Statics and Strength of Materials Formula Sheet
12/12/94 — A. Ruina
Not given here are the conditions under which the formulae are accurate or useful.
Basic Statics
Stress, strain, and Hooke’s Law
Stress
Free Body Diagram
The FBD is a picture of any system for which you would like to apply mechanics equations and
of all the external forces and torques which act on the system.
Normal:
Strain
σ = P⊥ /A
² = δ/L0 =
Hooke’s Law
L−L0
L0
σ = E²
[ ² = σ/E + α∆T ]
Action & Reaction
²tran = −ν²long
If
A
feels force
F and couple
M from
then
B
feels force
−F
F and couple
−M
M from
(With F and −F
F acting on the same line of action.)
B.
A.
Shear:
Force and Moment Balance
τ = Pk /A
γ =
change of
formerly right angle
τ = Gγ
2G =
E
1+ν
These equations apply to every FBD in equilibrium:
z X
}|
Stress and deformation of some things
M oment Balance about pt C
F orce Balance
{
z X}|
F = 0
{
M /C = 0
All external
forces
All external
torques
The torque M /C of a force depends on the reference point C. But, for a body in
equilibrium, and for any point C, the sum of all the torques relative to point C must
add to zero ).
•
Dotting the force balance equation with a unit vector gives a scalar equation,
e.g. {
•
P
F} · i = 0
P
⇒
Geometry
P = σA
² = δ/L
Tension
•
P
Equilibrium
M /C } · λ = 0
⇒
Torsion
T =
R
net moment about axis in direction λ through C = 0.
The moment of a force is unchanged if the force is slid along its line of action.
•
For many purposes the words ‘moment’, ‘torque’, and ‘couple’ have the same meaning.
•
Two-force body. If a body in equilibrium has only two forces acting on it then the
two forces must be equal and opposite and have a common line of action.
•
Three-force body. If a body in equilibrium has only three forces acting on it then the
three forces must be coplanar and have lines of action that intersect at one point.
•
truss: A collection of weightless two-force bodies connected with hinges (2D) or ball
and socket joints (3D).
Method of joints. Draw free body diagrams of each of the joints in a truss.
•
Method of sections. Draw free body diagrams of various regions of a truss. Try to
make the FBD cuts for the sections go through only three bars with unknown forces
(2D).
•
Caution: Machine and frame components are often not two-force bodies.
•
Hydrostatics: p = ρgh,
R
Bending
and
Shear in
Beams
M = −
R
dM = V ,
dx
V =
•
R
Pressure
Vessels
A=
R
J=
R
I=
R
π(c2 − c2 )
1
2
2πct
ρ2 dA
y 2 dA
y dA = A0 ȳ 0
P
P
Pyi Ai
A0 ȳ 0
i i
τ dA
τ =
pAgas = σAsolid
−M y
I
VQ
It
σ =
pr
(sphere)
2t
σl =
pr
(cylinder)
2t
σc =
pr
(cylinder)
t
clamped-free
clamped-clamped
clamped-pinned
Lef f = 2L
Lef f = L/2
Lef f = .7L
Mohr’s Circle
Rotating the surface of interest an angle θ in physical space corresponds to a rotation of 2θ on
the Mohr’s circle in the same direction.
rectangle
σ1 +σ2
=
2
bh
π (c4 − c4 )
2
1
2
2πc3 t
π (c4 − c4 )
4
2
1
πc3 t
bh3 /12
center
center
center
σ1 −σ2
=
2
τ
=
σ−C
σx + σy
r³
2
p
2
(σx − C)2 + τxy
=
σx − σy
2
´2
2
+ τxy
2τ
σx − σy
Miscellaneous
Ai
P
σ =
Lef f = L
tan 2θ =
(Ii + d2 Ai )
i
d2 u = 1 = κ
ρ
dx2
pinned-pinned
C =
Ai
u00 =
2
Critical buckling load = Pcrit = π EI .
L2
ef f
R =
dA
Q=
P
dA
R
y dA
R
ȳ =
Composite
thin-wall
annulus
(approx)
dV = −w
dx
Buckling
p dA
annulus
(circle: c1 = 0)
u00 = M
EI
² = −y/ρ = −yκ
yσ dA
Tρ
J
τ t∆x = ∆M Q/I
Cross Section Geometry
R
φ = TL
JG
γ = ρφ/L
τ =
•
Definition
ρτ dA
Fx = 0.
Some Statics Facts and Definitions
F =
δ = PL
AE
[δ = P L + αL∆T ]
AE
Dotting the moment balance equation with a unit vector gives a scalar equation,
e.g. {
Results
2
b( h −y 2 )
4
2
•
Power in a shaft:
•
Saint Venant’s Principle: Far from the region of loading, the stresses in a structure
would only change slightly if a load system were replaced with any other load system
having the same net force and moment.
P = T ω.