CE 2210 Quiz Name:

CE 2210 Quiz Name: _______________________

1. Rectangle 6. Circle

2. Right Triangle 7. Hollow Circle

3. Triangle 8. Parabola

4. Trapezoid 9. Parabolic Spandrel

5. Semicircle 10. General Spandrel

A bh=

y

h

2=

I

bh

12x

3

=

x

b

2=

I

hb

12y

3

=

Ibh

3x

3

=′

Ihb

3y =′

3

A

bh

2=

y

h

3=

I

bh

36x

3

=

x

b

3=

I

hb

36y

3

=

I

bh

12x

3

=′

Ihb

12y

3

=′

A

bh

2=

y

h

3=

I

bh

36x

3

=

x

a b( )

3=

+

I

bha ab b

36( )y

2 2= − +

I

bh

12x

3

=′

Aa b h( )

2=

+

y

a b

a bh

1

3

2=

++

Ih

a ba ab b

36( )( 4 )x

32 2=

++ +

C

b

h x

x′

x

yy′

y–

–

b

h

Cx

x′

x−yy′

y−

b

h

Cx

x

y

y′

y−

−

a

x′

b

h

Cx

y

a

−

Cx

x′

y, y′

yr −

d

Cx

y

r

d

D

C

rx

y

R

Zero slope

h Cx

bx′

x

yy′

y−

−

b

h

yCx

x′

x

yy′

Zeroslope

−

−

b

h

yCx

x′

x

yy′

Zeroslope

−

−

Table A.1 Properties of Plane Figures

A

r

2

2π=

y

r4

3π=

I r

8

8

9x

4ππ

= −

I Ir

8x y

4π= =′ ′

A r

d

42

2

π π= =

I Ir d

4 64x y

4 4π π= = =

A R r D d( )4

( )2 2 2 2π π= − = −

I I R r

4( )x y

4 4π= = −

D d

64( )4 4π

= −

yh

bx

22′ = ′

A

bh2

3=

x

b3

8=

y

h3

5=

y

h

bx

22′ = ′

Abh

3=

x

b3

4=

y

h3

10=

yh

bx

nn′ = ′

Abh

n 1=

+

xn

nb

1

2=

++

yn

nh

1

4 2=

++

791

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Common Greek letters

Alpha Mu

Beta Nu

Gamma Pi

, Delta Rho

Epsilon , Sigma

Theta TauKappa Phi

Lambda Omega

α µβ νγ π

δ ρε σθ τκ φλ ω

∆Σ

Basic definitionsAverage normal stress in an axial member

F

Aavgσ =

Average direct shear stressV

AVavgτ =

Average bearing stressF

Ab

b

σ =

Average normal strain in an axial member

L

L Ld

d

t

t

h

hor or

long

lat

ε δ

ε

=∆

=

=∆ ∆ ∆

Average normal strain caused by temperature change

TTε α= ∆

Average shear strain

change in angle from2

radγ π=

Hooke’s law (one-dimensional)

Eσ ε= and Gτ γ=

Poisson’s ratiolat

long

ν εε

= −

Relationship between E, G, and ν

GE

2(1 )ν=

+

Definition of allowable stress

FSor

FSallow

failureallow

failureσ σ τ τ= =

Factor of safety

FS or FSfailure

actual

failure

actual

σσ

ττ

= =

Axial deformationDeformation in axial members

FL

AEδ = or

F L

A Ei i

i ii∑δ =

Force-temperature-deformation relationship

FL

AET Lδ α= + ∆

TorsionMaximum torsion shear stress in a circular shaft

Tc

Jmaxτ =

where the polar moment of inertia J is defined as:

π π [ ][ ]= − = −J R r D d2 32

4 4 4 4

Angle of twist in a circular shaftTL

JGφ = or

T L

J Gi i

i ii∑φ =

Power transmission in a shaft

P Tω=Power units and conversion factors

1 W1 N m

s1 hp

550 lb ft

s

6,600 lb in.

s

1 Hz1 rev

s1 rev 2 rad

1 rpm2 rad

60 s

π

π

=⋅

=⋅

=⋅

= =

=

Fundamental Mechanics of Materials Equations

AppendixE

828

BMappendixE.indd Page 828 13/10/16 9:15 PM f-389 /208/WB01929/9781119227489/bmmatter/App_E/text_s

Gear relationships between gears A and B

T

R

T

RR R R R

R

R

D

D

N

NGear ratio

A

A

B

BA A B B A A B B

A

B

A

B

A

B

φ φ ω ω= = − =

= = =

Six rules for constructing shear-force and bending-moment diagrams

Rule 1: V P0∆ =

Rule 2: V V V w x dx( )x

x

2 11

2∫∆ = − =

Rule 3: dV

dxw x( )=

Rule 4: M M M V dxx

x

2 11

2∫∆ = − =

Rule 5: dM

dxV=

Rule 6: M M0∆ = −

FlexureFlexural strain and stress

y1

xερ

= − E

yxσρ

= −

Flexure FormulaMy

Ix

z

σ = − or Mc

I

M

SS

I

cwheremaxσ = = =

Transformed-section method for beams of two materials [where material (2) is transformed into an equivalent amount of material (1)]

nE

E2

1

= My

In

My

Ix x1

transformed2

transformed

σ σ= − = −

Bending due to eccentric axial loadF

A

My

Ix

z

σ = −

Unsymmetric bending of arbitrary cross sections

I z I y

I I IM

I y I z

I I IMx

z yz

y z yzy

y yz

y z yzz2 2

σ =−−

+

− +−

or

M I M I y

I I I

M I M I z

I I IM I M I

M I M I

( ) ( )

tan

xz y y yz

y z yz

y z z yz

y z yz

y z z yz

z y y yz

2 2σ

β

= −+

−+

+−

=++

Unsymmetric bending of symmetric cross sections

M z

I

M y

Ix

y

y

z

z

σ = − M I

M Itan

y z

z y

β =

Bending of curved bars

∫σ

( )( )= −

−−

=M r r

r A r rr

AdA

r

wherexn

c nn

A

Horizontal shear stress associated with bending

τ = = ΣVQ

I tQ y AwhereH i i

Shear flow formula

qVQ

I=

Shear flow, fastener spacing, and fastener shear relationship

q s n V n Af f f f fτ≤ =

For circular cross sections,

Q r d2

3

1

123 3= = (solid sections)

[ ][ ]= − = −Q R r D d2

3

1

123 3 3 3 (hollow sections)

Beam deflectionsElastic curve relations between w, V, M, θ, and v for constant EI

vdv

dx

M EId v

dx

VdM

dxEI

d v

dx

wdV

dxEI

d v

dx

Deflection

Slope (for small deflections)

Moment

Shear

Load

2

2

3

3

4

4

θ

=

= =

=

= =

= =

Plane stress transformations

n

t

x

y

n

n

t

t

nt

tn

nt

tn

σ

σ σ

σθ

θ

ττ

ττ

Stresses on an arbitrary plane

cos sin 2 sin cosn x y xy2 2σ σ θ σ θ τ θ θ= + +

sin cos 2 sin cost x y xy2 2σ σ θ σ θ τ θ θ= + −

( )sin cos (cos sin )nt x y xy2 2τ σ σ θ θ τ θ θ= − − + −

829


or

2 2cos 2 sin 2n

x y x yxyσ

σ σ σ σθ τ θ=

++

−+

2 2cos 2 sin 2t

x y x yxyσ

σ σ σ σθ τ θ=

+−

−−

2sin 2 cos 2nt

x yxyτ

σ σθ τ θ= −

−+

Principal stress magnitudes

2 2p p

x y x yxy1, 2

22σ

σ σ σ στ=

+±

−

+

Orientation of principal planes

tan 22

pxy

x y

θτ

σ σ=

−

Maximum in-plane shear stress magnitude

2or

2x y

xyp p

max

22

max1 2τ

σ στ τ

σ σ= ±

−

+ =−

2x y

avgσσ σ

=+

tan 22

note: 45sx y

xys pθ

σ στ

θ θ= −−

= ± °

Absolute maximum shear stress magnitude

2abs max

max minτ σ σ=

−

Normal stress invariance

x y n t p p1 2σ σ σ σ σ σ+ = + = +

Plane strain transformationsStrain in arbitrary directions

cos sin sin cosn x y xy2 2ε ε θ ε θ γ θ θ= + +

sin cos sin cost x y xy2 2ε ε θ ε θ γ θ θ= + −

2( )sin cos (cos sin )nt x y xy2 2γ ε ε θ θ γ θ θ= − − + −

or

2 2cos 2

2sin 2n

x y x y xyεε ε ε ε

θγ

θ=+

+−

+

2 2cos 2

2sin 2t

x y x y xyεε ε ε ε

θγ

θ=+

−−

−

( )sin 2 cos 2nt x y xyγ ε ε θ γ θ= − − +

Principal strain magnitudes

2 2 2p p

x y x y xy1, 2

2 2

εε ε ε ε γ

=+

±−

+

Orientation of principal strains

tan 2 pxy

x y

θγ

ε ε=

−

Maximum in-plane shear strain

22 2

orx y xy

p pmax

2 2

max 1 2γε ε γ

γ ε ε= ±−

+

= −

2x y

avgεε ε

=+

Normal strain invariance

x y n t p p1 2ε ε ε ε ε ε+ = + = +

Generalized Hooke’s lawNormal stress/normal strain relationships

E

E

E

1[ ( )]

1[ ( )]

1[ ( )]

x x y z

y y x z

z z x y

ε σ ν σ σ

ε σ ν σ σ

ε σ ν σ σ

= − +

= − +

= − +

E

E

E

(1 )(1 2 )[(1 ) ( )]

(1 )(1 2 )[(1 ) ( )]

(1 )(1 2 )[(1 ) ( )]

x x y z

y y x z

z z x y

σν ν

ν ε ν ε ε

σν ν

ν ε ν ε ε

σν ν

ν ε ν ε ε

=+ −

− + +

=+ −

− + +

=+ −

− + +

Shear stress/shear strain relationships

G G G

1;

1;

1xy xy yz yz zx zxγ τ γ τ γ τ= = =

where

GE

2(1 )ν=

+

Volumetric strain or Dilatation

eV

V E

1 2( )x y z x y zε ε ε ν σ σ σ=

∆= + + =

−+ +

Bulk modulus

KE

3(1 2 )ν=

−Normal stress/normal strain relationships for plane stress

E

E

E

1( )

1( )

( )

1( )

x x y

y y x

z x y

z x y

ε σ νσ

ε σ νσ

ε ν σ σ

ε νν

ε ε

= −

= −

= − +

= −−

+

or

E

E1

( )

1( )

x x y

y y x

2

2

σν

ε νε

σν

ε νε

=−

+

=−

+

Shear stress/shear strain relationships for plane stress

GG

1orxy xy xy xyγ τ τ γ= =

830


Thin-walled pressure vesselsTangential stress and strain in spherical pressure vessel

pr

t

pd

t

pr

tE2 4 2(1 )t tσ ε ν= = = −

Longitudinal and circumferential stresses in cylindrical pressure vessels

pr

t

pd

t

pr

tE2 4 2(1 2 )long longσ ε ν= = = −

pr

t

pd

t

pr

tE2 2(2 )hoop hoopσ ε ν= = = −

Thick-walled pressure vesselsRadial stress in thick-walled cylinder

a p b p

b a

a b p p

b a r

( )

( )r

i o i o2 2

2 2

2 2

2 2 2σ =

−−

−−

−or

a p

b a

b

r

b p

b a

a

r1 1r

i o2

2 2

2

2

2

2 2

2

2σ =

−−

−−

−

Circumferential stress in thick-walled cylinder

a p b p

b a

a b p p

b a r

( )

( )i o i o

2 2

2 2

2 2

2 2 2σ =

−−

+−

−θ

ora p

b a

b

r

b p

b a

a

r1 1i o

2

2 2

2

2

2

2 2

2

2σ =

−+

−−

+

θ

Maximum shear stress

a b p p

b a r

1

2( )

( )

( )r

i omax

2 2

2 2 2τ σ σ= − =

−−θ

Longitudinal normal stress in closed cylindera p b p

b ai o

long

2 2

2 2σ =

−−

Radial displacement for internal pressure only

δ ν ν[ ]=−

− + +a p

b a rEr b

( )(1 ) (1 )r

i2

2 22 2

Radial displacement for external pressure only

δ ν ν[ ]= −−

− + +b p

b a rEr a

( )(1 ) (1 )r

o2

2 22 2

Radial displacement for external pressure on solid cylinder

p r

E

(1 )r

oδ ν= −

−

Contact pressure for interference fit connection of thick cylinder onto a thick cylinder

δ ( )( )( )=

− −−

pE c b b a

b c a2c

2 2 2 2

3 2 2

Contact pressure for interference fit connection of thick cylinder onto a solid cylinder

δ ( )=

−p

E c b

bc2c

2 2

2

Failure theoriesMises equivalent stress for plane stress

σ σ σ σ σ σ σ σ σ τ= − + = − + + 3M p p p p x x y y xy12

1 2 22 1/2 2 2 2 1/2

Column bucklingEuler buckling load

PEI

KL( )cr

2

2

π=

Euler buckling stress

E

KL r( / )cr

2

2σ π

=

Radius of gyration

rI

A2 =

Secant formula

P

A

ec

r

KL

r

P

EA1 sec

2max 2

σ = +

831


822

Sim

ply

Su

pp

ort

ed B

eam

s Bea

mSl

ope

Defl

ecti

onE

last

ic C

urve

1

x

vP

v max

1θ2θ

L 2—L 2—

PL E

I16

12

2

θθ

=−

=−

vP

L EI

48m

ax

3

=−

vP

x EI

Lx

48(3

4)

22

=−

−

xL

for

02

≤≤

2

x

v

La

2θ1θ

P

b

Pb

Lb

LEI

()

61

22

θ=

−−

Pa

La

LEI

()

62

22

θ=

+−

vP

ab

LEI

3

22

=−

xa

at=

vP

bx LEI

Lb

x6

()

22

2=

−−

−

xa

for

0≤

≤

3

x

v

L

1θ2θ

M

ML EI

31θ

=−

ML EI

62θ

=+

vM

L EI

93

max

2

=−

xL

at1

3 3=

−

vM

x

LEI

LLx

x6

(23

)2

2=

−−

+

4

x

v

v max

1θ2θ

w

L 2—L 2—

wL E

I24

12

3

θθ

=−

=−

vw

L EI

5 384

max

4

=−

vw

x EI

LLx

x24

(2

)3

23

=−

−+

5

x

v

La

1θ2θ

ww

a LEI

La

24(2

)1

22

θ=

−−

wa LE

IL

a24

(2)

2

22

2θ

=+

−

vw

a LEI

LaL

a24

(47

3)

32

2=

−−

+

xa

at=

vw

x LEI

LxaL

xa

x

aL

aL

a24

(4

2

44

)

32

22

22

34

=−

−+

+−

+

xa

for

0≤

≤

vw

a LEI

xLx

ax L

xa

L24

(26

4)

23

22

22

=−

−+ +

−

ax

Lfo

r≤

≤

6

x

v

L

2θ1θ

w0

wL EI

7 360

10

3

θ=

−

wL EI

452

03

θ=

+

vw

L

EI

0.00

652

max

04

=−

xL

at0.

5193

=

vw

x LEI

LL

xx

360

(710

3)

04

22

4=

−−

+

BMappendixC.indd Page 822 12/10/16 3:44 PM f-389 /208/WB01929/9781119227489/bmmatter/App_C/text_s

823

Can

tile

ver

Bea

ms

Bea

mSl

ope

Defl

ecti

onE

last

ic C

urve

7

L

v max

max

θ

x

vP

PL E

I2

max

2

θ=

−v

PL EI

3m

ax

3

=−

vP

x EI

Lx

6(3

)2

=−

−

8x

v

v max

max

θ

P

L 2—L 2—

PL E

I8

max

2

θ=

−v

PL E

I

5 48m

ax

3

=−

vP

x EI

Lx

xL

12(3

2)

for

02

2

=−

−≤

≤

vP

L EI

xL

Lx

L48

(6)

for

2

2

=−

−≤

≤

9

L

v max

max

θ

x

v

MM

L

EI

max

θ=

−v

ML EI

2m

ax

2

=−

vM

x EI

2

2

=−

10

L

v max

max

θ

x

vw

wL EI

6m

ax

3

θ=

−v

wL EI

8m

ax

4

=−

vw

x EI

LLx

x24

(64

)2

22

=−

−+

11

L

v max

max

θ

x

vw

0

wL E

I24

max

03

θ=

−v

wL E

I30

max

04

=−

vw

x LEI

LL

xLx

x12

0(1

010

5)

02

32

23

=−

−+

−

BMappendixC.indd Page 823 12/10/16 3:44 PM f-389 /208/WB01929/9781119227489/bmmatter/App_C/text_s

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