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Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

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Gear relationships between gears A and B

TA

TB

= R φ = − R φ R ω = R ω

RA

RB

RA

DA

N A

Gear ratio = = =

R D N

B

A A B B A A B B

B

Six rules for constructing shear-force and

bending-moment diagrams

Rule 1: ∆ V = P 0

2

Rule 2: ∆ V = V − V = w( x)

dx

Rule 3: dV

dx

=

2 1

wx ( )

2

Rule 4: ∆ M = M − M = V dx

Rule 5: dM

dx

= V

Rule 6: ∆ M = −M 0

Flexure

2 1

x

x1

Flexural strain and stress

1

E

ε x =− y σ x =− y

ρ

ρ

Flexure Formula

My

Mc M

σ x =− or σ max = = where S =

Iz

I S

Transformed-section method for beams of two materials

[where material (2) is transformed into an equivalent

amount of material (1)]

E2

My

My

n = σx1

=− σx2

=−n

E

I

I

1

B

x

x1

transformed

Bending due to eccentric axial load

F My

σ x = −

A I

z

Unsymmetric bending of arbitrary cross sections

⎡ Iz z − Iyz

y⎤

⎡ Iyy Iyzz

σ x = ⎢ M y

M

2 ⎥ + − + ⎤

2 ⎥ z

⎣ II y z−

I yz ⎦ ⎣ II y z−

I yz ⎦

or

( MI z y + MI y yz ) y ( MI y z+

MI z yz ) z

σ x =−

+

2 2

II y z−

I yz II y z−

I yz

MI y z+

MI z yz

tan β =

MI + MI

z y y yz

Unsymmetric bending of symmetric cross sections

Mz y My z

σ x = −

I I

y

z

tan β =

MI

y z

MI

z

y

transformed

I

c

Bending of curved bars

σ

x

M r

=−

rA r

( n − r)

( − r )

c

n

where r

n

=

A

A

dA

r

Horizontal shear stress associated with bending

VQ

τ H = where Q = Σ yiAi

It

Shear flow formula

VQ

q =

I

Shear flow, fastener spacing, and fastener shear relationship

qs ≤ nV = n τ A

f f f f f

For circular cross sections,

2

3

1

3

Q = r = d (solid sections)

3 12

Q = 2 [ 3 3

1

R − r ] = [ 3 3

D − d ] (hollow sections)

3

12

Beam deflections

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

Deflection = v

dv

Slope = = θ (forsmall deflections)

dx

M EI d 2

Moment =

v

2

dx

dM

V EI dv 3

Shear = =

3

dx dx

dV

Load w EI d 4

= =

v

4

dx dx

Plane stress transformations

t

σt

σn

τtn

θ

τnt

y

τnt

τtn

θ

σt

σn

Stresses on an arbitrary plane

2 2

σ = σ cos θ + σ sin θ + 2τ sinθcosθ

n x y xy

2 2

σ = σ sin θ + σ cos θ − 2τ sinθcosθ

t x y xy

2 2

τ =−( σ − σ )sinθcos θ + τ (cos θ − sin θ)

nt x y xy

n

x

829

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