Stress Strain Young Modulus and Shear Stress

8/13/2019 Stress Strain Young Modulus and Shear Stress

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Normal stress

Intensityof force, or force per unit area, acting normal to

Symbol used fornormal stress, is (sigma)

( ) = ()

()

MUHAMMAD FIRDAUS BIN ZAKARIA, 2013


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(a)

(b)

(c)

Units

Nm-2

(1 Pa

( ) = ()

()



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Tensile StressApositive sign will be used to indicate a tensile stress

(member in tension),

: normal force pulls or stretches the area element A

Compressive StressAnegative sign will be used to indicate a compressive stress

(member in compression)

: normal force pushes or compresses area element A



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Rod diameter, =30mm, Force, P=15kNDetermine the tensile stress for the rod.



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Cross-sectional area, A


=

4 =

4 (0.03) = 7.0695 10

Therefore, the tensile stress in the rod is:-

=

A =

15000

7.0695 10= .


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P=240kN, L=1m, L = 0.55mm, 1=90mm, 2=130mm

A short post constructed from a hollow circular tube of

Aluminium supports a compressive load of 240 kN.

The inner and outer diameters of the tube are

1=90mm and 2=130mm, respectively, and its length

is 1m. The shortening of the post due to the load is

measured as 0.55 mm.

Determine the compressive stress in the post.



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Assuming that the compressive load acts at the center of the hollow

we can use the equation =P/A to calculate the normal stress. The fo

equals to 240kN and the cross-sectional area is:-


=

4 1 2

=

4[ 0.13 0.09 ] = 6.91 10

Therefore, the compressive stress in the post is:-

=

A =

240000

6.91 10= .


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Bar width = 35 mm, Thickness = 10 mmDetermine max. average normal stress in bar when subjected to

loading shown.



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Internal loading

Normal force diagram

By inspection, largest loading

area is BC, where PBC = 30 kN



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Average normal stress


BC=PBC

A

30(103) N

(0.035 m)(0.010 m)= = 85.7 MPa


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Normal strain

Deformation (elongation or contraction) per unit of lengtha member under axial loading.

Symbol used for normal strain, is (epsilon)

Strain can be expressed as a percentage strain

( ) = ()

()

( ) =

%



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Units (SI system)Normal strain is a

dimensionless quantity, asits a ratio of two lengths

But common practice tostate it in terms ofmeters/meter (m/m)

is small for mostengineering applications,

so is normally expressedas micrometers per meter(m/m) where 1m =106

Also expressed as apercentage,e.g., 0.001 m/m = 0.1 %



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Original Length, L= 0.6m, Change in length, L = 150x10-6

Determine the corresponding strain for the rod.



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Normal strain


( ) = () ()

=

150 10

0.6 = 2.5 10

Therefore, the tensile strain in the rod is:-


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P=240kN, L=1m, L = 0.55mm, 1=90mm, 2=130mm


Aluminium supports a compressive load of 240 kN.The inner and outer diameters of the tube are




Determine the compressive strain in the post.



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Compressive strain


( ) = () ()

=

5.5 10

1 = 5.5 10

Therefore, the compressive strain in the post is:-


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Youngs Modulus, EAlso known as Modulus of Elasticity, is a quantity thatrepresents the elasticity in length of a material.

It is defined as the ratio of stress to strain, that is

Since strain is dimensionless, E will have the same unit asstress such as N/m2 or Pascals.

The value of Youngs Modulus doesnot depend on thelength of the wire, but it depends on the wire material.

So, Young's Modulus of a wire does not change if theoriginal length is reduced nor increased.


, =,

, =

/

/=

.

.


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Youngs Modulus, E Values of E for other commonly used engineering materia

are often tabulated in engineering codes and referencebooks.

It should be noted that the modulus of elasticity is amechanical property that indicate the stiffness of amaterial.

Material that are very stiff, such as steel, have large value E (Esteel= 200GPa), whereas spongy materials such asvulcanized rubber may have low vaues (Erubber= 0.70MPa



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Rod diameter, =30mm, Force, P=15kN, Original Length, L=Change in length, L = 150x10-6m

Determine thevalue of Youngs Modulus for the rod.



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P=240kN, L=1m, L = 0.55mm, 1=90mm, 2=130mm


Aluminium supports a compressive load of 240 kN.The inner and outer diameters of the tube are




Determine theYoungs Modulus of the circular tube.



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, =,

, =

.

.

=,

, =

34.7MPa

5.5 10

= . /

=.

. =

(240(10) N)(1)

(6.91 10)(5.5 10)

= . /

Or,


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Ealuminium=70GPa, Ecopper=110GPa, Esteel = 200GPa

Aluminum bar AB, copper bar BC, and steel bar CD are seamlessly

with each other and rigidly fastened to the wall as shown in the fiCross-section area for AB, BC and CD are 200mm2, 150mm2, a

respectively. Determine the amount of elongation that occurs on th

as a result of the imposed loads. Modulus of elasticity of aluminum,

steel are 70GPa, 110GPa,and 200GPa respectively .


12 kN10 kN 8 kN

A BC

D

0.6m 0.2m0.4m


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12 kN10 kN 8 kN

A BC

D

0.6m 0.2m0.4m

123

12 kNPCD

1 +

= 0

12 = 0

= 12

=

=12000 0.2

(100 10)(200 10/

= 1.2 10

8kNPBC

2

+

12 8

=400

(150 10

= 9.69


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+

=

12 8 10

= 6

=

=6000

(200 10)(7

= 2.571 1

12 kN10 kN

8 kN

3

PAB

= = 2.571 10 9.697 10 1 . 2 1 0

= .

Total elongation that occurs on the bar ABCD;


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Shear Stress

Last time we talked about normal

stress (),which acts perpendicularto the cross-section.

Shear stress () acts tangential to the

surface of a material element.



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Where Do Shearing Stresses Occur?

Shearing stresses are commonlyfound in bolts, pins, rivets andwelded materials.

We do not assume is uniformover the cross-section, becausethis is not the case.

Therefore, is the average shear

stress. The maximum value of may be

considerably greater than avg.



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Shear stress is the stress componentthat act in the plane of the sectionedarea.

Consider a force F acting to the bar

For rigid supports, and F is largeenough, bar will deform and failalong the planes identified by ABand CD

Free-body diagram indicates thatshear force, V = F/2 be applied atboth sections to ensure equilibrium



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Average shear stress over each section is:

avg = average shear stress at section,

assumed to be same at each part on thesection

V = internal resultant shear force atsection determined from equations ofequilibrium

A = area of section

Case discussed above is example ofsimple or direct shear

Caused by the direct action of appliedload F



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Single shear

Steel and wood joints shown below are examples of

single-shear connections, also known as lap joints. Since we assume members are thin, there are no

moments caused by F



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Single shear

For equilibrium, x-sectional area of bolt and bonding

surface between the two members are subjected tosingle shear force, V = F

The average shear stress equation can be applied todetermine average shear stress acting on coloredsection in (d).



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Bolt is in single shear

Free body diagram of bolt

= =

Single shear


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Double shear

The joints shown below are examples of double-shea

connections, often called double lap joints. For equilibrium, x-sectional area of bolt and bonding

surface between two members subjected to doubleshear force, V = F/2

Apply average shear stress equation to determineaverage shear stress acting on colored section in (d).



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Bolt is in double shear

Free body diagram of bolt

=

= /

=

Free body diagram of center of b

Double shear


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What have we learn so far?

Axial forces in two-force member cause normal stresses

Transverse force exerted on the bolts and pins cause shestresses

However, axial forces cause both normal and shearing strplanes which are notperpendicular to the axis.

This is also the case for transverse forces exerted on a bo



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Procedure for analysis

Internal shear1. Section member at the pt where the avg is to be

determined

2. Draw free-body diagram

3. Calculate the internal shear force V

Average shear stress

1. Determine sectioned area A

2. Compute average shear stress avg = V/A



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Procedure for analysis

Consider an inclined section of a uniaxial bar.

The resultant force in the axial direction must equal P to sequilibrium

The force can be resolved into components perpendicula

section, F, and parallel to the section, V.

The area of the section is


= , =

= = /


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What have we learn so far?

We can formulate the average normal stress on the sectio

The average shear stress on the section is

Thus, a normal force applied to a bar on an inclined sectiproduces a combination of shear and normal stresses.


=

= /

=

=

=

/ =


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Depth and thickness = 40 mmDetermine average normal stress and average shear stress acting

along (a) section planes a-a, and (b) section plane b-b.



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Part (a)Internal loading

Based on free-body diagram, Resultant loading of axial force,P = 800 N



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Part (a)Average stress

Average normal stress,

=P

A

800 N

(0.04 m)(0.04 m)= 500 kPa=



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Part (a)Internal loading

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

avg = 0



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Part (b)Internal loading

+

Fx= 0; 800 N + N sin 60 + V cos 60 = 0+

Fy= 0; V sin 60N cos 60 = 0



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Part (b)

Internal loading

Or directly using x, y axes,

Fx= 0;

Fy= 0;

+

+

N 800 N cos 30 = 0

V 800 N sin 30 = 0



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Part (b)

Average normal stress

=N

A

692.8 N

(0.04 m)(0.04 m/sin 60)= 375 kPa=



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Part (b)

Average shear stress

avg =VA

400 N(0.04 m)(0.04 m/sin 60)

= 217 kPa=

Stress distribution as shown below;



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Stress Strain Young Modulus and Shear Stress