Strength Handout Wup

1/27/2011

We always ask if there’s still hope lef t or if there’s still time. But we never realize that hope only leaves when we doubt and time only runs out the moment we give up.

We never realize that hope only leaves when we doubt and time only runs out the moment we give up.


s t r e ng t h of m a t e r ial s


Strength of Materials deals with the nature and ef f ects of stresses in the parts of engineering structures Its principal ob ject is to determine the proper size and f orm of pieces which have to bear given loads, or, conversely, to determine the loads which can be saf ely applied to pieces whose dimensions and arrangement are already given.

Stress The mutual action between two bodies, or between two parts of a body, whereby each of the two exerts a f orce upon the other The ratio of the applied load to the cross-sectional area of an element in tension.



Categories

Categories

R esidual Stress due to the manuf acturing processes that leave stresses in a material.

Structural Stress produced in structural members because of the weights they support.

Pressure Stress stresses induced in vessels containing pressurized materials.

Flow Stress Occur when a mass of f lowing f luid induces a dynamic pressure on a conduit wall.

Thermal Stress Exists whenever temperature gradients are present in a material.

Fatigue Stress Due to cyclic application of a stress. Could be due to vibration or thermal cycling.

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T ypes of Stress A xia / l Normal Stress Force perpendicular to an area divided by the area

Shear Stress Force parallel to an area divided by the area


TYPES OF SIMPLE STR ESS 1. Nor mal / Axial


2. SHEAR ING STR ESS a. Single Shear

Str ess

P

With nor mal str ess, σ, the ar ea is nor mal to the for ce car r ied. P

P

σ= P/A P = Tensile / Com pr essive Load A = Cr oss Sectional Ar ea P

P


P

d Sh e ar e d a r e a

AS 



4

d 2

σ = P/A A = Total Shear ed Ar ea


Punching Shear

Double Shear P P

Shear ed Ar ea

Shear ed ar ea

A = 2 π D2 / 4

σ = P/A A = Total Shear ed Ar ea

As = π D t

σp = P / As

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II. THIN-WALLED CYLINDER S


b. Longitudinal Str ess

a. Tangential Str ess P L

t P

Di

St = ρD/2t

t

ρ = pr essur e in N/mm2 D = inside diameter t = thick ness in mm

SL = ρD/4t ρ = pr essur e in N/mm2 D = inside diameter t = thick ness in mm

BAR LOW FOR MULA



Thermal Stress

Stress due to Thermal Expansion

 Linear

Expansion ∆ L =

 Volumetric

α L

L ( ∆ T)

Expansion ∆ V =

α V V

( ∆ T)



Strain

Hook e’s Law

A measure of the def ormation of the material that is dimensionless. The change of shape produced by stress ε = ∆ L / L ε = strain ∆ L = change in length L = original length

“Stress is directly proportional to Strain”

σ = Eε ; E = σ/ ε named af ter the physicist R obert Hook e, 1676

σ = Str ess ε = Str ain E = Modulus of Elasticity (Young’s Modulus) (GPa)

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Young’s Modulus

Poisson’s R atio

a measure of the amount of f orce needed to produce a unit def ormation Value f or some materials Steel Aluminum Copper

E = 200 GPa E = 69 GPa E = 120 GPa

the ratio of transverse contraction strain to longitudinal extension strain in the direction of stretching f orce Tensile def ormation is considered positive and compressive def ormation is considered negative

μ = - εlateral / εlongitudinal



Positive Poisson’s R atio

Negative Poisson’s R atio

Re-entr ant polymer f oams Nor mal polymer f oams


Stress – Strain Graph


Elastic Limit the point at which permanent def ormation occurs, that is, af ter the elastic limit, if the force is tak en of f the sample, it will not return to its original size and shape, permanent def ormation has occurred.

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Proportional Limit

Ultimate Strength (Tensile)

The greatest stress at which a material is capable of sustaining the applied load without deviating f rom the proportionality of stress to strain.

The maximum stress a material withstands when sub jected to an applied load. The stress required to produce rupture.



Yield Strength

Work ing Stress

Stress at which material exceeds the elastic limit and will not return to its origin shape or length if the stress is removed.

the actual stress of a material under a given loading



Allowable Stress

Factor of Saf ety

The maximum saf e stress that a material can carry

The ratio of ultimate strength to allowable strength

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1. The condition under which the str ess is constant or unif or m is known as

2. The highest or dinate on the str ess-str ain cur ve is called

A.

Simple str ess

A.

ruptur e str ess

B.

Shear ing str ess

B.

elastic limit

C.

Tangential str ess

C.

ultimate str ess or ultimate str ength

D.

Nor mal str ess

D.

pr opor tional limit


3.


Shear ing str ess is also known as A.

Simple str ess

B.

Shear ing str ess

C.

Tangential str ess

D.

Nor mal str ess


5. W hat type of str ess is pr oduced whenever the applied load cause one section of a body to tend to slide past its ad jacent section?


4. Str ess caused by f or ces per pendicular to the ar eas on which they act is called A.

Simple str ess

B.

Shear ing str ess

C.

Tangential str ess

D.

Nor mal str ess


6. Str ess caused by for ces acting along or par allel to the ar ea resisting the for ces is known as

A.

nor mal str ess

A.

Simple str ess

B.

sliding str ess

B.

Shear ing str ess

C.

shear ing str ess

C.

Tangential str ess

D.

bear ing str ess

D.

Nor mal str ess

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7. The ratio of the unit later al def or mation to the unit longitude def or mation is called


8. It descr ibes the length elasticity of the mater ial.

A.

compr essibility

A.

Bulk modulus

B.

bulk modulus

B.

Young’s modulus or tensile modulus

C.

shear modulus

C.

Modulus of Compr essibility

D.

Poisson’s r atio

D.

Shear modulus


9. Deter mine the outside diameter of a hollow steel tube that will car r y a tensile load of 500 kN at a str ess of 140 MPa. Assume the wall thickness to be one-tenth of the outside diameter . A. 132 mm

C. 113 mm

B. 143 mm

D. 133 mm


11. A hole is to be punched out of a plate having an ultimate shear ing str ess of 300 MPa. If the compr essive str ess in the punch is limited to 400 MPa, deter mine the maximum thickness of plate fr om which a hole 100 mm in diameter can be punched.


10. What f orce is required to punch a 20-

mm diameter hole through a 10-mm thick plate having ultimate strength of 450 MPa? a.) 283 k N b.) 312 k N

c.) 382 k N d.) 293 k N


12. A spher ical pr essur e vessel 400-mm in diameter has a unif or m thickness of 6 mm. The vessel contains gas under a pr essur e of 8 MPa. If the ultimate tensile str ess of the mater ial is 420 MPa, what is the factor of saf ety with respect to tensile f ailur e?

A. 33.3 mm

C. 13.4 mm

A. 3.15

C. 3.4

B. 17.9 mm

D. 26.9 mm

B. 1.90

D. 2.6

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13. A steel rod with a cr oss-sectional ar ea of 150 mm2 is str etched between two fixed points. The tensile load at 20°C is 5000 N. W hat will be the str ess at -20°C? Assume α=11.7 µm/m °C and E=200x109 N/m2. A. 112.8 MPa

C. 132.4 MPa

B. 117.9 MPa

D. 126.9 MPa





P A

Elongation (δ)

δ = PL / AE P = For ce L = Or iginal Length A = Cr oss Sectional Ar ea E = Young’s Modulus

T E


Elongation due to Weight (δ)

δ = gρL2 / 2E ρ = unit mass g = gr avity E = Young’s Modulus


14. A steel wir e 10 m long, hanging ver tically suppor ts a tensile load of 2000 N. Neglecting the weight of the wir e, deter mine the r equir ed diameter if the str ess is not to exceed 140 MPa and the total elongation is not to exceed 5 mm. Assume E = 200 GPa. A. 4.26 mm

C. 5.05 mm

B. 3.12 mm

D. 2.46 mm



15. A steel rod having a cr oss-sectional ar ea of 300mm2 and length of 150 m is suspended ver tically fr om one end .It suppor ts a load of 20 kN at the lower end. If the unit mass of steel is 7850 kg/m3 and E=200 GPa, find the total elongation of the rod.

16. The str aight-line por tion of the str essstr ain diagr am has slope equal to the _ _ _ _ _ _ _ _ _ _ _ _ _ of the mater ial.

A.

modulus of r igidity

A. 33.45 mm

C. 53.44 mm

B.

compr essibility

B. 54.33 mm

D. 35.44 mm

C.

modulus of elasticity

D.

shear modulus

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17. The str ess beyond which the mater ial will not retur n to its or iginal shape when unloaded is called A.

elastic limit

B.

maximum str ess

C.

ultimate str ess

D.


18. The point on the str ess-str ain diagr am at which ther e is an appr eciable elongation or yielding of the mater ial without any cor r esponding incr ease of load is called A.

yield point

B.

elastic limit

C.

ultimate str ess or ultimate str ength

D.

pr opor tional limit

allowable str ess


Torsion In solid mechanics, it is the twisting of an ob ject due to an applied torque In circular sections, the resultant shearing stress is perpendicular to the radius.


Torsion τMAX = T*r / J

c

τMAX = maximum shear str ess J = polar moment of iner tia r = outer r adius of the shaf t



Polar Moment of Inertia

Shear Stress f or: Solid Shaf t:

Solid Shaf t: J = πD4 / 32

Ss = [T (D/2) ] / (π D4/32) = 16 T / πD3 Hollow Shaf t:

Hollow Shaf t: J = π (Do4 – Di4) / 32

Ss = [T (D/2) ] / [π (D4-d4) /32] = 16 T D / π (D4-d4)

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Angle of Twist θ = TL / J G T = Tor que L = Length J = Polar Moment of Iner tia G = Shear Modulus


19. What is the minimum diameter of a solid steel shaf t that will not twist thr ough mor e than 3° in a 6-m length when sub jected to a tor que of 14 kN-m? Use G=83 GN/m2.


Power Transmitted b y Shaf t P = Tω = T(2πf ) or P = 2π T n P = power (W) T = Tor que ( Nm) f = f r equency (Hz) n = angular s peed (r ev/s)


20. Deter mine the length of the shor test 2-mm diameter br onze wir e which can be twisted thr ough two complete tur ns without exceeding a shear ing str ess of 70 MPa. Use G=35 GPa.

A. 118 mm

A. 6280 mm

B. 145 mm

B. 3420 mm

C. 122 mm

C. 1280 mm

D. 113 mm

D. 1658 mm


21. A solid steel shaf t 5 m long is str essed to 60 Mpa when twisted thr ough 4°. Using G = 83 GPa, compute the power that can be tr ansmitted by the shaf t at 20 rev/s. A. 1.21 MW B. 1.67 MW C. 3.21 MW D. 1.26 MW


Spring a device that stores potential energy by straining the bonds between the atoms of an elastic material. device made of an elastic material that undergoes a signif icant change in shape, or def ormation, under an applied load

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Uses

Helical Spring

in spring balances f or weighing f or the storage of mechanical energy, as in watch and clock springs or doorclosing springs to absorb impact, as in coil or leaf springs used f or automobile suspensions, and to reduce vibration by the use of rubber block s

a spiral wound wire with a constant coil diameter and unif orm pitch Common Forms:


Compression Spring Tension Spring


Helical Spring Max Shearing Stress A p pr o ximat e: max  

E xact :

 m ax 

16PR

d3

 1 

  4R  d

16P R  4m  1 0.615   d3  4m  4 m 

Spring Def ormation L

δ = 64PR 3n / Gd4

R = mean r adius of s pr ing (D is mean diameter ) d = diameter of wir e m = D/d


22. A helical spr ing is made by wr apping steel wir e 20 mm in diameter ar ound a f or ming cylinder 150 mm in diameter . Compute the number of tur ns requir ed to per mit an elongation of 100 mm without exceeding a shear ing str ess of 140 MPa. Use G=83 GPa.

P


23. Deter mine the maximum shear ing str ess in a helical steel spr ing composed of 20 tur ns of 20-mm diameter wir e on a mean radius of 80 mm when the spr ing is suppor ting a load of 2kN. A. 120.6 MPa

A. 15.43 tur ns

C. 18.24 tur ns

B. 13.83 tur ns

D. 12.36 tur ns



B. 117.9 MPa C. 132.4 MPa D. 126.9 MPa

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CABLES

Parabolic Cable

I. Parabolic

Tension at the Su p por t T2

=

(ωL/2)2

+

L = Hor izontal S pan d = sag of ca ble ω = weight per unit length

H2

A p pr oximate Length of Ca ble S = L + 8d2/3L – 32d4/5L3 Tension at the Lowest Point H = ωL2 / 8d



Catenar y Cable

II. Catenary The theor etical sha pe of a hanging flexi ble chain or ca ble when su p por ted at its ends and acted u pon by a unif or m gr avitational f or ce (its own weight) and in equili br ium. The cur ve has a U sha pe that is similar in a p pear ance to the par a bola, though it is a dif f er ent cur ve.


Catenar y ω = weight per unit length y = height of the su p por t (r es pective) c = minimum clear ance f r om the gr ound S = ca ble length (r es pective)

Tension at the su p por t

Re lationship among S, y & c : 2 2 2  S1    y1    c 

TL = ωyL

2

2

 S2    y 2    c 

2

Tension at the sup por t T1 & T2 : Dis tan ce Between sup por ts :

 S1  y1    c 

x1  c ln 

 S2  y 2    c 

x 2  c ln 

T1  H2   S1  T2 

2

2

 H   S2 

2


24. What tension must be applied at the ends of an aluminum cable suppor ting a load of 0.5 kg per hor izontal meter in a span of 100 m if the sag is to be limited to 1.2 5 m? A. 423.42 kg B. 329.82 kg C. 500.62 kg D. 184.29 kg

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25. Calculate the allowable spacing of the two tower s to car r y a flexible wir e cable weighing 0.03 kg per hor izontal meter if the maximum tension at the lowest point is not to exceed 1150 kg at sag of 0.50 m.

26. A cer tain cable is suspended between two suppor ts at the same elevation and 100m apar t. The load is 50N per meter hor izontal length including the weight of the cable. The sag of the cable is 6m. Calculate the total length of the cable.

A. 248 m

C. 390 m

B. 408 m

D. 422 m


27. A cable 600 m long weighing 1500 N per meter has tension of 750 kN at each end. Compute the maximum sag of the cable. A. 200 m

A. 150.68

C. 100.67

B. 100.32

D. 110.53


28. The light cable suppor ts a mass of 12 kg per meter of hor izontal length and is suspended between the two points on the same level 300 m apar t. If the sag is 60 m, find the total length of the cable.

B. 150 m

A. 329 m

C. 139 m

C. 100 m

B. 239 m

D. 429 m

D. 220 m

Strength Handout Wup

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