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Determine the maximum normal stress in the blade

18/03/2021 Client: saad24vbs Deadline: 3 days

Fundamental Equations of Mechanics of Materials Axial Load

Normal Stress

s = P A

Displacement

d = L L

0 P(x)dx A (x)E

d = ! PL AE

dT = a "TL

Torsion

Shear stress in circular shaft

t = Tr

J

where

J =

p

2 c4 solid cross section

J =

p

2 (co

4 - ci 4) tubular cross section

Power

P = Tv = 2pf T

Angle of twist

f = L L

0 T(x)dx J(x)G

f = !

TL JG

Average shear stress in a thin-walled tube

tavg =

T 2tA m

Shear Flow

q = tavg t =

T 2Am

Bending

Normal stress

s =

My

I Unsymmetric bending

s = -

Mz y

Iz +

Myz

Iy , tan a =

Iz Iy

tan u

Shear

Average direct shear stress

tavg = V A

Transverse shear stress

t =

VQ It

Shear flow

q = tt =

VQ I

Stress in Thin-Walled Pressure Vessel

Cylinder

s1 =

pr

t s2 =

pr

2t

Sphere

s1 = s2 =

pr

2t

Stress Transformation Equations

sx# = sx + sy

2 +

sx - sy 2

cos 2u + txy sin 2u

tx#y# = - sx - sy

2 sin 2u + txy cos 2u

Principal Stress

tan 2up = txy

(sx - sy)>2 s1,2 =

sx + sy 2

{ A asx - sy2 b2 + txy2 Maximum in-plane shear stress

tan 2us = - (sx - sy)>2

txy

tmax = A asx - sy2 b2 + t2xy savg =

sx + sy 2

Absolute maximum shear stress

tabs max

= smax

2 for smax, smin same sign

tabs max

= smax - smin

2 for smax, smin opposite signs

Geometric Properties of Area Elements Material Property Relations

Poisson’s ratio

n = - Plat

Plong

Generalized Hooke’s Law

ex =

1 E

3sx - n(sy + sz)4

ey = 1 E

3sy - n(sx + sz)4

ez = 1 E

3sz - n(sx + sy)4

gxy = 1 G

txy, gyz = 1 G

tyz, gzx = 1 G

tzx

where

G =

E 2(1 + n)

Relations Between w, V, M

dV dx

= w(x), dM dx

= V

Elastic Curve

1 r

= M EI

EI

d4v

dx4 = w(x)

EI

d3v

dx3 = V (x)

EI

d2v

dx2 = M(x)

Buckling Critical axial load

Pcr =

p2EI

(KL)2

Critical stress

scr =

p2E

(KL >r)2 , r = 2I>A Secant formula

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