9Chapter2

CONCEPT OF STRESS2.1 Normal Stress2.2 Shearing Stress2.3 Bearing Stress in Connections2.4 Allowable Stress and Factor of Safety2.5 Stress Under General Loading2.6 Maximum Stresses

2.1 Normal StressDistributed forces within a load-carrying member can be represented by a staticallyequivalent system consisting of a force and a moment vector acting at any arbitrary point(usually the centroid) of a section. Relating stresses to external forces and moments is atwo step process as shown in Figure 2.1.

Normal Stress is the intensity of internal distributed force that is normal to the surface ofan imaginary cut surface. The normal stress acting in the direction of the axis of a slendermember (rods, cables, bars, columns, etc.) is called the axial stress. Generally, In theconsideration of materials view, stress is a variable that can be used as a measure ofstrength of a structural member.Consider a prismatic bar shown inFigure 2.2, the intensity of

internal distributed forces on animaginary cut surface of a body iscalled the stress on a surface. Theaxial forces produce a uniformstretching of the bar, it is calledthe bar is in tension. Plane aa is across section that is perpendicularto the longitudinal axis, and A iscross section area.

Assumptions:

The material of the bar is homogeneous (uniform density) and isotropic (samedirectional properties)

The bar is prismatic (uniform cross section), with no stress raisers such as holes,

9

Chapter2

CONCEPT OF STRESS2.1 Normal Stress2.2 Shearing Stress2.3 Bearing Stress in Connections2.4 Allowable Stress and Factor of Safety2.5 Stress Under General Loading2.6 Maximum Stresses

2.1 Normal StressDistributed forces within a load-carrying member can be represented by a staticallyequivalent system consisting of a force and a moment vector acting at any arbitrary point(usually the centroid) of a section. Relating stresses to external forces and moments is atwo step process as shown in Figure 2.1.

Normal Stress is the intensity of internal distributed force that is normal to the surface ofan imaginary cut surface. The normal stress acting in the direction of the axis of a slendermember (rods, cables, bars, columns, etc.) is called the axial stress. Generally, In theconsideration of materials view, stress is a variable that can be used as a measure ofstrength of a structural member.Consider a prismatic bar shown inFigure 2.2, the intensity of

internal distributed forces on animaginary cut surface of a body iscalled the stress on a surface. Theaxial forces produce a uniformstretching of the bar, it is calledthe bar is in tension. Plane aa is across section that is perpendicularto the longitudinal axis, and A iscross section area.

Assumptions:

The material of the bar is homogeneous (uniform density) and isotropic (samedirectional properties)

The bar is prismatic (uniform cross section), with no stress raisers such as holes,

9

Chapter2

CONCEPT OF STRESS2.1 Normal Stress2.2 Shearing Stress2.3 Bearing Stress in Connections2.4 Allowable Stress and Factor of Safety2.5 Stress Under General Loading2.6 Maximum Stresses

2.1 Normal StressDistributed forces within a load-carrying member can be represented by a staticallyequivalent system consisting of a force and a moment vector acting at any arbitrary point(usually the centroid) of a section. Relating stresses to external forces and moments is atwo step process as shown in Figure 2.1.

Normal Stress is the intensity of internal distributed force that is normal to the surface ofan imaginary cut surface. The normal stress acting in the direction of the axis of a slendermember (rods, cables, bars, columns, etc.) is called the axial stress. Generally, In theconsideration of materials view, stress is a variable that can be used as a measure ofstrength of a structural member.Consider a prismatic bar shown inFigure 2.2, the intensity of

internal distributed forces on animaginary cut surface of a body iscalled the stress on a surface. Theaxial forces produce a uniformstretching of the bar, it is calledthe bar is in tension. Plane aa is across section that is perpendicularto the longitudinal axis, and A iscross section area.

Assumptions:

The material of the bar is homogeneous (uniform density) and isotropic (samedirectional properties)

The bar is prismatic (uniform cross section), with no stress raisers such as holes,

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notches, or threads, etc.

The bar should have no residual stressesand should not be subjected to temperaturechanges

The axial force P acts through the centroidof the cross section (centric loading, toavoid buckling

The section (where stress is computed) is remote from a loaded end (Saint Venants principle)

The intensity of the force (force per unit area) is called stress, assuming that the stress hasuniform distribution, then

)1.....(....................APFlim aveA0A

The normal stress at a particular point may not be equal to the average stress but theresultant of the stress distribution must satisfy:

)2.(..........dAdFAPA

ave When the bar is stretched, the resulting stress are tensile stress, if the bar is compressed,the stress are compressive stress. Sign convention of the normal stress is: tensile stress as

positive and compressive stress as negative.

Stress Units:

SI unit: (Pa, Pascal), N mm (MPa), 1 MPa = 10 Pa, 1 GPa = 10 PaUSCS unit: (psi) , kips in (ksi), 1 ksi = 10 psi1 psi = 6 , 895 Pa & 1ksi = 6.895 MPa

Example 2.1

Given: a stepped bar fixed at end D and loading asshown in Figure 2.3. If 20 , =25 , = 30 , , = 35 .Find:

1. Internal forces in each segment?2. The largest stress in the bar?

SOLUTIONS:

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Example 2.2

Given: A solid circular post ABC (see figure) supports a load = 2500 acting at the top.A second load P2 is uniformly distributed around the shelf at B. The diameters of the upper andlower parts of the post are = 1.25 and = 2.25 ., respectively.Find:

a) The normal stress in the upper part of the post?b) If it is desired that the lower part of the post have the same compressive stress as the upper

part, what should be the magnitude of the load P?SOLUTION:

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Example 2.3

Given: Two solid cylindrical rods (1) and (2) are joined together at flange B and loaded, as shownin Fig.1.3.The diameter of rod(1) is d1 =24 mm and the diameter of rod(2) isd2 =42 mm.Find: Using the method of section,determine the normal stresses in rods(1)and(2)?Assumptions: The two rods are welded together B. the weight of the bar isneglected.

SOLUTION:

2.2 Shear Stress ():Shear stress: The intensity of internal distributed force that isparallel to the surface of an imaginary cut surface is called theshear stress. There are three kinds of shearing stresses:

Direct shear (mostly due to normal loads) discussed in thischapter

Torsional shear (mostly due to torsional loads) discussedlater.

Shear stress or flexural shear (due to transverse loads) discussed later.

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The resultant of the internal shear force distribution is defined as the shear of the section and is equalto the load P.

)3......(..........AF

AP

ave in the case of single shear

)4....(..........A2

FAP

ave in the case of double shear

)5......(..........nAF

AP

ave in the case of nth bolts

Example 2.4:

Given: For the connection shown in Fig. P1.19, if the diameter of each bolt is7/8 in, and the load isP is 45 kips.Find:

1. Determine the average shearstress in the bolts?2. The tensile stress in the plate (2x0.2 in2) at a section passes through threebolts?Assumptions: the weight of the bolted joint can be neglected.SOLUTIONS:

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2.3 Bearing Stress(b): Bearing stress: The compressive normal stress that isproduced when one surface presses against other is calledthe bearing stress. In other words, bearing stress is acompressive stress that occurs as a result of contact (pointor surface) between two loaded members. Bolts, rivets, and pins create stresses on the points ofcontact or bearing surfaces of the members they connect. The resultant of the force distribution on the surface isequal and opposite to the force exerted on the pin.Corresponding average force intensity is called the bearingstress,

)6.....(..........tdP

AP

,Aarea,Projected/Pload,Axial

pb

Pbearing

14

2.3 Bearing Stress(b): Bearing stress: The compressive normal stress that isproduced when one surface presses against other is calledthe bearing stress. In other words, bearing stress is acompressive stress that occurs as a result of contact (pointor surface) between two loaded members. Bolts, rivets, and pins create stresses on the points ofcontact or bearing surfaces of the members they connect. The resultant of the force distribution on the surface isequal and opposite to the force exerted on the pin.Corresponding average force intensity is called the bearingstress,

)6.....(..........tdP

AP

,Aarea,Projected/Pload,Axial

pb

Pbearing

14

2.3 Bearing Stress(b): Bearing stress: The compressive normal stress that isproduced when one surface presses against other is calledthe bearing stress. In other words, bearing stress is acompressive stress that occurs as a result of contact (pointor surface) between two loaded members. Bolts, rivets, and pins create stresses on the points ofcontact or bearing surfaces of the members they connect. The resultant of the force distribution on the surface isequal and opposite to the force exerted on the pin.Corresponding average force intensity is called the bearingstress,

)6.....(..........tdP

AP

,Aarea,Projected/Pload,Axial

pb

Pbearing

15

Example 2.5(P2.18):

Given: Two plates are joined by four rivets of 20-mm diameter, as shown in Figure P2.18. If the shearing,tensile, and bearing stresses are limited to 80,100, and140 MPa respectively.Find: The maximum load P.

2.4 Allowable Stress & Factor of Safety

Structural members or machines must be designed such that the working stresses are lessthan the ultimate strength of the material. The factor of safety, ns, is the ratio ofmaximum load that produces failure of the member to the load allowed under serviceconditions: = . . ( )The allowable load is also called the service load or working load. This ratio, mustalways greater than unity, ns >1. The factor of safety may also be defined by:= . . ( )

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The allowable stress is also known as the applied stress, working stress, or design stress,and it represent the required strength.

Factor of safety considerations: uncertainty in material properties

uncertainty of loadings uncertainty of analyses

number of loading cycles types of failure

maintenance requirements and deterioration effects importance of member to structures integrity risk to life and property influence on machine function

Selection of a factor of safety:1. 1:25 to 2 for known materials used under controllable conditions and subjected to

loads and stresses that can be readily determined with certainty. It is used wherelow weight is a particularly important consideration.

2. 2 to 3 for average materials operated in ordinary environments and subjected toloads and stresses that can be determined.

3. 3 to 4 for average materials used in uncertain environments or subjected touncertain stresses.

2.5 Stress Under General Loading

A member subjected to a general combination of loads is cut into two segments by a planepassing through Q.

, will be considered positive if the outward normal to the surface is in the positive idirection.

Stress component is positive if numerator and denominator have the same sign. Thus ij ispositive if: (1) are both positive. (2) , are both negative. The distribution of internal stress components may be defined as,

16

The allowable stress is also known as the applied stress, working stress, or design stress,and it represent the required strength.

Factor of safety considerations: uncertainty in material properties

uncertainty of loadings uncertainty of analyses

number of loading cycles types of failure

maintenance requirements and deterioration effects importance of member to structures integrity risk to life and property influence on machine function

Selection of a factor of safety:1. 1:25 to 2 for known materials used under controllable conditions and subjected to

loads and stresses that can be readily determined with certainty. It is used wherelow weight is a particularly important consideration.

2. 2 to 3 for average materials operated in ordinary environments and subjected toloads and stresses that can be determined.

3. 3 to 4 for average materials used in uncertain environments or subjected touncertain stresses.

2.5 Stress Under General Loading

A member subjected to a general combination of loads is cut into two segments by a planepassing through Q.

, will be considered positive if the outward normal to the surface is in the positive idirection.

Stress component is positive if numerator and denominator have the same sign. Thus ij ispositive if: (1) are both positive. (2) , are both negative. The distribution of internal stress components may be defined as,

16

The allowable stress is also known as the applied stress, working stress, or design stress,and it represent the required strength.

Factor of safety considerations: uncertainty in material properties

uncertainty of loadings uncertainty of analyses

number of loading cycles types of failure

maintenance requirements and deterioration effects importance of member to structures integrity risk to life and property influence on machine function

Selection of a factor of safety:1. 1:25 to 2 for known materials used under controllable conditions and subjected to

loads and stresses that can be readily determined with certainty. It is used wherelow weight is a particularly important consideration.

2. 2 to 3 for average materials operated in ordinary environments and subjected toloads and stresses that can be determined.

3. 3 to 4 for average materials used in uncertain environments or subjected touncertain stresses.

2.5 Stress Under General Loading

A member subjected to a general combination of loads is cut into two segments by a planepassing through Q.

, will be considered positive if the outward normal to the surface is in the positive idirection.

Stress component is positive if numerator and denominator have the same sign. Thus ij ispositive if: (1) are both positive. (2) , are both negative. The distribution of internal stress components may be defined as,

17

dAdF

AF&

dAdF

AF

dAdF

Azz

0Axz

yy

0Axy

xx

0Ax

limlim

Flim

.(9)

The double subscript notation is interpreted as follows: The first subscript indicates thedirection of a normal to the plane or face on which the stress component acts; the secondsubscript relates to the direction of the stress itself.

Symmetric Shear Stresses: Represented by a pair of symmetric shear stress points towardsthe corner or away from the corner. Thus, we now examine properties of shearing stress bystudying the equilibrium of forces. For equilibrium in 2-D as shown in the Figure 2. , an equaland opposite internal force and stress distribution must be exerted on the other segment of themember.

Stress components are defined for the planes cut parallel to the x, y and z axes. Forequilibrium, equal and opposite stresses are exerted on the hidden planes.

The combination of forces generated by the stresses must satisfy the

conditions for equilibrium: 0MMM0FFF

zyx

zyx

.(10) Consider the moments about the z axis:

y xx y

y xx yz a.Aa.A0M

(11)

z xx zz yy z a n ds i m i l a r l y , Stress cube showing all positive stress components. It followsthat only 6 components of stress are required to define thecomplete state of stress, written as:

, , = , , Sign Convention: See section 1.6. It can be summarize as:

i. When both the outer normal and the stress component face in a positivedirection relative to the coordinate axes, the stress is positive.

ii. When both the outer normal and the stress component face in a negativedirection relative to the coordinate axes, the stress is positive.

iii. When the normal points in a positive direction while the stress points in a

17

dAdF

AF&

dAdF

AF

dAdF

Azz

0Axz

yy

0Axy

xx

0Ax

limlim

Flim

.(9)

The double subscript notation is interpreted as follows: The first subscript indicates thedirection of a normal to the plane or face on which the stress component acts; the secondsubscript relates to the direction of the stress itself.

Symmetric Shear Stresses: Represented by a pair of symmetric shear stress points towardsthe corner or away from the corner. Thus, we now examine properties of shearing stress bystudying the equilibrium of forces. For equilibrium in 2-D as shown in the Figure 2. , an equaland opposite internal force and stress distribution must be exerted on the other segment of themember.

Stress components are defined for the planes cut parallel to the x, y and z axes. Forequilibrium, equal and opposite stresses are exerted on the hidden planes.

The combination of forces generated by the stresses must satisfy the

conditions for equilibrium: 0MMM0FFF

zyx

zyx

.(10) Consider the moments about the z axis:

y xx y

y xx yz a.Aa.A0M

(11)

z xx zz yy z a n ds i m i l a r l y , Stress cube showing all positive stress components. It followsthat only 6 components of stress are required to define thecomplete state of stress, written as:

, , = , , Sign Convention: See section 1.6. It can be summarize as:

i. When both the outer normal and the stress component face in a positivedirection relative to the coordinate axes, the stress is positive.

ii. When both the outer normal and the stress component face in a negativedirection relative to the coordinate axes, the stress is positive.

iii. When the normal points in a positive direction while the stress points in a

17

dAdF

AF&

dAdF

AF

dAdF

Azz

0Axz

yy

0Axy

xx

0Ax

limlim

Flim

.(9)

The double subscript notation is interpreted as follows: The first subscript indicates thedirection of a normal to the plane or face on which the stress component acts; the secondsubscript relates to the direction of the stress itself.

Symmetric Shear Stresses: Represented by a pair of symmetric shear stress points towardsthe corner or away from the corner. Thus, we now examine properties of shearing stress bystudying the equilibrium of forces. For equilibrium in 2-D as shown in the Figure 2. , an equaland opposite internal force and stress distribution must be exerted on the other segment of themember.

Stress components are defined for the planes cut parallel to the x, y and z axes. Forequilibrium, equal and opposite stresses are exerted on the hidden planes.

The combination of forces generated by the stresses must satisfy the

conditions for equilibrium: 0MMM0FFF

zyx

zyx

.(10) Consider the moments about the z axis:

y xx y

y xx yz a.Aa.A0M

(11)

z xx zz yy z a n ds i m i l a r l y , Stress cube showing all positive stress components. It followsthat only 6 components of stress are required to define thecomplete state of stress, written as:

, , = , , Sign Convention: See section 1.6. It can be summarize as:

i. When both the outer normal and the stress component face in a positivedirection relative to the coordinate axes, the stress is positive.

ii. When both the outer normal and the stress component face in a negativedirection relative to the coordinate axes, the stress is positive.

iii. When the normal points in a positive direction while the stress points in a

18

negative direction (or vice versa), the stress is negative.iv. In accordance with this sign convention, tensile stresses are always positive

and compressive stresses always negative. Figure 1.2 depicts a system ofpositive normal and shear stresses.

Typical Cases of Stress:1. Triaxial Stress: when the element is subjected only to stresses, , = = = 0 , acting in mutually perpendicular

directions is said to be in a state of triaxial stress. Such a state of stress can bewritten as,

= 0 00 00 0An example of equal triaxial compression is found in a small element of liquidunder static pressure.

2. 2-D or Plane stress: in this case only the x, and y faces

are subjected to stresses , , = == 0. The plane stress matrix is written as,=3. Pure Shear: In this case, the element is subjected to

plane shearing stresses only Fig. Typical pure shear

occurs over the cross sections and on longitudinal planesof a circular shaft subjected to torsion.

4. Uniaxial Stress: When normal stresses act along one direction only, the onedimensional state of stress is referred to as a uniaxial tension or compression.

2.6 Design of Bars For Axial Loading:

1. Evaluate the mode of possible failure.2. Determine the relationship between load and stress.3. Determine the material strength.4. Select the factor of safety, = , and the cross sectional dimensions are

obtained as, =

19

Example 2.6:

What is the maximum possible value of the clamping force C

in the jaws of the pliers shown in the figure, if =3.75 , & = 1.60 , and the ultimate shear stress in the0.20-in. diameter pin is 50 ksi? What is the maximumpermissible value of the applied load P if a factor of safety of3.0 with respect to failure of the pin is to be maintained?

20

Example 2.7: