Chap01-A

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Chapter 1

Chapter 1

Stress

Stress

1.1 Introduction

1.1 Introduction





The

The three

three

fundamental areas of engineering

fundamental areas of engineering

mechanics are

mechanics are statics

statics

,, dynamics

dynamics

,, and mechanics of

and mechanics of

materials

materials

..





Statics

Statics

and

and dynamics

dynamics

:: the external effects

the external effects

upon

upon

rigid 

rigid 

 bodies, bodies for whic

 bodies, bodies for whic

h the change in shape

h the change in shape

(deformation) can be neglected.

(deformation) can be neglected.





mechanics of materials

mechanics of materials

:: the internal effects

the internal effects

and

and

deformations

deformations

that are caused by the applied loads.

that are caused by the applied loads.





A mechanic part or structure must be

A mechanic part or structure must be strong

strong

enough

enough

to carry the applied load without

to carry the applied load without breaking

 breaking

and the

and the

same time, the

1.1 Introduction

1.1 Introduction





The

The three

three

fundamental areas of engineering

fundamental areas of engineering

mechanics are

mechanics are statics

statics

,, dynamics

dynamics

,, and mechanics of

and mechanics of

materials

materials

..





Statics

Statics

and

and dynamics

dynamics

:: the external effects

the external effects

upon

upon

rigid 

rigid 

 bodies, bodies for whic

 bodies, bodies for whic

h the change in shape

h the change in shape

(deformation) can be neglected.

(deformation) can be neglected.





mechanics of materials

mechanics of materials

:: the internal effects

the internal effects

and

and

deformations

deformations

that are caused by the applied loads.

that are caused by the applied loads.





A mechanic part or structure must be

A mechanic part or structure must be strong

strong

enough

enough

to carry the applied load without

to carry the applied load without breaking

 breaking

and the

and the

same time, the





The differences between

The differences between right-body mechanics

right-body mechanics

and

and

mechanics of materials

mechanics of materials

can be appreciated.

can be appreciated.





In right-body

In right-body

mechanics

mechanics

,

,

the force

the force

P

P

can be found

can be found

easily from

easily from equilibrium analysis

equilibrium analysis

. We assume that the

. We assume that the

 bar is both

 bar is both rigid 

rigid 

(the

(the deformation

deformation

s of the bar is

s of the bar is

neglecte

neglecte

d) and strong

d) and

strong

enough to support the load

enough to support the load

W.

W.





In mechanics of

In mechanics of

materials

materials

, t

, the

he sta

static

ticss sol

soluti

ution i

on iss

extended to include an analysis of

extended to include an analysis of the forces acting

the forces acting

inside

inside

the bar to be certain that the bar will neither

the bar to be certain that the bar will neither

 break 

 break 

nor

nor deform excessively

deform excessively

..

Figure

Figure 1.1 1.1 EquilibriumEquilibrium analysis will determine the analysis will determine the force

force P P, but not the strength, but not the strength or the rigidity of the bar. or the rigidity of the bar.

1.2 Analysis of Internal Forces



The equilibrium analysis of a rigid body is concerned

 primarily with the calculation of external forces and

internal forces that act internal connections.



In mechanics of materials, this analysis determines

internal forces

－

that is, forces that act on cross

sections that are internal to the body itself.



We must investigate the

manner

in which these

internal forces are distributed within the body. Only

after these computations have been made can the

design engineer select

the proper dimensions

for a

member and select

the material

from which the

member should be fabricated.

Figure 1.2 External forces acting on a body.

Figure 1.3(a) Free body diagram (FBD) for determining the internal force system acting on section (1).



the external forces that hold a body in equilibrium are

known : the internal forces by equilibrium analysis.



The resultant force R , acting at the centroid C of the

cross section and C

R 

_{, the resultant couple C}

R 

_{, the}

Figure 1.3(b) Resolving the internal force R into the axial force P and the Shear force V .

Figure 1.3 (c) Resolving the internal couple C R _{into the}

torque T and the bending moment M .



Both R and C

R 

_{in terms of}

_two

_components

_：

_one

 perpendicular to the cross section and the other lying

in the cross section.

Following physically meaningful names

：

P

：

The force component that is perpendicular to the

cross section, tending to elongate or shorten the

 bar, is called

the normal force

.

V 

：

The force component lying in the plane of the cross

section, tending to shear (slide) one segment of the

 bar relative to the other segment, is called

the

shear force.

T 

：

The component of the resultant couple that tends to

twist (rotate) the bar is called

the twisting

moment or torque.

 M 

：

The component of the resultant couple that tends to

Figure 1.4 Deformations produced by the components of internal forces and couples.

the normal force

the shear force



In design, the manner in which

the internal forces are distributed

is equally important.



This consideration leads us to

introduce the force intensity at a

 point, called stress, which plays a

central role in the design of

load- bearing members.

Figure 1.5 Normal and shear stresses acting on the section at point O are defined in Eq. (2).



The stress vector acting on the

cross section at point

O

is

defined as

(1.1)

 A

 R

t 

o  A

Δ

Δ

=

→ Δ

lim



Its normal component

σ

(lowercase Greek sigma ) and

shear component

τ

(lowercase Greek tau) ,shown in

Fig. 1.5

（

 b

）

, are

(1.2)



The dimension of stress is [F/L

2

_]

_－

_{that is, force}

divided by area.



In SI units, force is measured in newtons

（

 N

）

and 

area in square meters, from pascals

(

Pa)

：

1.0 Pa

＝

1.0

 N/m

2

_{, 1.0 MPa}

_＝

_1.0

_×

₁₀

6

_Pa.



In U.S. Customary units, force is measured in pounds

and area in square inches, so that the unit of stress is

 pounds per square inch (lb/in.

2

_{), frequently}

abbreviated as psi. (1.0 ksi

＝

1000 psi), where “kip”is

the abbreviation for kilopound.

dA

dP

 A

P

o  A

=

Δ

Δ

=

→ Δ

lim

σ 

dA

dV 

 A

V 

o  A

=

Δ

Δ

=

→ Δ

lim

τ 



The commonly used sign convention for axial force

is to define tensile forces as positive and compressive

forces as negative. This convention is carried over to

normal stresses: Tensile stresses are considered to be

 positive, compressive stresses negative.



A simple sign convention for

shear stresses

does not

exist

；

a convention that depends on a coordinate

system will be introduced later in the text.



If the stresses are

uniformly distributed 

, they can be

computed form

(1.3)

Where A is the area of the cross section.

 A

P

=

σ 

 A

V 

=

τ 

1.3  Axially Loaded Bars

 a. Centroidal (axial) loading When the loading is uniform, its resultant passes through the

centroid of the loaded area. Therefore, the resultant P = pA

of each end load acts along the centroidal axis of the bar . The loads are called axial or

 centroidal loads.

 In Fig. 1.6 (a), the internal forces acting on all cross-sections are also uniformly distributed. the normal stress

acting at any point on a cross section is

(1.4)

Figure 1.6 A bar loaded axially by (a) uniformly distributed load of

intensity p (Pa or psi); and (b) a statically equivalent centroidal force P = pA.

 A P =

 The stress distribution caused by the concentrated loading in Fig. 1.6(b) is more complicated. The maximum stress is

considerably higher than the average stress P/ A. As we move away from the ends, the stress becomes more uniform, The

stress distribution is approximately uniform in the bar, except in the regions close to the ends.

Figure 1.7 Normal stress distribution in a strip caused by a concentrated load.

b . Saint Venant’s Principle

 About 150 year ago the French mathematician Saint Venant studied the effects of statically equivalent loads on the twisting of bars, called Saint Venant’s principle:

The difference between the effects of two different but statically equivalent loads becomes very small at sufficiently large

Figure 1.8 Normal stress distribution in a grooved bar.

 Most analysis in mechanics of materials is based on

simplifications that can be  justified with Saint

Venant

´

s principle.

 We often replace loads

(including support reactions ) by their resultants and  ignore the effects of holes, groove, and fillets on

stresses and deformations.

 Consider, as an example, the grooved cylindrical bar of radius

 R shown in Fig. 1.8(a). Introduction of the groove disturbs the uniformity of the stress, but this effect is confined to the

c.

Stresses on inclined planes

 Consider the stresses that act on plane

α

-

α

that is inclined at the angle

θ

to the cross section, the area of the inclined plane is A/cos

θ

.

 Because the segment is a two-force body, the resultant internal force acting on the inclined plane must  be resolved the

normal component

Pcos

θ

and the shear component Psin

θ

.

Figure 1.9 Determining the stresses acting on an inclined section of a bar.

 the corresponding stresses, shown in Fig.1.9(c), are (1.5a) (1.5b) θ  θ  θ  σ 

_cos

2

cos

/

cos

 A P  A P

₌

=

θ  θ  θ  θ  θ  τ  _{sin 2} 2 cos sin cos / sin  A P  A P  A P

₌

₌

=

θ  θ  θ  σ  _cos2 cos / cos  A P  A P

₌

=

θ  θ  θ  θ  θ  τ  _{sin 2} 2 cos sin cos / sin  A P  A P  A P

=

=

=

 the maximum normal stress _max is P /  A, and it acts on the cross section of the bar (on the plane

θ

= 0). The shear stress

τ

is zero when

θ

= 0.

 The maximum shear stress _max is P /2 A, which acts on the planes inclined at

θ

= 45

°

to the cross section. The normal stress _max is P /2 A when

θ

= 45

°

.

 In summary, axial load causes not only normal stress but also shear stress. The magnitudes of  both stresses depend on the orientation of the  plane on which they act.

 By replacing

θ

with

θ

+90

°

in Eqs. (1.5), we obtain the

stresses acting on plane a

´

-a

´

, which is perpendicular to a-a):

(1.6)

cos(

θ＋

90

°

)=

－

sin

θ

and sin2(

θ＋

90

°

)=

－

sin2

θ

. θ  σ  _sin2  A P

=

′

θ  τ  _{sin 2} A 2 P

−

=

′

_{Figure 1.10 Stresses acting}

on two mutually perpendicular inclined sections of a bar. θ  θ  θ  σ  _cos2 cos / cos  A P  A P

₌

=

θ  θ  θ  θ  θ  τ  _sin2 2 cos sin cos / sin  A P  A P  A P

₌

₌

=

 The complementary stresses on a small (infinitesimal) element of the material, the sides of which are parallel to the complementary  planes, as in Fig.10(b).

 When labeling the stresses, we made use of the following

important result that follows from Eqs.(1.5) and (1.6):

(1.7)

τ  τ 

′

=

−

 Because the stresses in Eqs. (1.5) and (1.6) act on mutually

 perpendicular , or

“complementary”planes, they are called complementary stresses.

θ  σ  _sin 2  A P = ′ θ  θ  θ  σ  _cos2 cos / cos  A P  A P ₌ = θ  θ  θ  θ  θ  τ  _sin2 2 cos sin cos / sin  A P  A P  A P _{= =} = θ  τ  _sin2 A 2 P − = ′ τ  = _{sin 2}θ  2 A P θ  τ  _sin2 A 2 P − = ′

 The shear stresses that act on

complementary planes have the same magnitude but opposite sense.

 The design of axially loaded bars is usually based on the maximum normal stress in the bar. The stress is commonly called simply the normal stress and denoted by

design criterion thus is that

σ＝

P/A must not exceed the working stress of the material from which the bar is to be fabricated. The working stress, _w also called the allowable  stress , is the largest value of stress that can be safely carried  by the material.

τ 

τ 

′

=

−

(1.7)

d.

Procedure for stress analysis

Equilibrium Analysis

•

If necessary, find the external reactions using a free-body diagram (FBD) of the entire structure.

• Compute the axial force P in the member using the method of sections has been found by equilibrium analysis.

Computation of Stresses

• The average normal stress in the member can be obtained from

σ

＝

P/A, where A is the cross-sectional area of the member at the cutting plane.

• In slender bars,

σ＝

P/A is the actual normal stress if the

section is sufficiently far from applied loads and abrupt changes in the cross section(Saint Venant

´

s principle).

Note on the Analysis of Trusses

 The usual assumptions made in the analysis of trusses are

：

(1) weights of the member are negligible in comparison to the

applied loads

；

(2) joints behave as smooth pins

；

and (3) all loads are applied at the joints.

 Under these assumptions, each member of the truss is an axially loaded bar . The internal forces in the bars can be obtained by the method of sections or the method of joints (utilizing the free-body diagrams of joints).

Sample Problem 1.1

The bar ABCD in Fig.(a) consists of three cylindrical steel

segments, each with a different cross-sectional area. Axial loads are applied as shown. Calculate the normal stress in each segment

.

© 2003 Brooks/Cole Publishing / Thomson Learning™ © 2003 Brooks/Cole Publishing / Thomson Learning™

Solution

The required free-body diagrams (FBDs), shown in Fig. (b), The normal stresses in the three segments are

) ( 3330 . 2 . 1 4000 2  psi T  in lb  A P  AB  AB  AB = = = σ  ) ( 2780 . 8 . 1 5000 2  psi C  in lb  A P  BC   BC   BC  = = = σ  ) ( 4380 . 6 . 1 7000 2  psi C  in lb  A P CD CD CD = = = σ 

Observe that the lengths of the segment do not affect the calculation of stresses. In addition, the fact that bar is made of steel is irrelevant.

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Sample Problem 1.2

For the truss shown in Fig. (a), calculate the normal stresses in (1) member AC 

；

and (2) member BD. The cross-sectional area of each member is 900mm2_.

Solution

Equilibrium analysis using the FBD of the entire in Fig. (a) give the following values for the external reactions

：

 A _y=40 kN, H  _y=60 kN, and  H  _x=0.

Because the force system is concurrent and

coplanar , there are two independent equilibrium equations From the FBD in Fig. (b).

Solving the equations gives P _AC =53.33 kN

(tension). Thus, the normal stress in member AC is

＝

59.3

×

106 _N/m2

_＝

_{59.3 MPa (T) Answer} 

∑

_Fy

=

₀

+

↑

∑

_Fx = _0→+ 0 5 3 40

+

P _AB

=

0 5 4

₌

+

 _AB  AC  P P 2 6 3 2 10 900 10 33 . 53 900 33 . 53 m  N  mm KN   A P  AC   AC   AC 

_×

−

×

=

=

=

σ  Part2

In truss analysis, each member of the truss is an axially loaded  bar . Note that we have assumed both of these forces to be tensile.

Part2

We have again assumed that the forces in the members are tensile. To calculate the force in member BD, we use the equilibrium

equation

P _BD= -66.67KN= 66.67 kN(C) the normal stress in member BD is

=-74.1

×

106 _N/m2_{=74.1 MPa}

_（

_C

_）

_Answer + 2 6 3 2 10 900 10 33 . 53 900 33 . 53 m  N  mm kN   A P  AC   AC   AC 

_×

−

×

=

=

=

σ  0 ) 3 ( ) 4 ( 30 ) 8 ( 40 0

−

+

−

=

=

∑

 M  E  P BD

Sample Problem 1.3

Figure(a) shows a two-member truss

supporting a block of weight W. The cross-sectional areas of the member are 800 mm 2

for AC . Determine the maximum safe value of W if the working stresses are 110MPa for

 AB and 120 MPa for AC .

Solution

The forces in the bars can be obtained by

analyzing the free-body diagram (FBD) of pin

 A in Fig.(b) .The equilibrium equations are

0 40 cos 60 cos 0

−

°

=

=

° →

+

∑

F  x P AC  P AB 0 40 sin 60 sin 0

+

↑

°

+

°

−

=

=

Solving simultaneously, we get

P _AB= 0.5077 W P _AC= 0.7779W For bar AB P_AB= _ABA_AB

0.5077 W = (110

×

106 _N/m2₎₍₈₀₀

_×

₁₀-6_m2₎

W = 173.3

×

103 _{N= 173.3 kN}

The value of W that will cause the normal stress in bar AC to equal its working stress is found from

For bar AC P_AC= _AC A_AC

0.7779 W = (120

×

106 _N/m2_{) (400}

_×

₁₀-6_m2₎

W = 61.7

×

103 _{N = 61.7 kN}

The maximum safe value of W is the smaller of the preceding two value. W = 61.7 kN

1.4.Shear stress

By definition, normal stress acting on interior surface is directed  perpendicular to that surface. Shear stress, on the other hand, is

tangent to the surface on which it acts.

 Three other examples of shear stress are illustrated in Fig. 1.11.

Figure 1.11 Examples of direct shear single shear in a rivet double shear in a bolt shear in a metal sheet produced by a punch. (a) (b) and (c)

 The distribution of direct shear stress is usually complex and not easily determined. It is common practice to assume that the shear force V is uniformly distributed over the shear area A, so that the shear stress can be computed from

(1.8)

 Eq. (1.8) must be interpreted as the average shear stress. It is often used in design to evaluate the strength of connectors, such as rivets, bolt, and welds.

 The loads shown in Fig. 1.11 are sometimes referred to as

direct shear , to distinguish them from the induced shear 

illustrated in Fig.1.9.  A V 

=

τ  Fig. 1.11 Fig. 1.9

1.5 Bearing Stress

 If two bodies are pressed against each other , compressive forces are developed on the area of contact. The pressure caused by these surface loads is called bearing stress.

Figure 1.12 Example of bearing stress (a) a rivet in a lap joint

stress is not constant

 Examples of bearing stress are the soil  pressure beneath a  pier and the contact  pressure between a

rivet and the side of its hole. shown in Fig. 1.12(a). The

 bearing stress caused  by the rivet is not

constantt as

illustrated in Fig. 1.12(b).

 The reduced area A_b is taken to be the projected area of the rivet

：

 A_b= td 

Where t is the thickness of the plate and d represents the diameter of the rivet in Fig. 1.12 (c) . The bearing force P_b

equals the applied load P. So that the bearing stress becomes (1.9) td  P  A P b b b = = σ 

Figure 1.12 (c) bearing stress

caused by the bearing force P _b

is assumed to be uniform on projected area td.

 assuming that the bearing stress

σ

 _b is uniformly distributed over a reduced area.

Sample Problem 1.4

The lap joint shown in Fig.(a) is fastened by four rivets of 3/4-in.

diameter. Find the maximum load P that can be applied if the working stresses are 14 ksi for shear in the rivet and 18 ksi for bearing in the plate. Assume that the applied load is distributed evenly among the four rivets, and neglect friction between the plates.

Solution

Calculate P using each of the two design criteria. Figure (b) shows the FBD of the lower plate, the lower halves of the rivets are in the  plate. This cut exposes the shear forces V that act on the cross