11-15-14 strength of materials

8/10/2019 11-15-14 Strength of Materials

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Engr. Antonio Valente Macarilay

STRENGTH OF MATERIALS


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Strength of Materialsdeals with the nature and effects ofstresses in the parts of engineeringstructuresIts principal object is to determine theproper size and form of pieces which haveto bear given loads, or, conversely, todetermine the loads which can be safelyapplied to pieces whose dimensions andarrangement are already given.


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StressThe ratio of the applied load to thecross-sectional area of an elementexperiencing the force.


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TYPES OF SIMPLE STRESS

With normal stress, , the area is normal

to the force carried. P

P

P

P

1. Normal / Axial Stress

= P / A

P = Tensile / Compressive LoadA = Cross Sectional Area


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P

P d Sheared area

2. SHEARING STRESSa. Single Shear

2

4S

A d

= P/A A = Total Sheared Area


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Double Shear

P

P

Sheared area

= P/A A = Total Sheared Area

A = 2 D2 / 4


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Punching Shear

Sheared Area

p = P / A s A s = D t


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II. THIN-WALLED CYLINDERS

P

P t i D

t

a. Tangential Stress

St = D/2t

= pressure in N/mm 2

D = inside diameter

t = thickness in mm

BARLOW FORMULA


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b. Longitudinal Stress

SL = D/4t

= pressure in N/mm 2 D = inside diameter

t = thickness in mm


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Thermal Stress

Linear ExpansionL = L L (T)

Volumetric ExpansionV = V V (T)


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Stress due to Thermal Expansion


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Strain

A measure of the deformation of thematerial that is dimensionless.

The change of shape produced by

stress

= L / LWhere: = strain

L = change in length L = ori inal len th


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Hookes Law

Stress is directly proportional to Strain

named after the physicist Robert Hooke, 1676 = E ; E = /

= Stress = StrainE = Modulus of Elasticity(Youngs Modulus) (GPa)


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Poissons Ratio the ratio of transverse contraction strain tolongitudinal extension strain in the direction ofstretching forceTensile deformation is considered positive andcompressive deformation is considered negative

= - lateral / longitudinal


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Stress Strain Graph


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Elastic Limit

the point at which permanent

deformation occurs, that is, after theelastic limit, if the force is taken off thesample, it will not return to its originalsize and shape, permanentdeformation has occurred.


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Proportional Limit

The greatest stress at which a

material is capable of sustaining theapplied load without deviating fromthe proportionality of stress to strain.


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Ultimate Strength (Tensile)

The maximum stress a material

withstands when subjected to anapplied load.The stress required to produce

rupture.


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Yield Strength

Stress at which material exceeds

the elastic limit and will not return toits origin shape or length if thestress is removed.


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Working Stress

the actual stress of a material undera given loading


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Allowable Stress

The maximum safe stress that amaterial can carry


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Factor of Safety

The ratio of ultimate strength toallowable strength


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The condition under which the stress is

constant or uniform is known as A. Simple stress

B. Shearing stress

C. Tangential stress

D. Normal stress


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The highest ordinate on the stress-strain curve

is called A. rupture stress

B. elastic limit

C. ultimate stress or ultimate strength

D. proportional limit


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Shearing stress is also known as

A. Simple stress

B. Shearing stress


D. Normal stress


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Stress caused by forces perpendicular to the

areas on which they act is called A. Simple stress

B. Shearing stress


D. Normal stress


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What type of stress is produced whenever the

applied load cause one section of a body totend to slide past its adjacent section?

A. normal stress

B. sliding stress

C. shearing stress

D. bearing stress


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The ratio of the unit lateral deformation to theunit longitude deformation is called

A. compressibility

B. bulk modulus

C. shear modulus

D. Poissons ratio


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It describes the length elasticity of the material.

A. Bulk modulus

B. Youngs modulus or tensile modulus

C. Modulus of Compressibility

D. Shear modulus


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Determine the outside diameter of a hollowsteel tube that will carry a tensile load of 500 kNat a stress of 140 MPa. Assume the wallthickness to be one-tenth of the outsidediameter.

A. 132 mm C. 113 mm

B. 143 mm D. 133 mm


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What force is required to punch a 20-mm diameter hole through a 10-mm thick plate having ultimatestrength of 450 MPa?

a.) 283 kN c.) 382 kN

b.) 312 kN d.) 293 kN


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A spherical pressure vessel 400-mm indiameter has a uniform thickness of 6 mm. Thevessel contains gas under a pressure of 8 MPa.If the ultimate tensile stress of the material is420 MPa, what is the factor of safety with

respect to tensile failure?

A. 3.15 C. 3.4

B. 1.90 D. 2.6


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Elongation ( )

P = ForceL = Original Length

A = Cross Sectional AreaE = Youngs Modulus

= PL / AE


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Elongation due to Weight ( )

= unit mass g = gravity

E = Youngs Modulus

= g L2 / 2E


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A steel rod having a cross-sectional area of300mm 2 and length of 150 m is suspendedvertically from one end .It supports a load of 20kN at the lower end. If the unit mass of steel is7850 kg/m 3 and E=200 GPa, find the total

elongation of the rod.

A. 33.45 mm C. 53.44 mm

B. 54.33 mm D. 35.44 mm


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TorsionIn solid mechanics, it is the twistingof an object due to an appliedtorque


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Polar Moment of Inertia

Solid Shaft:J = D4 / 32

Hollow Shaft:J = (D o4 D i4) / 32


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Angle of Twist

= TL / JGT = TorqueL = LengthJ = Polar Moment of InertiaG = Shear Modulus


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Power Transmitted by Shaft

P = T = T(2 f)or

P = 2 T nP = power (W)T = Torque (Nm)f = frequency (Hz)n = angular speed (rev/s)


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What is the minimum diameter of a solid steelshaft that will not twist through more than 3 in a6-m length when subjected to a torque of 14kN-m? Use G=83 GN/m2.

A. 118 mm

B. 145 mm

C. 122 mm

D. 113 mm


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Spring

device made of an elastic materialthat undergoes a significant changein shape, or deformation, under anapplied load


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Helical Spring

Max Shearing Stress Approximate:

Exact:

R = mean radius of spring (D is mean diameter)d = diameter of wirem = D/d

max 3

16PR d1

d 4R

max 3

16PR 4m 1 0.615d 4m 4 m


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Spring Deformation

= 64PR 3n / Gd 4

P

L


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Determine the maximum shearing stress in ahelical steel spring composed of 20 turns of 20-mm diameter wire on a mean radius of 80 mmwhen the spring is supporting a load of 2kN.

A. 120.6 MPaB. 117.9 MPa

C. 132.4 MPa

D. 126.9 MPa


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CABLESI. Parabolic


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Parabolic CableTension at the SupportT2 = ( L/2) 2 + H 2

Approximate Length of CableS = L + 8d 2/3L 32d 4/5L 3

Tension at the Lowest PointH = L2 / 8d

L = Horizontal Spand = sag of cable = weight per unit length


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II. CatenaryThe theoretical shape of ahanging flexible chain orcable when supported at its

ends and acted upon by auniform gravitational force(its own weight) and inequilibrium. The curve hasa U shape that is similar inappearance to the

parabola, though it is adifferent curve.


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Catenary Cable

Tension at the support

TL = yL

1 1 2 21 2

Dis tance Between supports :

S y S yx c ln x c ln

c c

1 2

22

1 1

2 2

2 2

Tension at the support T & T :

T H S

T H S

2 2 2

1 1

2 2 22 2

Relationship among S,y & c :

S y c

S y c


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Catenary

= weight per unit lengthy = height of the support (respective)

c = minimum clearance from the groundS = cable length (respective)


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Calculate the allowable spacing of the twotowers to carry a flexible wire cable weighing0.03 kg per horizontal meter if the maximumtension at the lowest point is not to exceed 1150kg at sag of 0.50 m.

A. 248 m C. 390 m

B. 408 m D. 422 m


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TO GOD BE THE GLORY!

End of session.

11-15-14 strength of materials

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