing an imaginary section perpendicular tothe axis of the shaft through C yields the free-body diagram of segmentAC shown in Fig. 1–5c.


Ans.

Ans.

Ans.

What do the negative signs for and indicate? As an exercise,calculate the reaction at B and try to obtain the same results usingsegment CBD of the shaft.

MCVC

MC = -5.69 N # m

MC + 40 N10.025 m2 + 18.75 N10.250 m2 = 0d+ © MC = 0;

VC = -58.8 N

-18.75 N - 40 N - VC = 0+q © Fy = 0;

NC = 0:+ © Fx = 0;

AyAy

Ay = -18.75 N

-Ay10.400 m2 + 120 N10.125 m2 - 225 N10.100 m2 = 0d+ © MB = 0;

(c)

40 N18.75 N

0.250 m

0.025 m

MC

VC

CA

NC


Fig. 1–6a


E X A M P L E 1.3

The hoist in Fig. 1–6a consists of the beam AB and attached pulleys,the cable, and the motor. Determine the resultant internal loadingsacting on the cross section at C if the motor is lifting the 500-lb loadW with constant velocity. Neglect the weight of the pulleys and beam.

(b)

4.5 ft

C

0.5 ft

500 lb

500 lb

VC

MC

NCA

Fig. 1–6

4 ft 2 ft6 ft

0.5 ft

0.5 ft

A CD

B

W (a)

SolutionThe most direct way to solve this problem is to section both the cableand the beam at C and then consider the entire left segment.

Free-Body Diagram. See Fig. 1–6b.


Ans.

Ans.

Ans.

As an exercise, try obtaining these same results by considering justthe beam segment AC, i.e., remove the pulley at A from the beam andshow the 500-lb force components of the pulley acting on the beamsegment AC. Also, this problem can be worked by first finding thereactions at B, ( ) and thenconsidering segment CB.

MB = 7000 lb # ftBy = 1000 lb,Bx = 0,

MC = -2000 lb # ft

500 lb 14.5 ft2 - 500 lb 10.5 ft2 + MC = 0d+ © MC = 0;

-500 lb - VC = 0 VC = -500 lb+q © Fy = 0;

500 lb + NC = 0 NC = -500 lb:+ © Fx = 0;


Fig. 1–7d

Fig. 1–7c

Fig. 1–7b


6200 lb

= 4650 lb= 7750 lbFBA FBD

34

5

(c)

B

(d)

MG

NG

VG

2 ft

34

5

7750 lb1500 lb

A G

E X A M P L E 1.4

Determine the resultant internal loadings acting on the cross sectionat G of the wooden beam shown in Fig. 1–7a. Assume the joints at A,B, C, D, and E are pin connected.

(a)

300 lb/ft

2 ft 2 ft 6 ft

1500 lb

A

B

G D

C

3 ft

E

Fig. 1–7

3 ft

6 ft (6 ft) = 4 ft

(6 ft)(300 lb/ft) = 900 lb12

1500 lb

= 2400 lb

= 6200 lb

= 6200 lb

Ey

Ex

FBC

(b)

23

Solution

Support Reactions. Here we will consider segment AG for theanalysis. A free-body diagram of the entire structure is shown in Fig.1–7b. Verify the computed reactions at E and C. In particular, notethat BC is a two-force member since only two forces act on it. For thisreason the reaction at C must be horizontal as shown.

Since BA and BD are also two-force members, the free-bodydiagram of joint B is shown in Fig. 1–7c. Again, verify the magnitudesof the computed forces and

Free-Body Diagram. Using the result for the left section AGof the beam is shown in Fig. 1–7d.

Equations of Equilibrium. Applying the equations of equilibriumto segment AG, we have

Ans.

Ans.

Ans.

As an exercise, compute these same results using segment GE.

MG = 6300 lb # ft

MG - 17750 lb2A35 B 12 ft2 + 1500 lb 12 ft2 = 0d+ © MG = 0;

VG = 3150 lb

-1500 lb + 7750 lb A35 B - VG = 0+q © Fy = 0;

7750 lb A45 B + NG = 0 NG = -6200 lb:+ © Fx = 0;

FBA,

FBD.FBA


Fig. 1–8a


E X A M P L E 1.5

Determine the resultant internal loadings acting on the cross sectionat B of the pipe shown in Fig. 1–8a. The pipe has a mass of 2 kg/mand is subjected to both a vertical force of 50 N and a couple momentof at its end A. It is fixed to the wall at C.

SolutionThe problem can be solved by considering segment AB, which doesnot involve the support reactions at C.

Free-Body Diagram. The x, y, z axes are established at B and thefree-body diagram of segment AB is shown in Fig. 1–8b. The resultantforce and moment components at the section are assumed to act inthe positive coordinate directions and to pass through the centroid ofthe cross-sectional area at B. The weight of each segment of pipe iscalculated as follows:

These forces act through the center of gravity of each segment.

Equations of Equilibrium. Applying the six scalar equations ofequilibrium, we have*

Ans.Ans.

Ans.

Ans.

Ans.

Ans.

What do the negative signs for and indicate? Note thatthe normal force whereas the shear force is

Also, the torsional moment is

and the bending moment is

*The magnitude of each moment about an axis is equal to the magnitude of each forcetimes the perpendicular distance from the axis to the line of action of the force. Thedirection of each moment is determined using the right-hand rule, with positivemoments (thumb) directed along the positive coordinate axes.

30.3 N # m.MB = 2130.322 + 102 =1MB2y = 77.8 N # m

TB =21022 + 184.322 = 84.3 N.

VB =NB = 1FB2y = 0,1MB2y1MB2x

1MB2z = 0©1MB2z = 0;

1MB2y = -77.8 N # m

1MB2y + 24.525 N 10.625 m2 + 50 N 11.25 m2 = 0©1MB2y = 0;

1MB2x = -30.3 N # m

- 9.81 N 10.25 m2 = 0

1MB2x + 70 N # m - 50 N 10.5 m2 - 24.525 N 10.5 m2©1MB2x = 0;

1FB2z = 84.3 N

1FB2z - 9.81 N - 24.525 N - 50 N = 0© Fz = 0;

1FB2y = 0© Fy = 0;

1FB2x = 0© Fx = 0;

WAD = 12 kg>m211.25 m219.81 N>kg2 = 24.525 N

WBD = 12 kg>m210.5 m219.81 N>kg2 = 9.81 N

70 N # m

0.625 m

70 N·m

(b)

y0.625 m

A

50 N

0.25 m0.25 m

x

z

9.81 N

24.525 NB

(MB)z

(MB)y

(MB)x

(FB)x

(FB)y

(FB)z

Fig. 1–8

0.75 m

50 N

1.25 m

B

A

0.5 m

C

D

70 Nm(a)



1-2. Determine the resultant internal torque acting onthe cross sections through points C and D. The supportbearings at A and B allow free turning of the shaft.

1-3. Determine the resultant internal torque acting onthe cross sections through points B and C.

1-5. Determine the resultant internal loadings acting onthe cross section through point D of member AB.

8 kN

3 m

1 m

6 kN6 kN

4.5 kN4.5 kN

200 mm200 mm

A

(b)

200 mm200 mm

Prob. 1–1

3 ft

2 ft

2 ft

1 ft

B

A

C

500 lbft

350 lbft

600 lbft

Prob. 1–3

A

BD

C300 mm

200 mm

150 mm200 mm

250 mm

150 mm

400 Nm

150 Nm

250 Nm

Prob. 1–2

3 kip3 kip

5 kip

10 ft

4 ft

4 ft

8 in.8 in.

A

C

D

(a)

B

Prob. 1–1

P R O B L E M S

1-1. Determine the resultant internal normal force actingon the cross section through point A in each column. In (a),segment BC weighs 180 lb/ft and segment CD weighs 250lb/ft. In (b), the column has a mass of 200 kg/m.

*1-4. Determine the resultant internal normal and shearforce in the member at (a) section a–a and (b) sectionb–b, each of which passes through point A. The 500-lbload is applied along the centroidal axis of the member.

30°

A

ba

b a

500 lb500 lb

Prob. 1–4

200 mm

50 mm

150 mm

50 mm

300 mm

C

A D

B

70 N m

Prob. 1–5


PROBLEMS • 17

1-6. The beam AB is pin supported at A and supportedby a cable BC. Determine the resultant internal loadingsacting on the cross section at point D.

1-7. Solve Prob. 1–6 for the resultant internal loadingsacting at point E.

*1-8. The beam AB is fixed to the wall and has a uniformweight of 80 lb/ft. If the trolley supports a load of 1500lb, determine the resultant internal loadings acting on thecross sections through points C and D.

1-9. Determine the resultant internal loadings acting onthe cross section at point C. The cooling unit has a totalweight of 52 kN and a center of gravity at G.

1-10. Determine the resultant internal loadings actingon the cross sections through points D and E of the frame.

1-11. Determine the resultant internal loadings actingon the cross sections through points F and G of the frame.

4 ft

B

C

6 ft

8 ft

3 ft

A

D

1200 lb

E

3 ft

Probs. 1–6/7

D

5 ft20 ft

3 ft10 ft

C

BA

1500 lb

Prob. 1–8

3 ft

F

3 ft

30° 30°

0.2 ft

G

A B

ED

C

Prob. 1–9

4 ft

A

D2 ft

C

B

75 lb/ft

1 ft

150 lb

1 ft

E

30°

2 ft1 ft

G

1 ft

F

Probs. 1–10/11



A

BC

12 ft3 ft

60 lb/ ft

Prob. 1–17

*1-12. Determine the resultant internal loadings actingon (a) section a–a and (b) section b–b. Each section islocated through the centroid, point C.

1-13. Determine the resultant internal loadings actingon the cross section through point C in the beam. Theload D has a mass of 300 kg and is being hoisted by themotor M with constant velocity.

1-14. Determine the resultant internal loadings acting onthe cross section through point E of the beam in Prob. 1–13.

1-15. The 800-lb load is being hoisted at a constantspeed using the motor M, which has a weight of 90 lb.Determine the resultant internal loadings acting on thecross section through point B in the beam. The beam hasa weight of 40 lb/ft and is fixed to the wall at A.

*1-16. Determine the resultant internal loadings actingon the cross section through points C and D of the beamin Prob. 1–21.

1-17. Determine the resultant internal loadings actingon the cross section at point B.

45°

8 ft

4 ft45°

A

C

B

b

a

a b

600 lb/ft

Prob. 1–12

M

2 m

B

CE

D

A

0.1 m

2 m 2 m

1.5 m

0.1 m

1 m

2

Probs. 1–13/14

M

4 ft 3 ft 4 ft

C B

1.5 ft

A

0.25 ft

4 ft 3 ft

D

Probs. 1–15/16


PROBLEMS • 19

1-18. The beam supports the distributed load shown.Determine the resultant internal loadings acting on thecross section through point C. Assume the reactions atthe supports A and B are vertical.

1-19. Determine the resultant internal loadings actingon the cross section through point D in Prob. 1–18.

*1-20. The serving tray T used on an airplane issupported on each side by an arm. The tray is pinconnected to the arm at A, and at B there is a smoothpin. (The pin can move within the slot in the arms topermit folding the tray against the front passenger seatwhen not in use.) Determine the resultant internalloadings acting on the cross section of the arm throughpoint C when the tray arm supports the loads shown.

1-21. The metal stud punch is subjected to a force of 120N on the handle. Determine the magnitude of the reactiveforce at the pin A and in the short link BC. Also,determine the internal resultant loadings acting on thecross section passing through the handle arm at D.

1-22. Solve Prob. 1–21 for the resultant internal loadingsacting on the cross section passing through the handlearm at E and at a cross section of the short link BC.

1-23. The pipe has a mass of 12 kg/m. If it is fixed to thewall at A, determine the resultant internal loadings actingon the cross section at B. Neglect the weight of the wrenchCD.

3 m 3 m

DCA B

0.5 kN/m

1.5 kN/m

3 m

Probs. 1–18/19

9 N

500 mm

12 N

15 mm 150 mm

60°

AB

C

T

CV

CM

CN

100 mm

Prob. 1–20

60° 50 mm

100 mm

200 mm

300 mm

B

C

D

120 N

50 mm 100 mm

E

30°A

Probs. 1–21/22

300 mm

200 mm

150 mm

60 N

60 N400 mm

150 mm

B

A

x

y

z

C

D

Prob. 1–23



1-27. A hand crank that is used in a press has thedimensions shown. Determine the resultant internalloadings acting on the cross section at A if a vertical forceof 50 lb is applied to the handle as shown. Assume thecrank is fixed to the shaft at B.

B

30°

z

x

y

50 lb

3 in.

7 in.

7 in.

A

Prob. 1–27

*1-24. The main beam AB supports the load on the wingof the airplane. The loads consist of the wheel reaction of35,000 lb at C, the 1200-lb weight of fuel in the tank of thewing, having a center of gravity at D, and the 400-lb weightof the wing, having a center of gravity at E. If it is fixed tothe fuselage at A, determine the resultant internal loadingson the beam at this point. Assume that the wing does nottransfer any of the loads to the fuselage, except throughthe beam.

1-25. Determine the resultant internal loadings actingon the cross section through point B of the signpost. Thepost is fixed to the ground and a uniform pressure of

acts perpendicular to the face of the sign.7 lb>ft2

1-26. The shaft is supported at its ends by two bearingsA and B and is subjected to the forces applied to thepulleys fixed to the shaft. Determine the resultant internalloadings acting on the cross section located at point C.The 300-N forces act in the direction and the 500-Nforces act in the direction. The journal bearings at Aand B exert only x and z components of force on the shaft.

+x-z

6 ft

4 ft2 ft

1.5 ft

D E

1 ft

A

B

C

35,000 lb

z

x

y

Prob. 1–24

4 ft

z

y

6 ft

x

B

A

3 ft

2 ft

3 ft

7 lb/ft2

Prob. 1–25

60° 50 mm

100 mm

200 mm

300 mm

B

C

D

120 N

50 mm 100 mm

E

30°A

Prob. 1–26

Wrong ArtAADFKFO0


PROBLEMS • 21

*1-28. Determine the resultant internal loadings actingon the cross section of the frame at points F and G. Thecontact at E is smooth.

1-29. The bolt shank is subjected to a tension of 80 lb.Determine the resultant internal loadings acting on thecross section at point C.

1-30. Determine the resultant internal loadings acting onthe cross section at points B and C of the curved member.

1-31. The curved rod AD of radius r has a weight perlength of w. If it lies in the vertical plane, determine theresultant internal loadings acting on the cross sectionthrough point B. Hint: The distance from the centroid Cof segment AB to point O is OC = [2r sin 1u>22]>u.

4 ft

1.5 ft 1.5 ft

3 ft

A

B

C

E

D

G

80 lb

5 ft

2 ft

2 ft

30°

F

Prob. 1–28

A B

C

90°6 in.

Prob. 1–29

C

A

B

30°

45°2 ft

500 lb

3

45

Prob. 1–30

*1-32. The curved rod AD of radius r has a weight perlength of w. If it lies in the horizontal plane, determinethe resultant internal loadings acting on the cross sectionthrough point B. Hint: The distance from the centroid Cof segment AB to point O is CO = 0.9745r.

1-33. A differential element taken from a curved bar isshown in the figure. Show that

and dT>du = M.dM>du = -T,dV>du = -N,dN>du = V,

θ

O

r

C

B

D

Aθ

Prob. 1–31

A

B

C 45°90°

D

O

r

22.5°

Prob. 1–32

M V

N

+V dV

dθ

+M dM

+N dN

T+dT

T

Prob. 1–33



F1 F2

F∆

∆A

F∆F∆ z

z

y

F∆ y

x

F∆ x

yz

yxy

xy

xz

x

z

(c)x y

F1

(b)

zz

x y(a)

x y

Fig. 1–10

1.3 Stress

It was stated in Section 1.2 that the force and moment acting at aspecified point on the sectioned area of a body, Fig. 1–9, represents theresultant effects of the actual distribution of force acting over thesectioned area, Fig. 1–9b. Obtaining this distribution of internal loadingis of primary importance in mechanics of materials.To solve this problemit is necessary to establish the concept of stress.

Consider the sectioned area to be subdivided into small areas, such asshown dark shaded in Fig. 1–10a. As we reduce to a smaller and

smaller size, we must make two assumptions regarding the properties ofthe material. We will consider the material to be continuous, that is, toconsist of a continuum or uniform distribution of matter having no voids,rather than being composed of a finite number of distinct atoms ormolecules. Furthermore, the material must be cohesive, meaning that allportions of it are connected together, rather than having breaks, cracks,or separations. A typical finite yet very small force acting on itsassociated area is shown in Fig. 1–10a. This force, like all the others,will have a unique direction, but for further discussion we will replace itby its three components, namely, and which are takentangent and normal to the area, respectively.As the area approacheszero, so do the force and its components; however, the quotient ofthe force and area will, in general, approach a finite limit. This quotientis called stress, and as noted, it describes the intensity of the internal forceon a specific plane (area) passing through a point.

¢F¢A

¢Fz,¢Fy,¢Fx,

¢A,¢F,

¢A¢A

F1 F2

O

(a)

MRO FR

Fig. 1–9


SECTION 1.3 Stress • 23

Normal Stress. The intensity of force, or force per unit area, actingnormal to is defined as the normal stress, (sigma). Since isnormal to the area then

(1–4)

If the normal force or stress “pulls” on the area element as shownin Fig. 1–10a, it is referred to as tensile stress, whereas if it “pushes” on

it is called compressive stress.

Shear Stress. The intensity of force, or force per unit area, actingtangent to is called the shear stress, (tau). Here we have shear stresscomponents,

(1–5)

Note that the subscript notation z in is used to reference thedirection of the outward normal line, which specifies the orientation ofthe area Fig. 1–11. Two subscripts are used for the shear-stresscomponents, and The z axis specifies the orientation of the area,and x and y refer to the direction lines for the shear stresses.

General State of Stress. If the body is further sectioned by planesparallel to the x–z plane, Fig. 1–10b, and the y–z plane, Fig. 1–10c, we canthen “cut out” a cubic volume element of material that represents thestate of stress acting around the chosen point in the body, Fig. 1–12. Thisstate of stress is then characterized by three components acting on eachface of the element. These stress components describe the state of stressat the point only for the element orientated along the x, y, z axes. Had thebody been sectioned into a cube having some other orientation, then thestate of stress would be defined using a different set of stress components.

Units. In the International Standard or SI system, the magnitudes ofboth normal and shear stress are specified in the basic units of newtonsper square meter This unit, called a pascal israther small, and in engineering work prefixes such as kilo- symbolized by k, mega- symbolized by M, or giga- symbolized by G, are used to represent larger, more realistic values ofstress. *Likewise, in the U.S. Customary or Foot-Pound-Second systemof units, engineers usually express stress in pounds per square inch (psi)or kilopounds per square inch (ksi), where 1 kilopound 1kip2 = 1000 lb.

11092,11062, 11032,11 Pa = 1 N>m221N>m22.

tzy.tzx

¢A,

sz

txy = lim¢A:0

¢Fy

¢A

tzx = lim¢A:0

¢Fx

¢A

t¢A

¢A

¢A

sz = lim¢A:0

¢F2

¢A

¢Fzs¢A

z

zyzx

x y

z

Fig. 1–11

xy

z

zσzx

zy

yz

yx

xz

xσ xyyσ

Fig. 1–12

*Sometimes stress is expressed in units of where However, in theSI system, prefixes are not allowed in the denominator of a fraction and therefore it isbetter to use the equivalent 1 N>mm2 - 1 MN>m2 = 1 MPa.

1 mm - 10-3 m.N>mm2,


Fig. 1–13b

P

P

External force

Cross-sectionalarea

Internal force

(b)

Fig. 1–13a


1.4 Average Normal Stress in an Axially Loaded Bar

Frequently structural or mechanical members are made long and slender.Also, they are subjected to axial loads that are usually applied to the endsof the member. Truss members, hangers, and bolts are typical examples. Inthis section we will determine the average stress distribution acting on thecross section of an axially loaded bar, such as the one having the generalform shown in Fig. 1–13a. This section defines the cross-sectional area ofthe bar, and since all such cross sections are the same, the bar is referredto as being prismatic. If we neglect the weight of the bar and section it asindicated, then, for equilibrium of the bottom segment, Fig. 1–13b, theinternal resultant force acting on the cross-sectional area must be equal inmagnitude, opposite in direction, and collinear to the external force actingat the bottom of the bar.

Assumptions. Before we determine the average stress distribution actingover the bar’s cross-sectional area, it is necessary to make two simplifyingassumptions concerning the material description and the specific applicationof the load.

1. It is necessary that the bar remains straight both before and after theload is applied, and also, the cross section should remain flat or planeduring the deformation, that is, during the time the bar changes itsvolume and shape. If this occurs, then horizontal and vertical grid linesinscribed on the bar will deform uniformly when the bar is subjectedto the load, Fig. 1–13c. Here we will not consider regions of the barnear its ends, where application of the external loads can causelocalized distortions. Instead we will focus only on the stressdistribution within the bar’s midsection.

2. In order for the bar to undergo uniform deformation, it is necessary thatP be applied along the centroidal axis of the cross section,and the materialmust be homogeneous and isotropic.Homogeneous material has the samephysical and mechanical properties throughout its volume, and isotropicmaterial has these same properties in all directions. Many engineeringmaterials may be approximated as being both homogeneous and isotropicas assumed here. Steel, for example, contains thousands of randomlyoriented crystals in each cubic millimeter of its volume, and since mostproblems involving this material have a physical size that is much largerthan a single crystal, the above assumption regarding its materialcomposition is quite realistic. It should be mentioned, however, that steelcan be made anisotropic by cold-rolling, i.e., rolling or forging it atsubcritical temperatures. Anisotropic materials have different propertiesin different directions, and although this is the case, if the anisotropy isoriented along the bar’s axis, then the bar will also deform uniformly whensubjected to an axial load. For example, timber, due to its grains or fibersof wood, is an engineering material that is homogeneous and anisotropicand is therefore suited for the following analysis.

P

P

(a)

(c)

Region ofuniformdeformationof bar

P

P

⎧⎪⎪⎪⎨⎪⎪⎪⎩

Fig. 1–13


SECTION 1.4 Average Normal Stress in an Axially Loaded Bar • 25

Average Normal Stress Distribution. Provided the bar is subjected toa constant uniform deformation as noted, then this deformation is the resultof a constant normal stress Fig. 1–13d.As a result,each area on the crosssection is subjected to a force and the sum of these forces actingover the entire cross-sectional area must be equivalent to the internal resultantforce P at the section. If we let and therefore then,recognizing is constant, we have

(1–6)

Here

average normal stress at any point on the cross-sectional area

internal resultant normal force, which is applied through the centroidof the cross-sectional area.P is determined using the method of sectionsand the equations of equilibrium.

cross-sectional area of the bar

The internal load P must pass through the centroid of the cross-sectionsince the uniform stress distribution will produce zero moments about anyx and y axes passing through this point, Fig. 1–13d. When this occurs,

These equations are indeed satisfied, since by definition of the centroid,and (See Appendix A.)

Equilibrium. It should be apparent that only a normal stress exists onany volume element of material located at each point on the cross sectionof an axially loaded bar. If we consider vertical equilibrium of the element,Fig. 1–14, then applying the equation of force equilibrium,

In other words, the two normal stress components on the element mustbe equal in magnitude but opposite in direction. This is referred to asuniaxial stress.

s = s¿s1¢A2 - s¿1¢A2 = 0© Fz = 0;

x dA = 0.y dA = 0

0 = - A

x dF = - A

xs dA = -sA

x dA1MR2y = © My;

0 = A

y dF = A

ys dA = sA

y dA1MR2x = © Mx;

A =

P =s =

s =P

A

P = s A

dF = A

© dA+q FRz = © Fz;

s

¢F : dF,¢A : dA

¢F = s ¢A,¢As,

(d)

P

F∆ = ∆σ A

P

y

x

x

z

σ

y

A∆

Fig. 1–13 (cont.)

σ

σ ¿

A

Fig. 1–14



The previous analysis applies to members subjected to either tensionor compression, as shown in Fig. 1–15. As a graphical interpretation, themagnitude of the internal resultant force P is equivalent to the volumeunder the stress diagram; that is,Furthermore, as a consequence of the balance of moments, this resultantpasses through the centroid of this volume.

Although we have developed this analysis for prismatic bars, thisassumption can be relaxed somewhat to include bars that have a slighttaper. For example, it can be shown, using the more exact analysis of thetheory of elasticity, that for a tapered bar of rectangular cross section,for which the angle between two adjacent sides is 15°, the average normalstress, as calculated by is only 2.2% less than its value foundfrom the theory of elasticity.

Maximum Average Normal Stress. In our analysis both the internalforce P and the cross-sectional area A were constant along the longitudinalaxis of the bar, and as a result the normal stress is also constantthroughout the bar’s length. Occasionally, however, the bar may besubjected to several external loads along its axis, or a change in its cross-sectional area may occur. As a result, the normal stress within the barcould be different from one section to the next, and, if the maximumaverage normal stress is to be determined, then it becomes important tofind the location where the ratio P>A is a maximum. To do this it isnecessary to determine the internal force P at various sections along thebar. Here it may be helpful to show this variation by drawing an axial ornormal force diagram. Specifically, this diagram is a plot of the normalforce P versus its position x along the bar’s length. As a sign convention,P will be positive if it causes tension in the member, and negative if itcauses compression. Once the internal loading throughout the bar isknown, the maximum ratio of P>A can then be identified.

s = P>A

s = P>A,

P = s A 1volume = height * base2.

This steel tie rod is used to suspend aportion of a staircase, and as a result it issubjected to tensile stress.

= P_A

P

PP

P

Tension Compression

Fig. 1–15



IMPORTANT POINTS• When a body that is subjected to an external load is sectioned,

there is a distribution of force acting over the sectioned area whichholds each segment of the body in equilibrium. The intensity ofthis internal force at a point in the body is referred to as stress.

• Stress is the limiting value of force per unit area, as the areaapproaches zero. For this definition, the material at the point isconsidered to be continuous and cohesive.

• In general, there are six independent components of stress at eachpoint in the body, consisting of normal stress, and shearstress,

• The magnitude of these components depends upon the type ofloading acting on the body, and the orientation of the element atthe point.

• When a prismatic bar is made from homogeneous and isotropicmaterial, and is subjected to axial force acting through thecentroid of the cross-sectioned area, then the material within thebar is subjected only to normal stress. This stress is assumed to beuniform or averaged over the cross-sectional area.

txz.tyz,txy,sz,sy,sx,

PROCEDURE FOR ANALYSISThe equation gives the average normal stress on the cross-sectional area of a member when the section is subjected to aninternal resultant normal force P. For axially loaded members,application of this equation requires the following steps.

Internal Loading.

• Section the member perpendicular to its longitudinal axis at thepoint where the normal stress is to be determined and use thenecessary free-body diagram and equation of force equilibriumto obtain the internal axial force P at the section.

Average Normal Stress.

• Determine the member’s cross-sectional area at the section andcompute the average normal stress

• It is suggested that be shown acting on a small volume elementof the material located at a point on the section where stress iscalculated. To do this, first draw on the face of the elementcoincident with the sectioned area A. Here acts in the samedirection as the internal force P since all the normal stresses on thecross section act in this direction to develop this resultant. Thenormal stress acting on the opposite face of the element can bedrawn in its appropriate direction.

s

s

s

s

s = P>A.

s = P>A


Fig. 1–16a


(d)

30 kN

85.7 MPa35 mm

10 mm

Fig. 1–16

E X A M P L E 1.6

The bar in Fig. 1–16a has a constant width of 35 mm and a thicknessof 10 mm. Determine the maximum average normal stress in the barwhen it is subjected to the loading shown.

(b)

9 kN

9 kN

12 kN

12 kN

= 12 kNPAB

= 30 kNPBC

22 kN= 22 kNPCD

P(kN)

x122230

(c)

12 kN 22 kN9 kN

9 kN

4 kN

4 kN35 mm

A DB C

(a)

Solution

Internal Loading. By inspection, the internal axial forces in regions AB,BC, and CD are all constant yet have different magnitudes. Using themethod of sections, these loadings are determined in Fig. 1–16b; and thenormal force diagram which represents these results graphically is shownin Fig. 1–16c. By inspection, the largest loading is in region BC, where

Since the cross-sectional area of the bar is constant, thelargest average normal stress also occurs within this region of the bar.

Average Normal Stress. Applying Eq. 1–6, we have

Ans.

The stress distribution acting on an arbitrary cross section of thebar within region BC is shown in Fig. 1–16d. Graphically the volume(or “block”) represented by this distribution of stress is equivalent tothe load of 30 kN; that is, 30 kN = 185.7 MPa2135 mm2110 mm2.

sBC =PBC

A=

3011032N10.035 m210.010 m2 = 85.7 MPa

PBC = 30 kN.


Fig. 1–17c

Fig. 1–17b


E X A M P L E 1.7

The 80-kg lamp is supported by two rods AB and BC as shown inFig. 1–17a. If AB has a diameter of 10 mm and BC has a diameterof 8 mm, determine the average normal stress in each rod.

(b)

60°

FBA FBC

y

x

80(9.81) = 784.8 N

B

34

5

A

60° B

C

34

5

(a)

Fig. 1–17

Solution

Internal Loading. We must first determine the axial force in eachrod.A free-body diagram of the lamp is shown in Fig. 1–17b.Applyingthe equations of force equilibrium yields

By Newton’s third law of action, equal but opposite reaction, theseforces subject the rods to tension throughout their length.

Average Normal Stress. Applying Eq. 1–6, we have

Ans.

Ans.

The average normal stress distribution acting over a cross sectionof rod AB is shown in Fig. 1–17c, and at a point on this cross section,an element of material is stressed as shown in Fig. 1–17d.

sBA =FBA

ABA=

632.4 N

p10.005 m22 = 8.05 MPa

sBC =FBC

ABC=

395.2 N

p10.004 m22 = 7.86 MPa

FBC = 395.2 N, FBA = 632.4 N

FBC A35 B + FBA sin 60° - 784.8 N = 0+q © Fy = 0;

FBC A43 B - FBA cos 60° = 0:+ © Fx = 0;

632.4 N

8.05 MPa

8.05 MPa

(c)(d)


Fig. 1–18b


E X A M P L E 1.8

The casting shown in Fig. 1–18a is made of steel having a specificweight of Determine the average compressive stressacting at points A and B.

gst = 490 lb>ft3.

0.75 ft

0.75 ft

2.75 ft

y

z

x

(a)

A

B0.75 ft0.4 ft

Fig. 1–18

2.75 ft

(b)

A

P

(c)

9.36 psi

B

Wst

Solution

Internal Loading. A free-body diagram of the top segment of thecasting where the section passes through points A and B is shown inFig. 1–18b.The weight of this segment is determined from Thus the internal axial force P at the section is

Average Compressive Stress. The cross-sectional area at the sectionis and so the average compressive stress becomes

Ans.

The stress shown on the volume element of material in Fig. 1–18cis representative of the conditions at either point A or B. Notice thatthis stress acts upward on the bottom or shaded face of the elementsince this face forms part of the bottom surface area of the cut section,and on this surface, the resultant internal force P is pushing upward.

= 9.36 psi

= 1347.5 lb>ft2 = 1347.5 lb>ft2 11 ft2>144 in22 s =

P

A=

2381 lb

p10.75 ft22A = p10.75 ft22,

P = 2381 lb

P - 1490 lb>ft3212.75 ft2p10.75 ft22 = 0

P - Wst = 0+q © Fz = 0;

Wst = gstVst.


Fig. 1–19b


E X A M P L E 1.9

Member AC shown in Fig. 1–19a is subjected to a vertical force of3 kN. Determine the position x of this force so that the averagecompressive stress at the smooth support C is equal to the averagetensile stress in the tie rod AB. The rod has a cross-sectional areaof and the contact area at C is 650 mm2.400 mm2

x

A

B

C

200 mm

(a)

3 kN

Fig. 1–19(b)

x

3 kN

A

200 mm

FAB

FC

Solution

Internal Loading. The forces at A and C can be related byconsidering the free-body diagram for member AC, Fig. 1–19b. Thereare three unknowns, namely, and x. To solve this problem wewill work in units of newtons and millimeters.

(1)

(2)

Average Normal Stress. A necessary third equation can be writtenthat requires the tensile stress in the bar AB and the compressivestress at C to be equivalent, i.e.,

Substituting this into Eq.1, solving for then solving for we obtain

The position of the applied load is determined from Eq. 2,

Ans.

Note that as required.0 6 x 6 200 mm,

x = 124 mm

FC = 1857 N

FAB = 1143 N

FC,FAB,

FC = 1.625FAB

s =FAB

400 mm2 =FC

650 mm2

-3000 N1x2 + FC 1200 mm2 = 0d+ © MA = 0;

FAB + FC - 3000 N = 0+q © Fy = 0;

FC,FAB,



(b)

(c)

F

F

VV

avg

F

(a)

BD

AC

Fig. 1–20

1.5 Average Shear Stress

Shear stress has been defined in Section 1–3 as the stress componentthat acts in the plane of the sectioned area. In order to show how thisstress can develop, we will consider the effect of applying a force F tothe bar in Fig. 1–20a. If the supports are considered rigid, and F is largeenough, it will cause the material of the bar to deform and fail along theplanes identified by AB and CD. A free-body diagram of theunsupported center segment of the bar, Fig. 1–20b, indicates that theshear force must be applied at each section to hold the segmentin equilibrium. The average shear stress distributed over each sectionedarea that develops this shear force is defined by

(1–7)

Here

average shear stress at the section, which is assumed to be thesame at each point located on the section

internal resultant shear force at the section determined from theequations of equilibrium

area at the section

The distribution of average shear stress is shown acting over thesections in Fig. 1–20c. Notice that is in the same direction as V, sincethe shear stress must create associated forces all of which contribute tothe internal resultant force V at the section.

The loading case discussed in Fig. 1–20 is an example of simple ordirect shear, since the shear is caused by the direct action of the appliedload F. This type of shear often occurs in various types of simpleconnections that use bolts, pins, welding material, etc. In all these cases,however, application of Eq. 1–7 is only approximate. A more preciseinvestigation of the shear-stress distribution over the critical section oftenreveals that much larger shear stresses occur in the material than thosepredicted by this equation. Although this may be the case, applicationof Eq. 1–7 is generally acceptable for many problems in engineeringdesign and analysis. For example, engineering codes allow its use whenconsidering design sizes for fasteners such as bolts and for obtaining thebonding strength of joints subjected to shear loadings. In this regard, twotypes of shear frequently occur in practice, which deserve separatetreatment.

tavg

A =

V =

tavg =

tavg =V

A

V = F>2


SECTION 1.5 Average Shear Stress • 33

Single Shear. The steel and wood joints shown in Figs. 1–21a and 1–21c,respectively,are examples of single-shear connections and are often referredto as lap joints. Here we will assume that the members are thin and that thenut in Fig. 1–21a is not tightened to any great extent so friction between themembers can be neglected. Passing a section between the members yieldsthe free-body diagrams shown in Figs. 1–21b and 1–21d. Since the membersare thin, we can neglect the moment created by the force F. Hence forequilibrium the cross-sectional area of the bolt in Fig.1–21b and the bondingsurface between the members in Fig.1–21d are subjected only to a single shearforce This force is used in Eq. 1–7 to determine the average shearstress acting on the colored section of Fig. 1–21d.

Double Shear. When the joint is constructed as shown in Fig. 1–22a or1–22c, two shear surfaces must be considered.These types of connectionsare often called double lap joints. If we pass a section between each of themembers, the free-body diagrams of the center member are shown in Figs.1–22b and 1–22d. Here we have a condition of double shear. Consequently,

acts on each sectioned area and this shear must be consideredwhen applying tavg = V>A.V = F>2

V = F.

F

F

(a)

F

(b)

= FV

F

(c)F

(d)

= FV

F

Fig. 1–21

F

(d)

=V F––2

=V F––2

F

(c)

F––2

F––2

F

(b)

=V F––2

=V F––2

F

(a)

F––2

F––2

Fig. 1–22

The pin on this tractor is subjected to doubleshear.

Reduce photo to 72%.AADFKFV0



Pure shear

=

Section plane

x

y

z

yz

yz

zy

zy

y

x

z

¿

¿

Fig. 1–23

Equilibrium. Consider a volume element of material taken at a pointlocated on the surface of any sectioned area on which the average shearstress acts, Fig. 1–23a. If we consider force equilibrium in the y direction,then

force| |

stress area| | | |

And in a similar manner, force equilibrium in the z direction yieldsFinally, taking moments about the x axis,

moment| |

force arm| |

stress area| | | | | |

so that

In other words, force and moment equilibrium requires the shear stressacting on the top face of the element, to be accompanied by shear stressacting on three other faces, Fig. 1–23b. Here all four shear stresses musthave equal magnitude and be directed either toward or away fromeach other at opposite edges of the element. This is referred to as thecomplementary property of shear, and under the conditions shown inFig. 1–23, the material is subjected to pure shear.

Although we have considered here a case of simple shear as causedby the direct action of a load, in later chapters we will show that shearstress can also arise indirectly due to the action of other types of loading.

tzy = tœzy = tyz = tœ

yz = t

tzy = tyz

-tzy1¢x ¢y2 ¢z + tyz1¢x ¢z2 ¢y = 0© Mx = 0;

tyz = tœyz.

tzy = tœzy

tzy1¢x ¢y2 - tœzy ¢x ¢y = 0© Fy = 0;



IMPORTANT POINTS• If two parts which are thin or small are joined together, the

applied loads can cause shearing of the material with negligiblebending. If this is the case, it is generally suitable for engineeringanalysis to assume that an average shear stress acts over the cross-sectional area.

• Oftentimes fasteners, such as nails and bolts, are subjected toshear loads. The magnitude of a shear force on the fastener isgreatest along a plane which passes through the surfaces beingjoined. A carefully drawn free-body diagram of a segment of thefastener will enable one to obtain the magnitude and direction ofthis force.

PROCEDURE FOR ANALYSISThe equation is used to compute only the average shearstress in the material. Application requires the following steps.

Internal Shear.

• Section the member at the point where the average shear stress isto be determined.

• Draw the necessary free-body diagram, and calculate the internalshear force V acting at the section that is necessary to hold the partin equilibrium.

Average Shear Stress.

• Determine the sectioned area A, and compute the average shearstress

• It is suggested that be shown on a small volume element ofmaterial located at a point on the section where it is determined.To do this, first draw on the face of the element, coincidentwith the sectioned area A. This shear stress acts in the samedirection as V. The shear stresses acting on the three adjacentplanes can then be drawn in their appropriate directions followingthe scheme shown in Fig. 1–23.

tavg

tavg

tavg = V>A.

tavg = V>A


Fig. 1–24a


E X A M P L E 1.10

The bar shown in Fig. 1–24a has a square cross section for which thedepth and thickness are 40 mm. If an axial force of 800 N is appliedalong the centroidal axis of the bar’s cross-sectional area, determinethe average normal stress and average shear stress acting on thematerial along (a) section plane a–a and (b) section plane b–b.

a

a

b

b

800 N

20 mm60°

(a)

20 mm

(b)

800 N P = 800 N

(c)

500 kPa

500 kPa

Fig. 1–24

Solution

Part (a)Internal Loading. The bar is sectioned, Fig. 1–24b, and the internalresultant loading consists only of an axial force for which

Average Stress. The average normal stress is determined from Eq.1–6.

Ans.

No shear stress exists on the section, since the shear force at thesection is zero.

Ans.

The distribution of average normal stress over the cross section isshown in Fig. 1–24c.

tavg = 0

s =P

A=

800 N10.04 m210.04 m2 = 500 kPa

P = 800 N.


Fig. 1–24e

Fig. 1–24d


V

800 N

60°

(d)

30°

y y¿

x¿

x30°

60°N

800 N

Part (b)Internal Loading. If the bar is sectioned along b–b, the free-bodydiagram of the left segment is shown in Fig. 1–24d. Here both a normalforce (N) and shear force (V) act on the sectioned area. Using x, yaxes, we require

or, more directly, using axes,

Solving either set of equations,

Average Stresses. In this case the sectioned area has a thicknessand depth of 40 mm and respectively,Fig. 1–24a. Thus the average normal stress is

Ans.

and the average shear stress is

Ans.

The stress distribution is shown in Fig. 1–24e.

tavg =V

A=

400 N10.04 m210.04619 m2 = 217 kPa

s =N

A=

692.8 N10.04 m210.04619 m2 = 375 kPa

40 mm>sin 60° = 46.19 mm,

N = 692.8 NV = 400 N

V - 800 N sin 30° = 0+˚© Fy¿ = 0;

N - 800 N cos 30° = 0+Ω© Fx¿ = 0;

y¿x¿,

V sin 60° - N cos 60° = 0+q © Fy = 0;

-800 N + N sin 60° + V cos 60° = 0:+ © Fx = 0;

(e)375 kPa

217 kPa

375 kPa


Fig. 1–25eFig. 1–25d

Fig. 1–25b

Fig. 1–25a


E X A M P L E 1.11

The wooden strut shown in Fig. 1–25a is suspended from a 10-mm-diameter steel rod, which is fastened to the wall. If the strut supportsa vertical load of 5 kN, compute the average shear stress in the rodat the wall and along the two shaded planes of the strut, one of whichis indicated as abcd.

Solution

Internal Shear. As shown on the free-body diagram in Fig. 1–25b,the rod resists a shear force of 5 kN where it is fastened to the wall.A free-body diagram of the sectioned segment of the strut that is incontact with the rod is shown in Fig. 1–25c. Here the shear force actingalong each shaded plane is 2.5 kN.

Average Shear Stress. For the rod,

Ans.

For the strut,

Ans.

The average-shear-stress distribution on the sectioned rod and strutsegment is shown in Figs. 1–25d and 1–25e, respectively. Also shownwith these figures is a typical volume element of the material takenat a point located on the surface of each section. Note carefully howthe shear stress must act on each shaded face of these elements andthen on the adjacent faces of the elements.

tavg =V

A=

2500 N10.04 m210.02 m2 = 3.12 MPa

tavg =V

A=

5000 N

p10.005 m22 = 63.7 MPa

c

5 kN

(a)

20 mm

40 mm

b

ad

(b)

5 kNV = 5 kN

force of strut on rod

a

d

c

5 kN

V = 2.5 kN

V = 2.5 kN

force ofrod on strut

(c)

b

Fig. 1–25

(d)

5 kN

63.7 MPa

5 kN

(e)

3.12 MPa


Fig. 1–26e

Fig. 1–26d

Fig. 1–26c

Fig. 1–26b


E X A M P L E 1.12

The inclined member in Fig. 1–26a is subjected to a compressive forceof 600 lb. Determine the average compressive stress along the smoothareas of contact defined by AB and BC, and the average shear stressalong the horizontal plane defined by EDB.

(b)

3

45

600 lb

FAB

FBC

(c)V

360 lb

Solution

Internal Loadings. The free-body diagram of the inclined memberis shown in Fig. 1–26b. The compressive forces acting on the areas ofcontact are

Also, from the free-body diagram of the top segment of the bottommember, Fig. 1–26c, the shear force acting on the sectioned horizontalplane EDB is

Average Stress. The average compressive stresses along the horizontaland vertical planes of the inclined member are

Ans.

Ans.

These stress distributions are shown in Fig. 1–26d.The average shear stress acting on the horizontal plane defined by

EDB is

Ans.

This stress is shown distributed over the sectioned area in Fig. 1–26e.

tavg =360 lb

13 in.211.5 in.2 = 80 psi

sBC =480 lb

12 in.211.5 in.2 = 160 psi

sAB =360 lb

11 in.211.5 in.2 = 240 psi

V = 360 lb:+ © Fx = 0;

FBC - 600 lb A45 B = 0 FBC = 480 lb+q© Fy = 0;

FAB - 600 lb A35 B = 0 FAB = 360 lb:+ © Fx = 0;

3

45

600 lb

160 psi

240 psi

(d)

360 lb

(e)80 psi

Fig. 1–26

1 in.

3

45

600 lb

(a)1.5 in. 3 in.

2 in.

AC

B

DE



*1-36. While running the foot of a 150-lb man ismomentarily subjected to a force which is 5 times his weight.Determine the average normal stress developed in the tibiaT of his leg at the mid section a–a. The cross section can beassumed circular, having an outer diameter of 1.75 in. andan inner diameter of 1 in. Assume the fibula F does notsupport a load.

1-37. The small block has a thickness of 0.5 in. If the stressdistribution at the support developed by the load varies asshown, determine the force F applied to the block, and thedistance d to where it is applied.

750 lb

a

T F

a

Prob. 1–36

1.5 in.

40 ksi

F

d

20 ksi

Prob. 1–37

P R O B L E M S

1-34. The column is subjected to an axial force of 8 kN atits top. If the cross-sectional area has the dimensions shownin the figure, determine the average normal stress acting atsection a–a. Show this distribution of stress acting over thearea’s cross section.

1-35. The anchor shackle supports a cable force of 600 lb.If the pin has a diameter of 0.25 in., determine the averageshear stress in the pin.

8 kN

aa

75 mm

10 mm

10 mm10 mm

75 mm

70 mm

70 mm

Prob. 1–34

0.25 in.

600 lbProb. 1–35


PROBLEMS • 41

1-38. The small block has a thickness of 5 mm. If the stressdistribution at the support developed by the load varies asshown, determine the force F applied to the block, and thedistance d to where it is applied.

1-39. The lever is held to the fixed shaft using a taperedpin AB, which has a mean diameter of 6 mm. If a couple isapplied to the lever, determine the average shear stress inthe pin between the pin and lever.

60 mm

120 mm

40 MPa

60 MPa

F

d

180 mm

Prob. 1–38

20 N 20 N

250 mm 250 mm

12 mm

A

B

Prob. 1–39

*1-40. The supporting wheel on a scaffold is held in placeon the leg using a 4-mm-diameter pin as shown. If the wheelis subjected to a normal force of 3 kN,determine the averageshear stress developed in the pin. Neglect friction betweenthe inner scaffold puller leg and the tube used on the wheel.

1-41. A 175-lb woman stands on a vinyl floor wearingstiletto high-heel shoes. If the heel has the dimensionsshown, determine the average normal stress she exerts onthe floor and compare it with the average normal stressdeveloped when a man having the same weight is wearingflat-heeled shoes. Assume the load is applied slowly, so thatdynamic effects can be ignored. Also, assume the entireweight is supported only by the heel of one shoe.

3 kN

Prob. 1–40

1.2 in.

0.5 in.

0.1 in.0.3 in.

Prob. 1–41



1-47. The built-up shaft consists of a pipe AB and solid rodBC. The pipe has an inner diameter of 20 mm and outerdiameter of 28 mm. The rod has a diameter of 12 mm.Determine the average normal stress at points D and E andrepresent the stress on a volume element located at each ofthese points.

*1-48. The board is subjected to a tensile force of 85 lb.Determine the average normal and average shear stressdeveloped in the wood fibers that are oriented along sectiona–a at 15° with the axis of the board.

3 in. 6 in.x

y

500 lb

P(x,y)12 in.

Prob. 1–45

30 mm

8 kN

25 mm

60

8 kN

Prob. 1–46

C

ED

A4 kN

8 kN

B 6 kN

6 kN

Prob. 1–47

153 in.

a

1 in.

85 lb85 lb

a

Prob. 1–48

1-42. The 50-lb lamp is supported by three steel rodsconnected by a ring at A. Determine which rod is subjectedto the greater average normal stress and compute its value.Take The diameter of each rod is given in the figure.

1-43. Solve Prob. 1–42 for

*1-44. The 50-lb lamp is supported by three steel rodsconnected by a ring at A.Determine the angle of orientation

of AC such that the average normal stress in rod AC istwice the average normal stress in rod AD. What is themagnitude of stress in each rod? The diameter of each rodis given in the figure.

u

u = 45°.

u = 30°.

1-45. The pedestal has a triangular cross section as shown.If it is subjected to a compressive force of 500 lb, specify thex and y coordinates for the location of point P(x, y), wherethe load must be applied on the cross section, so that theaverage normal stress is uniform. Compute the stress andsketch its distribution acting on the cross section at a locationremoved from the point of load application.

1-46. The two steel members are joined together using a60° scarf weld. Determine the average normal and averageshear stress resisted in the plane of the weld.

B

A

CD

0.25 in.0.3 in.

0.35 in.

θ45°

Probs. 1–42/43/44

Reduce art to80%

AAELBNS0

Reduce art to90%

AADFKGE0


PROBLEMS • 43

1-49. The plastic block is subjected to an axial compressiveforce of 600 N.Assuming that the caps at the top and bottomdistribute the load uniformly throughout the block,determine the average normal and average shear stressacting along section a-a.

1-50. The specimen failed in a tension test at an angle of52° when the axial load was 19.80 kip. If the diameter of thespecimen is 0.5 in., determine the average normal andaverage shear stress acting on the area of the inclined failureplane. Also, what is the average normal stress acting on thecross section when failure occurs?

1-51. A tension specimen having a cross-sectional area Ais subjected to an axial force P. Determine the maximumaverage shear stress in the specimen and indicate theorientation of a section on which it occurs.u

50 mm 50 mm

50 mm

150 mm

a

a

600 N

600 N

30°

Prob. 1–49

52°

0.5 in.

Prob. 1–50

θ

PP

A

Prob. 1–51

40 mm

A

CB

50 mm

60°

45°

5 kN

Prob. 1–52

60°

20°4.5 in.

2 in.

A

B

C

6 kip

Prob. 1–53

800 lb 800 lb

30°

1 in.1 in.

1.5 in. 30°

Prob. 1–54

*1-52. The joint is subjected to the axial member forceof 5 kN. Determine the average normal stress acting onsections AB and BC. Assume the member is smooth andis 50 mm thick.

1-53. The joint is subjected to the axial member forceof 6 kip. Determine the average normal stress acting onsections AB and BC. Assume the member is smooth andis 1.5 in. thick.

1-54. The two members used in the construction of anaircraft fuselage are joined together using a 30° fish-mouthweld. Determine the average normal and average shearstress on the plane of each weld.Assume each inclined planesupports a horizontal force of 400 lb.



*1-56. Rods AB and BC have diameters of 4 mm and 6mm, respectively. If the load of 8 kN is applied to the ringat B, determine the average normal stress in each rod if

1-57. Rods AB and BC have diameters of 4 mm and 6 mm,respectively. If the vertical load of 8 kN is applied to the ringat B, determine the angle of rod BC so that the averagenormal stress in each rod is equivalent. What is this stress?

u

u = 60°.

1-58. The bars of the truss each have a cross-sectional areaof Determine the average normal stress in eachmember due to the loading State whether thestress is tensile or compressive.

1-59. The bars of the truss each have a cross-sectional areaof If the maximum average normal stress in anybar is not to exceed 20 ksi, determine the maximummagnitude P of the loads that can be applied to the truss.

1.25 in2.

P = 8 kip.1.25 in2.

*1-60. The truss is made from three pin-connectedmembers having the cross-sectional areas shown in thefigure. Determine the average normal stress developed ineach member when the truss is subjected to the load shown.State whether the stress is tensile or compressive.

400 lb

Prob. 1–55

8 kN

A

C

θ

B

Probs. 1–56/57

3 ft

4 ft 4 ft

P0.75 P

E DA

B C

Probs. 1–58/59

3 ft

4 ft

B

A

C

ABC2

A AC

= 0

.6 in

.

= 0.8 in.

2

A AB=

1.5

in.2

500 lb

Prob. 1–60

1-55. The driver of the sports car applies his rear brakesand causes the tires to slip. If the normal force on each reartire is 400 lb and the coefficient of kinetic friction betweenthe tires and the pavement is determine theaverage shear stress developed by the friction force on thetires.Assume the rubber of the tires is flexible and each tireis filled with an air pressure of 32 psi.

mk = 0.5,


PROBLEMS • 45

*1-64. The two-member frame is subjected to thedistributed loading shown. Determine the average normalstress and average shear stress acting at sections a–a andb–b. Member CB has a square cross section of 35 mm oneach side. Take w = 8 kN>m.

1-63. The railcar docklight is supported by thepin at A. If the lamp weighs 4 lb, and the

extension arm AB has a weight of 0.5 lb/ft, determine theaverage shear stress in the pin needed to support the lamp.Hint: The shear force in the pin is caused by the couplemoment required for equilibrium at A.

18-in.-diameter

1-61. The uniform beam is supported by two rods AB andCD that have cross-sectional areas of and respectively. If determine the average normalstress in each rod.

1-62. The uniform beam is supported by two rods AB andCD that have cross-sectional areas of and respectively. Determine the position d of the 6-kN load sothat the average normal stress in each rod is the same.

8 mm2,12 mm2

d = 1 m,8 mm2,12 mm2

3 m

d

6 kN

B

A

D

C

Probs. 1–61/62

3 ft

1.25 in.

BA

Prob. 1–63

4 m

BA

C

3 m

b

b

a

a

w

Probs. 1–64/65

P2P1Pn

AmA2A1

d1

d2

dn

L1 L2 Lm

x

Prob. 1–66

1-65. The two-member frame is subjected to thedistributed loading shown. Determine the intensity w of thelargest uniform loading that can be applied to the framewithout causing either the average normal stress or theaverage shear stress at section b–b to exceed and respectively. Member CB has a squarecross section of 35 mm on each side.t = 16 MPa,

s = 15 MPa

1-66. Consider the general problem of a bar made fromm segments, each having a constant cross-sectional area and length If there are n loads on the bar as shown,write a computer program that can be used to determinethe average normal stress at any specified location x. Showan application of the program using the values

A2 = 1 in2.P2 = -300 lb,d2 = 6 ft,L2 = 2 ft,A1 = 3 in2,P1 = 400 lb,d1 = 2 ft,L1 = 4 ft,

Lm.Am



30

C

BA

0.5m1 m 1.5 m 1.5 m

0.5 mP 4P 4P 2P

Probs. 1–67/68

14 in.2 in.

0.8 in.

B

A

CGFB

FC

FA

75°

H

Prob. 1–69

C

10 ft

x

A

B

30

D

1500 lb

Prob. 1–70

300.4 m

30

0.2 m

1.2 m

A C

E DB

20

Prob. 1–71

1-67. The beam is supported by a pin at A and a short linkBC. If determine the average shear stressdeveloped in the pins at A, B, and C. All pins are in doubleshear as shown, and each has a diameter of 18 mm.

*1-68. The beam is supported by a pin at A and a shortlink BC. Determine the maximum magnitude P of the loadsthe beam will support if the average shear stress in each pinis not to exceed 80 MPa. All pins are in double shear asshown, and each has a diameter of 18 mm.

P = 15 kN,

1-69. When the hand is holding the 5-lb stone, thehumerus H, assumed to be smooth, exerts normal forces

and on the radius C and ulna A, respectively, asshown. If the smallest cross-sectional area of the ligamentat B is determine the greatest average tensilestress to which it is subjected.

0.30 in2,

FAFC

1-70. The jib crane is pinned at A and supports a chainhoist that can travel along the bottom flange of the beam,

If the hoist is rated to support a maximumof 1500 lb, determine the maximum average normal stressin the tie rod BC and the maximumaverage shear stress in the pin at B.???-in.-diameter

???-in.-diameter

1 ft … x … 12 ft.

1-71. Determine the average normal stress developed inlinks AB and CD of the two-tine grapple that supports thelog having a mass of 3 Mg. The cross-sectional area of eachlink is 400 mm2.

Reduce art to 85%AADFKGO0


PROBLEMS • 47

1-72. Determine the average shear stress developed in pinsA and B of the two-tine grapple that supports the log havinga mass of 3 Mg. Each pin has a diameter of 25 mm and issubjected to double shear.

1-73. The pedestal in the shape of a frustum of a cone ismade of concrete having a specific weight of Determine the average normal stress acting in the pedestalat its base. Hint: The volume of a cone of radius r and heighth is V = 1

3 pr2h.

150 lb>ft3.

1-74. The pedestal in the shape of a frustum of a cone ismade of concrete having a specific weight of Determine the average normal stress acting in the pedestalat its midheight, Hint: The volume of a cone ofradius r and height h is V = 1

3 pr2h.z = 4 ft.

150 lb>ft3.

1-75. The column is made of concrete having a densityof At its top B it is subjected to an axialcompressive force of 15 kN. Determine the averagenormal stress in the column as a function of the distancez measured from its base. Note: The result will be usefulonly for finding the average normal stress at a sectionremoved from the ends of the column, because oflocalized deformation at the ends.

2.30 Mg>m3.

300.4 m

30

0.2 m

1.2 m

A C

E DB

20

Prob. 1–72

z

y

x

8 ft

= 4 ftz

1 ft

1.5 ft

Prob. 1–73

z

y

x

8 ft

1 ft

1.5 ft

z = 4 ft

Prob. 1–74

z

x y

15 kN

4 m

180 mm

z

B

Prob. 1–75

Reduce art to 90%AAELBOI0

Reduce art to 90%AADFKGQ0



z

L

P

w w

w1

w1

Prob. 1–76

z

r

P

r1

Prob. 1–77

r

y

3 m

0.5 m

r = 0.5e−0.08y2

y

Prob. 1–78

*1-76. The pier is made of material having a specific weightIf it has a square cross section, determine its width w as

a function of z so that the average normal stress in the pierremains constant. The pier supports a constant load P at itstop where its width is w1.

g.

1-77. The pedestal supports a load P at its center. If thematerial has a mass density determine the radialdimension r as a function of z so that the average normalstress in the pedestal remains constant. The cross sectionis circular.

r,

1-78. The radius of the pedestal is defined bywhere y is given in meters. If the

material has a density of determine the averagenormal stress at the support.

2.5 Mg>m3,r = 10.5e-0.08y22 m,

1-79. Determine the greatest constant angular velocity of the flywheel so that the average normal stress in its rimdoes not exceed Assume the rim is a thin ringhaving a thickness of 3 mm, width of 20 mm, and a mass of30 kg/m. Rotation occurs in the horizontal plane. Neglectthe effect of the spokes in the analysis. Hint: Consider a free-body diagram of a semicircular portion of the ring. Thecenter of mass for a semicircular segment is located at

from the diameter.rN = 2r>p

s = 15 MPa.

v

0.8 mm

ω

Prob. 1–79


SECTION 1.6 Allowable Stress • 49

1.6 Allowable Stress

Appropriate factors of safety must beconsidered when designing cranes andcables used to transfer heavy loads.

An engineer in charge of the design of a structural member or mechanicalelement must restrict the stress in the material to a level that will be safe.Furthermore, a structure or machine that is currently in use may, onoccasion, have to be analyzed to see what additional loadings its membersor parts can support. So again it becomes necessary to perform thecalculations using a safe or allowable stress.

To ensure safety, it is necessary to choose an allowable stress thatrestricts the applied load to one that is less than the load the membercan fully support. There are several reasons for this. For example, theload for which the member is designed may be different from actualloadings placed on it. The intended measurements of a structure ormachine may not be exact due to errors in fabrication or in the assemblyof its component parts. Unknown vibrations, impact, or accidentalloadings can occur that may not be accounted for in the design.Atmospheric corrosion, decay, or weathering tend to cause materials todeteriorate during service. And lastly, some materials, such as wood,concrete, or fiber-reinforced composites, can show high variability inmechanical properties.

One method of specifying the allowable load for the design or analysisof a member is to use a number called the factor of safety. The factor ofsafety (F.S.) is a ratio of the failure load divided by the allowableload, Here is found from experimental testing of the material,and the factor of safety is selected based on experience so that the abovementioned uncertainties are accounted for when the member is usedunder similar conditions of loading and geometry. Stated mathematically,

(1–8)

If the load applied to the member is linearly related to the stress developedwithin the member, as in the case of using and thenwe can express the factor of safety as a ratio of the failure stress (or

) to the allowable stress (or ); *that is,

(1–9)

or

(1–10)F.S. =tfail

tallow

F.S. =sfail

sallow

tallowsallowtfail

sfail

tavg = V>A,s = P>A

F.S. =Ffail

Fallow

FfailFallow.Ffail

*In some cases, such as columns, the applied load is not linearly related to stress and thereforeonly Eq. 1–8 can be used to determine the factor of safety. See Chapter 13.



In any of these equations, the factor of safety is chosen to be greaterthan 1 in order to avoid the potential for failure. Specific values dependon the types of materials to be used and the intended purpose of thestructure or machine. For example, the F.S. used in the design of aircraftor space-vehicle components may be close to 1 in order to reduce theweight of the vehicle. On the other hand, in the case of a nuclear powerplant, the factor of safety for some of its components may be as high as3 since there may be uncertainties in loading or material behavior. Ingeneral, however, factors of safety and therefore the allowable loads orstresses for both structural and mechanical elements have become wellstandardized, since their design uncertainties have been reasonablyevaluated. Their values, which can be found in design codes andengineering handbooks, are intended to form a balance of ensuringpublic and environmental safety and providing a reasonable economicsolution to design.

1.7 Design of Simple Connections

By making simplifying assumptions regarding the behavior of thematerial, the equations and can often be used toanalyze or design a simple connection or a mechanical element. Inparticular, if a member is subjected to a normal force at a section, itsrequired area at the section is determined from

(1–11)

On the other hand, if the section is subjected to a shear force, then therequired area at the section is

(1–12)

As discussed in Sec. 1.6, the allowable stress used in each of theseequations is determined either by applying a factor of safety to a specifiednormal or shear stress or by finding these stresses directly from anappropriate design code.

We will now discuss four common types of problems for which theabove equations can be used for design.

A =Vtallow

A =Psallow

tavg = V>As = P>A


Fig. 1–28a

Fig. 1–28b

Cross-Sectional Area of a Connector Subjected to Shear. Oftenbolts or pins are used to connect plates, boards, or several memberstogether. For example, consider the lap joint shown in Fig. 1–28a. If the boltis loose or the clamping force of the bolt is unknown, it is safe to assumethat any frictional force between the plates is negligible. As a result, thefree-body diagram for a section passing between the plates and through thebolt is shown in Fig. 1–28b. The bolt is subjected to a resultant internalshear force of at this cross section. Assuming that the shear stresscausing this force is uniformly distributed over the cross section, the bolt’scross-sectional area A is determined as shown in Fig. 1–28c.

Required Area to Resist Bearing. A normal stress that is producedby the compression of one surface against another is called a bearingstress. If this stress becomes large enough, it may crush or locally deformone or both of the surfaces. Hence, in order to prevent failure it isnecessary to determine the proper bearing area for the material using anallowable bearing stress. For example, the area A of the column base plateB shown in Fig. 1–29 is determined from the allowable bearing stress ofthe concrete using This assumes, of course, that theallowable bearing stress for the concrete is smaller than that of the baseplate material, and furthermore the bearing stress is uniformly distributedbetween the plate and the concrete as shown in the figure.

A = P>1sb2allow.

V = P

Fig. 1–27b

SECTION 1.7 Design of Simple Connections • 51

Cross-Sectional Area of a Tension Member. The cross-sectional areaof a prismatic member subjected to a tension force can be determinedprovided the force has a line of action that passes through the centroid of thecross section.For example,consider the “eye bar”shown in Fig. 1–27a.At theintermediate section a–a, the stress distribution is uniform over the crosssection and the shaded area A is determined, as shown in Fig. 1–27b.

P

P

(a)

(a)

P

a

a

P

Fig. 1–27

Uniform normal stressσallow

P

(b)A P= σallow

———

P

(b)

= PV

(c)

Uniform shear stress

P allow

allow———P=A

B

A= —————( b)allow

P

Uniform normalstress distribution

( b)allow

P

Fig. 1–29



(a)P

Fig. 1–30a

d

Uniform shear stress

(b)

l = —————Pallow d

allow

P

Fig. 1–30

Required Area to Resist Shear Caused by Axial Load. Occasionallyrods or other members will be supported in such a way that shear stresscan be developed in the member even though the member may besubjected to an axial load. An example of this situation would be a steelrod whose end is encased in concrete and loaded as shown in Fig. 1–30a.A free-body diagram of the rod, Fig. 1–30b, shows that shear stress actsover the area of contact of the rod with the concrete. This area is where d is the rod’s diameter and l is the length of embedment.Althoughthe actual shear-stress distribution along the rod would be difficult todetermine, if we assume it is uniform, we can use to calculatel, provided we know d and Fig. 1–30b.tallow,

A = V>tallow

1pd2l,

IMPORTANT POINTS• Design of a member for strength is based on selecting an

allowable stress that will enable it to safely support its intendedload. There are many unknown factors that can influence theactual stress in a member and so, depending upon the intendeduses of the member, a factor of safety is applied to obtain theallowable load the member can support.

• The four cases illustrated in this section represent just a few ofthe many applications of the average normal and shear stressformulas used for engineering design and analysis. Wheneverthese equations are applied, however, it is important to be awarethat the stress distribution is assumed to be uniformly distributedor “averaged” over the section.

PROCEDURE FOR ANALYSISWhen solving problems using the average normal and shear stressequations, a careful consideration should first be made as to thesection over which the critical stress is acting. Once this section ismade, the member must then be designed to have a sufficient areaat the section to resist the stress that acts on it. To determine thisarea, application requires the following steps.

Internal Loading.

• Section the member through the area and draw a free-body diagramof a segment of the member. The internal resultant force at thesection is then determined using the equations of equilibrium.

Required Area.

• Provided the allowable stress is known or can be determined, therequired area needed to sustain the load at the section is thencomputed from or A = V>tallow.A = P>sallow


Fig. 1–31b


E X A M P L E 1.13

The two members are pinned together at B as shown in Fig. 1–31a.Top views of the pin connections at A and B are also given in thefigure. If the pins have an allowable shear stress of and the allowable tensile stress of rod CB is determine to the nearest the smallest diameter of pins A and Band the diameter of rod CB necessary to support the load.

116 in.

1st2allow = 16.2 ksi,tallow = 12.5 ksi

B

4 ft

3 kip3

5

4

C

2 ft

A

BA

(a)

Fig. 1–31

SolutionRecognizing CB to be a two-force member, the free-body diagram ofmember AB along with the computed reactions at A and B is shownin Fig. 1–31b. As an exercise, verify the computations and notice thatthe resultant force at A must be used for the design of pin A, sincethis is the shear force the pin resists.

4 ftA

(b)

3 kip

B2 ft

1 kip

3.33 kip

2.85 kip

2.67 kip

20.6°35

4


Fig. 1–31c


(c)

1.425 kip

1.425 kip

2.85 kip

3.33 kip

3.33 kip

Pin at BPin at A

Diameter of the Pins. From Fig. 1–31a and the free-body diagramsof the sectioned portion of each pin in contact with member AB,Fig. 1–31c, it is seen that pin A is subjected to double shear, whereaspin B is subjected to single shear. Thus,

Although these values represent the smallest allowable pin diameters,a fabricated or available pin size will have to be chosen.We will choosea size larger to the nearest as required.

Ans.

Ans.

Diameter of Rod. The required diameter of the rod throughout itsmidsection is thus,

We will choose

Ans.dBC = 916 in. = 0.5625 in.

dBC = 0.512 in.

ABC =P

1st2allow=

3.333 kip

16.2 kip>in2 = 0.2058 in2 = padBC2

4b

dB = 58 in. = 0.625 in.

dA = 716 in. = 0.4375 in.

116 in.

AB =VB

Tallow=

3.333 kip

12.5 kip>in2 = 0.2667 in2 = padB2

4b dB = 0.583 in.

AA =VA

Tallow=

1.425 kip

12.5 kip>in2 = 0.1139 in2 = padA2

4b dA = 0.381 in


Fig. 1–32aFig. 1–32b


E X A M P L E 1.14

The control arm is subjected to the loading shown in Fig. 1–32a.Determine to the nearest the required diameter of the steel pinat C if the allowable shear stress for the steel is Notein the figure that the pin is subjected to double shear.

tallow = 8 ksi.

14 in.

35

42 in.3 in.

8 in.

Cx

3 kip5 kip

FAB

Cy

(b)

C

(c)

3.041 kip

3.041 kip

6.082 kip

Pin at C

Fig. 1–32

(a)

C

35

42 in.3 in.

8 in.

A

C

3 kip5 kip

B

Solution

Internal Shear Force. A free-body diagram of the arm is shown inFig. 1–32b. For equilibrium we have

The pin at C resists the resultant force at C. Therefore,

Since the pin is subjected to double shear, a shear force of 3.041 kipacts over its cross-sectional area between the arm and each supportingleaf for the pin, Fig. 1–32c.

Required Area. We have

Use a pin having a diameter of

Ans.d = 34 in. = 0.750 in.

d = 0.696 in.

pad

2b2

= 0.3802 in2

A =Vtallow

=3.041 kip

8 kip>in2 = 0.3802 in2

FC = 211 kip22 + 16 kip22 = 6.082 kip

Cy - 3 kip - 5 kip A35 B = 0 Cy = 6 kip+q© Fy = 0;

-3 kip - Cx + 5 kip A45 B = 0 Cx = 1 kip:+ © Fx = 0;

FAB = 3 kip

FAB18 in.2 - 3 kip13 in.2 - 5 kip A35 B15 in.2 = 0d+ © MC = 0;


Fig. 1–33b


E X A M P L E 1.15

The suspender rod is supported at its end by a fixed-connected circulardisk as shown in Fig. 1–33a. If the rod passes through a 40-mm-diameter hole, determine the minimum required diameter of the rodand the minimum thickness of the disk needed to support the 20-kNload. The allowable normal stress for the rod is andthe allowable shear stress for the disk is tallow = 35 MPa.

sallow = 60 MPa,

20 kN

A allow

(b)

40 mm

20 kN

t

d

(a)

40 mm

Fig. 1–33Solution

Diameter of Rod. By inspection, the axial force in the rod is 20 kN.Thus the required cross-sectional area of the rod is

So that

Ans.

Thickness of Disk. As shown on the free-body diagram of the coresection of the disk, Fig. 1–33b, the material at the sectioned area mustresist shear stress to prevent movement of the disk through the hole.If this shear stress is assumed to be distributed uniformly over thesectioned area, then, since we have

Since the sectioned area the required thicknessof the disk is

Ans.t =0.5714110-32 m2

2p10.02 m2 = 4.55110-32 m = 4.55 mm

A = 2p10.02 m21t2,A =

Vtallow

=2011032 N

3511062 N>m2 = 0.571110-32 m2

V = 20 kN,

d = 0.0206 m = 20.6 mm

A = pad2

4b = 0.3333110-22 m2

A =Psallow

=2011032 N

6011062 N>m2 = 0.3333110-32 m2


Fig. 1–34d

Fig. 1–34bFig. 1–34a


E X A M P L E 1.16

An axial load on the shaft shown in Fig. 1–34a is resisted by the collarat C, which is attached to the shaft and located on the right side ofthe bearing at B. Determine the largest value of P for the two axialforces at E and F so that the stress in the collar does not exceed anallowable bearing stress at C of and the averagenormal stress in the shaft does not exceed an allowable tensile stressof 1st2allow = 55 MPa.

1sb2allow = 75 MPa

PP2

(b)

P380 mm

60 mmP

A

FP2

CE

B

(a)

20 mm

(c)

AxialLoad

Position

3P2P

Fig. 1–34

SolutionTo solve the problem we will determine P for each possible failurecondition. Then we will choose the smallest value. Why?

Normal Stress. Using the method of sections, the axial load withinregion FE of the shaft is 2P, whereas the largest axial load, 3P, occurswithin region EC, Fig. 1–34b. The variation of the internal loading isclearly shown on the normal-force diagram, Fig. 1–34c. Since the cross-sectional area of the entire shaft is constant, region EC will be subjectedto the maximum average normal stress. Applying Eq. 1–11, we have

Bearing Stress. As shown on the free-body diagram in Fig. 1–34d,the collar at C must resist the load of 3P, which acts over a bearingarea of Thus,

By comparison, the largest load that can be applied to the shaft issince any load larger than this will cause the allowable

normal stress in the shaft to be exceeded.P = 51.8 kN,

P = 55.0 kN

7511062 N>m2 =3P

2.199110-32 m2A =Psallow

;

Ab = [p10.04 m22 - p10.03 m22] = 2.199110-32 m2.

P = 51.8 kN

sallow =P

A 5511062 N>m2 =

3P

p10.03 m22

P3

(d)C


Fig. 1–35a


E X A M P L E 1.17

The rigid bar AB shown in Fig. 1–35a is supported by a steel rod AChaving a diameter of 20 mm and an aluminum block having a cross-sectional area of The 18-mm-diameter pins at A and C aresubjected to single shear. If the failure stress for the steel and aluminumis and respectively, and thefailure shear stress for each pin is determine the largestload P that can be applied to the bar.Apply a factor of safety of

SolutionUsing Eqs. 1–9 and 1–10, the allowable stresses are

The free-body diagram for the bar is shown in Fig. 1–35b. There arethree unknowns. Here we will apply the equations of equilibrium so asto express and in terms of the applied load P. We have

(1)

(2)

We will now determine each value of P that creates the allowable stressin the rod, block, and pins, respectively.Rod AC. This requires

Using Eq. 1,

Block B. In this case,

Using Eq. 2,

Pin A or C. Here

From Eq. 1,

By comparison, when P reaches its smallest value (168 kN), it developsthe allowable normal stress in the aluminum block. Hence,

Ans.P = 168 kN

P =114.5 kN12 m2

1.25 m= 183 kN

V = FAC = tallowA = 45011062 N>m2[p10.009 m22] = 114.5 kN

P =163.0 kN212 m2

0.75 m= 168 kN

FB = 1sal2allow AB = 3511062 N>m2[1800 mm2110-62 m2>mm2] = 63.0 kN

P =1106.8 kN212 m2

1.25 m= 171 kN

FAC = 1sst2allow1AAC2 = 34011062 N>m2[p10.01 m22] = 106.8 kN

FB12 m2 - P10.75 m2 = 0d+ © MA = 0;

P11.25 m2 - FAC12 m2 = 0d+ © MB = 0;

FBFAC

tallow =tfail

F.S.=

900 MPa2

= 450 MPa

1sal2allow =1sal2fail

F.S.=

70 MPa2

= 35 MPa

1sst2allow =1sst2fail

F.S.=

680 MPa2

= 340 MPa

F.S. = 2.tfail = 900 MPa,

1sal2fail = 70 MPa,1sst2fail = 680 MPa

1800 mm2.

2 m

A

0.75 m

(a)

Aluminum

Steel P

B

C

A

0.75 m

(b)

P

1.25 m

B

FB

FAC

Fig. 1–35


PROBLEMS • 59

P R O B L E M S

*1-80. Member B is subjected to a compressive force of800 lb. If A and B are both made of wood and are thick, determine to the nearest the smallestdimension h of the support so that the average shearstress does not exceed tallow = 300 psi.

14 in.

38 in.

1-81. The oak post is supported on thepine block. If the allowable bearing stresses for thesematerials are and determine the greatest load P that can be supported. If arigid bearing plate is used between these materials,determine its required area so that the maximum load Pcan be supported. What is this load?

spine = 25 MPa,soak = 43 MPa

60 mm * 60 mm

1-82. The joint is fastened together using two bolts.Determine the required diameter of the bolts if theallowable shear stress for the bolts is Assume each bolt supports an equal portion of the load.

tallow = 110 MPa.

800 lb

hA

B

12

513

Prob. 1–80

P

Prob. 1–81

80 kN

40 kN

30 mm

30 mm

40 kN

Prob. 1–82

500 mm

20 mm

daa

A

200 N

Prob. 1–83

1-83. The lever is attached to the shaft A using a keythat has a width d and length of 25 mm. If the shaft isfixed and a vertical force of 200 N is appliedperpendicular to the handle, determine the dimension dif the allowable shear stress for the key is tallow = 35 MPa.



*1-84. The fillet weld size a is determined by computingthe average shear stress along the shaded plane, whichhas the smallest cross section. Determine the smallest sizea of the two welds if the force applied to the plate is

The allowable shear stress for the weldmaterial is tallow = 14 ksi.P = 20 kip.

1-85. The fillet weld size If the joint isassumed to fail by shear on both sides of the block alongthe shaded plane, which is the smallest cross section,determine the largest force P that can be applied to theplate. The allowable shear stress for the weld material istallow = 14 ksi.

a = 0.25 in.

1-86. The tension member is fastened together using twobolts, one on each side of the member as shown. Eachbolt has a diameter of 0.3 in. Determine the maximumload P that can be applied to the member if the allowableshear stress for the bolts is and theallowable average normal stress is sallow = 20 ksi.

tallow = 12 ksi

1-87. The steel swivel bushing in the elevator control ofan airplane is held in place using a nut and washer asshown in Fig. (a). Failure of the washer A can cause thepush rod to separate as shown in Fig. (b). If the maximumaverage normal stress for the washer is andthe maximum average shear stress is determine the force F that must be applied to the bushingthat will cause this to happen. The washer is thick.1

10 in.

tmax = 21 ksi,smax = 60 ksi

*1-88. The two steel wires AB and AC are used tosupport the load. If both wires have an allowable tensilestress of determine the requireddiameter of each wire if the applied load is

1-89. The two steel wires AB and AC are used to supportthe load. If both wires have an allowable tensile stress of

and wire AB has a diameter of 6 mmand AC has a diameter of 4 mm, determine the greatestforce P that can be applied to the chain before one of thewires fails.

sallow = 180 MPa,

P = 5 kN.sallow = 200 MPa,

a

45

a

P 4 in.

Prob. 1–84

a

45

a

P 4 in.

Prob. 1–85

60

PP

Prob. 1–86

(a)

F

(b)

A

0.75 in.

F

Prob. 1–87

604

35

P

B

A

C

Probs. 1–88/89

Reduce art to90%

AAELBOO0


PROBLEMS • 61

1-90. The boom is supported by the winch cable that hasa diameter of 0.25 in. and an allowable normal stress of

Determine the greatest load that can besupported without causing the cable to fail when and Neglect the size of the winch.

1-91. The boom is supported by the winch cable that hasan allowable normal stress of If it isrequired that it be able to slowly lift 5000 lb, from to determine the smallest diameter of the cableto the nearest The boom AB has a length of 20 ft.Neglect the size of the winch.

116 in.

u = 50°,u = 20°

sallow = 24 ksi.

f = 45°.u = 30°

sallow = 24 ksi.

12 ft

20 ftθ

φA

B

Probs. 1–90/91

800 lb

E

C

F400 lb

D

6 ft 6 ft

6 ft 6 ft

45

30

60

B

A

Prob. 1–92

6 kip200 lb/ft

5 ft 3 ft5 ft

AB

Prob. 1–93

A B

7.5 ft

2 kip 2 kip

3 kip

2 kipP

5 ft 5 ft 5 ft

A¿ B¿

Probs. 1–94/95

*1-92. The truss is used to support the loading shown.Determine the required cross-sectional area of memberBC if the allowable normal stress is sallow = 24 ksi.

1-93. The beam is made from southern pine and issupported by base plates resting on brick work. If theallow-able bearing stresses for the materials are

determinethe required length of the base plates at A and B to thenearest in order to support the load shown. Theplates are 3 in. wide.

14 inch

1spine2allow = 2.81 ksi 1sbrick2allow = 6.70 ksi,

1-94. If the allowable bearing stress for the materialunder the supports at A and B is determine the size of square bearing plates and required to support the loading. Take Dimension the plates to the nearest Thereactions at the supports are vertical.

1-95. If the allowable bearing stress for the materialunder the supports at A and B is determine the maximum load P that can be applied tothe beam.The bearing plates and have square crosssections of and respectively.4 in. * 4 in.,2 in. * 2 in.

B¿A¿

1sb2allow = 400 psi,

12 in.

P = 1.5 kip.B¿A¿

1sb2allow = 400 psi,



C

60

2 ft 4 ft 2 ft

B

1500 lb 1500 lb

A

Prob. 1–96

P

B

AC 45

Prob. 1–97

P

B

AC 45

Prob. 1–98

*1-96. Determine the required cross-sectional area ofmember BC and the diameter of the pins at A and B ifthe allowable normal stress is and theallowable shear stress is tallow = 4 ksi.

sallow = 3 ksi

1-97. The two aluminum rods support the vertical forceof Determine their required diameters if theallowable tensile stress for the aluminum issallow = 150 MPa.

P = 20 kN.

1-98. The two aluminum rods AB and AC havediameters of 10mm and 8mm, respectively. Determine thelargest vertical force P that can be supported. Theallowable tensile stress for the aluminum issallow = 150 MPa.

1-99. The hangers support the joist uniformly, so that itis assumed the four nails on each hanger carry an equalportion of the load. If the joist is subjected to the loadingshown, determine the average shear stress in each nail ofthe hanger at ends A and B. Each nail has a diameter of0.25 in. The hangers only support vertical loads.

*1-100. The hangers support the joists uniformly, so thatit is assumed the four nails on each hanger carry an equalportion of the load. Determine the smallest diameter ofthe nails at A and at B if the allowable shear stress forthe nails is The hangers only supportvertical loads.

tallow = 4 ksi.

18 ft

A B

30 lb/ft40 lb/ft

Probs. 1–99/100


PROBLEMS • 63

1-101. The hanger assembly is used to support adistributed loading of Determine theaverage shear stress in the 0.40-in.-diameter bolt at A andthe average tensile stress in rod AB, which has a diameterof 0.5 in. If the yield shear stress for the bolt is and the yield tensile stress for the rod is determine the factor of safety with respect to yielding ineach case.

1-102. Determine the intensity w of the maximumdistributed load that can be supported by the hangerassembly so that an allowable shear stress of

is not exceeded in the 0.40-in.-diameterbolts at A and B, and an allowable tensile stress of

is not exceeded in the 0.5-in.-diameterrod AB.sallow = 22 ksi

tallow = 13.5 ksi

sy = 38 ksi,ty = 25 ksi,

w = 0.8 kip>ft.

4 ft 2 ft

3 ft

w

A

B

C

Probs. 1–101/102

C

B

A

4 ft

3 ft

1 ft

w

Probs. 1–103/104

1.5 m1.5 m1.5 m1.5 m2 m2 m

BC D

A

3 kN 1.5 kN2 kN

Prob. 1–105

1-103. The assembly is used to support the distributedloading of Determine the factor of safetywith respect to yielding for the steel rod BC and the pinsat B and C if the yield stress for the steel in tension is

and in shear The rod has adiameter of 0.4 in., and the pins each have a diameterof 0.30 in.

ty = 18 ksi.sy = 36 ksi

w = 500 lb>ft.

*1-104. If the allowable shear stress for each of the 0.3-in.-diameter steel pins at A, B, and C is and the allowable normal stress for the 0.40-in.-diameterrod is determine the largest intensity w ofthe uniform distributed load that can be suspended fromthe beam.

sallow = 22 ksi,

tallow = 12.5 ksi,

1-105. The compound wooden beam is connectedtogether by a bolt at B. Assuming that the connections atA, B, C, and D exert only vertical forces on the beam,determine the required diameter of the bolt at B and therequired outer diameter of its washers if the allowabletensile stress for the bolt is and theallowable bearing stress for the wood is

Assume that the hole in the washershas the same diameter as the bolt.1sb2allow = 28 MPa.

1st2allow = 150 MPa



A

2 m

x

B

w

2 m

Probs. 1–106/107/108

1-106. The bar is held in equilibrium by the pin supportsat A and B. Note that the support at A has a single leafand therefore it involves single shear in the pin, and thesupport at B has a double leaf and therefore it involvesdouble shear. The allowable shear stress for both pins is

If a uniform distributed load ofis placed on the bar, determine its minimum

allowable position x from B. Pins A and B each have adiameter of 8 mm. Neglect any axial force in the bar.

1-107. The bar is held in equilibrium by the pinsupports at A and B. Note that the support at A has asingle leaf and therefore it involves single shear in thepin, and the support at B has a double leaf and thereforeit involves double shear. The allowable shear stress forboth pins is If determine themaximum distributed load w the bar will support. PinsA and B each have a diameter of 8 mm. Neglect anyaxial force in the bar.

*1-108. The bar is held in equilibrium by the pinsupports at A and B. Note that the support at A has asingle leaf and therefore it involves single shear in thepin, and the support at B has a double leaf and thereforeit involves double shear. The allowable shear stress forboth pins is If and

determine the smallest required diameterof pins A and B. Neglect any axial force in the bar.w = 12 kN>m,

x = 1 mtallow = 125 MPa.

x = 1 m,tallow = 125 MPa.

w = 8 kN>mtallow = 150 MPa.

1-109. The pin is subjected to double shear since it isused to connect the three links together. Due to wear, theload is distributed over the top and bottom of the pin asshown on the free-body diagram. Determine the diameterd of the pin if the allowable shear stress is and the load Also, determine the loadintensities and w2.w1

P = 8 kip.tallow = 10 ksi

1-110. The pin is subjected to double shear since it isused to connect the three links together. Due to wear, theload is distributed over the top and bottom of the pin asshown on the free-body diagram. Determine themaximum load P the connection can support if theallowable shear stress for the material is andthe diameter of the pin is 0.5 in. Also, determine the loadintensities and w2.w1

tallow = 8 ksi

1-111. The thrust bearing consists of a circular collar Afixed to the shaft B. Determine the maximum axial forceP that can be applied to the shaft so that it does not causethe shear stress along a cylindrical surface a or b to exceedan allowable shear stress of tallow = 170 MPa.

P

P—2

1 in.

d

1.5 in.

1 in.

w2 w2

w1

P—2

Probs. 1–109/110

20 mm

30 mm 58 mm

b

b35 mm

B A

C

a

a

P

Prob. 1–111


CHAPTER REVIEW • 65

CHAPTER REVIEW• The internal loadings in a body consist of a normal force, shear

force, bending moment, and torsional moment.They represent theresultants of both a normal and shear stress distribution that actsover the cross section. To obtain these resultants, use the methodof sections and the equations of equilibrium.

• If a bar is made from homogeneous isotropic material and it issubjected to a series of external axial loads that pass through thecentroid of the cross section, then a uniform normal stressdistribution will act over the cross section. This average normalstress can be determined from where P is the internalaxial load at the section.

• The average shear stress can be determined using where V is the resultant shear force on the cross-sectional areaA. This formula is often used to find the average shear stress infasteners or in parts used for connections.

• The design of any simple connection requires that the averagestress along any cross section not exceed a factor of safety or anallowable value of or These values are reported incodes or standards and have been deemed safe on the basis ofexperiments or through experience.

tallow.sallow

t = V>A,

s = P>A,



3 m 1.5 m

15 kN/mC

B A

Prob. 1–115

1-114. Determine the resultant internal loadings actingon the cross sections located through points D and E ofthe frame.

1-115. The rod BC is made of steel having an allowabletensile stress of Determine its smallestdiameter so that it can support the load shown. The beamis assumed to be pin-connected at A.

sallow = 155 MPa.

4 ft

1.5 ftA

D

5 ft3 ftC

2.5 ft

E

B

150 lb/ft

Prob. 1–114

R E V I E W P R O B L E M S

*1-112. The long bolt passes through the 30-mm-thickplate. If the force in the bolt shank is 8 kN, determine theaverage normal stress in the shank, the average shearstress along the cylindrical area of the plate defined bythe section lines a–a, and the average shear stress in thebolt head along the cylindrical area defined by the sectionlines b–b.

1-113. The two-member frame is subjected to theloading shown. Determine the average normal stressand the average shear stress acting at sections a–a andb–b. Member CB has a square cross section of 2 in. oneach side.

8 kN18 mm

7 mm

30 mm

8 mm a

a

b

b

Prob. 1–112

4 ft 4 ft

300 lbft

80 lb

60

30

A

B

C

b

ba

a

Prob. 1–113


REVIEW PROBLEMS • 67

*1-116. The column has a cross-sectional area ofIt is subjected to an axial force of 50 kN. If

the base plate to which the column is attached has alength of 250 mm, determine its width d so that theaverage bearing stress under the plate at the ground isone-third of the average compressive stress in the column.Sketch the stress distributions acting over the column’scross-sectional area and at the bottom of the base plate.

1211032 mm2.

1-117. The beam AB is pin supported at A andsupported by a cable BC. A separate cable CG is used tohold up the frame. If AB weighs 120 lb>ft and the columnFC has a weight of 180 lb>ft, determine the resultantinternal loadings acting on cross sections located at pointsD and E. Neglect the thickness of both the beam andcolumn in the calculation.

1-118. The pulley is held fixed to the 20-mm-diametershaft using a key that fits within a groove cut into thepulley and shaft. If the suspended load has a mass of50 kg, determine the average shear stress in the keyalong section a–a. The key is 5 mm by 5 mm square and12 mm long.

1-119. The yoke-and-rod connection is subjected to atensile force of 5 kN. Determine the average normal stressin each rod and the average shear stress in the pin Abetween the members.

50 kN

250 mmd

Prob. 1–116

4 ft

12 ft

4 ft

8 ft

12 ft

6 ft

A

E

G

B

F

D

C

Prob. 1–117

25 mm

40 mm

30 mm

A

5 kN

5 kN

Prob. 1–119

75 mm

a a

Prob. 1–118


the bolts used for the connections of this steel framework...

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