CIV102 Formula Sheet
Meaning of Variables
d=distance
k=spring constant
m=mass(kg)
M=moment
F=Force (N)
w=load
P=point load
E=Yonge’s Modulus
N
σ = Stress (MPa= mm
2)
ϵ = Strain (Unitless)
g=gravitational constant=9.81 sm2
FOS=Factor of safety
Unit conversions
1J = 1N m = 1Kn ∗ mm
1T onne = 1000kg = 106 g
1KN = 1000N
KN
6
6 N
1M P a = 1 mm
2 = 10 P a = 10 m2
1km = 1000m = 106 mm
Basics
P
Statics:
F =0
Forces: F = m ∗ a
2
Forces in 2D: Fvec
= Fx2 + Fy2
Moment: N m = F ∗ d
Equilibrium
equations:
P
F
=
0
x
P
P Fy = 0
M = 0 Uniform load: F = w ∗ x
Moment for uniform load: M = F ∗ d where d is distance of lever arm
´B
Nonuniform load: F = A w(x) dx
Example: For a triangular uniform load, F = 12 ∗ W ∗ L
´B
Moment for non uniform load: F = d ∗ A w(x) dx
where d is distance to lever arm
Suspension Bridges
Cable hanging under its own weight (Catenary): y = a ∗ cosh( xa ) + b
−x
x
Catenary= a2 ∗ (e a + e a ) + b
Parabolic: cosh(x)
Load on top of cables (Vertical): Fyb = Fya = w∗L
2
2
Load on top of cables (Horizontal): Fyb = Fya = w∗L
8∗h
Stress and Strain
Hooke’s Law: F = k ∗ ∆L
Stress: σ = Fa = E ∗ ϵ
Where a is cross-sectional area
∆L
F
a = E ∗ Lo
E∗a
k = Lo
Energy:
W =F ∗d
E = 21 ∗ k ∗ (∆L)2
Strain energy density: Strainenergy
=
volume
J
m3
Oscillating systems
D’alambert’s principle: An accelerating system where a!=0 may be placed in
dynamic equilibrium by introducing an inertial force which equals Fi = m ∗ a and
acts opposite to the acceleration
qdirection
Natural Period: Wn =
rad
s
=
k
m
√k
√ g
∆o
Wn
m
= 2∗π
= 2∗π
Natural Frequency (Hz): Fn = 2∗π
Dampened oscillation: x = A ∗ sin(Wn ∗ t + ϕ)
Note: A and ϕ depend on initial conditions
Dampened oscillation with self weight: x = A ∗ sin(Wn ∗ t + ϕ) +
m∗g
k
Safety
To avoid failure: σapplied < σf ail
To be safe: σapplied ≤ σsaf e
σ ail
Allowable stress: σapplied ≤ FfOS
σapplied
Demand to capacity ratio: σallowable
σ
Allowable stress: σallowable = Fyield
OS
Structures
Pin: Can resist horizontal and vertical force, cannot resist moment
Roller: Can resist vertical force, cannot resist moment and horizontal force
Fixed end: Can resist all forces
Trusses
Always start with analyzing the reactions at the pin and roller
Method of joints: Analyze each joint; cannot have more than 3 unknowns
Method of sections: Cut the structure such that it cuts through 3 members
Force coming out of joint = Tension
Force going into joint = Compression
If hinges: split the truss into two sections
Truss deflections:
1) Solve the forces in each member using given forces on truss
2) Solve the forces in each member using a phantom force in required direction
P ∗L
3) Solve for ∆L for every member using ∆L = E∗A
4) For each member, multiply results from steps 2 and 3
5) Sum up all values from step 4
6) Divide sum from step 5 with phantom force to get deflection (Check units)
idk what to call this section
PN
2
Mass moment of inertia: Im =
´ 2 i=1 mi ∗ yi
Second moment of area: I = y dA
3
Second moment of area of a rectangle: I = b∗h
12
4
Second P
moment of area of a circle: I = π∗d
64
n
i=1 yi ∗Ai
ybar = P
n
A
i
i=1
Where yi is the distance from the bottom to the centroidal of shape n.
Second moment of area=I
If all local centroidal axis of a cross section line up:
Pn b ∗h3
I = i=1 i12 i
Note: Can also subtract (For example: HSS)
If local centroidal axis DO NOT match up:
Pn b ∗h3
I = i=1 i12 i + Ai ∗ d2i
Where di is the distance from the centroidal axis to the centroidal axis of Ai
Beams
dθ
Curvature: ϕ = dL
If ∆Θ is small, or ∆Θ < 5◦ , LAB = Lo
B
Curvature for large distances: ϕ = ∆AL−∆
AB
Curvature for small distances: Same formula as large distances and
ϕ = ∆a−∆b
Lab
o
Radius from where ends of beams curvature meet (R): R = ΘBL−ΘA
= ϕ1
σ = ϕ ∗ y ∗ E where y is the distance from centroid of beam to top of the beam
Moment: M = ϕ ∗ E ∗ I
Euler Buckling
2
d y
Curvature: ϕ = dx
2
Loadqon end of beam: P = E ∗ I ∗ ω 2
ω=
P
E∗I
2
2
Euler Buckling: Pe = n πL∗E∗I
where n is any positive integer
2
2
The critical case, when n=1: Pcrit = π ∗E∗I
L2
Pcrit is the force required to hold the beam statically in the buckled shape
HSS
In the form B*H*T. For example, HSS 305x305x13 is 305 high, 305 wide, 13 thick
(all measurements in mm).
For tension: FOS=2
A∗σ
Fsaf e = F OSy
F OS∗Fdemand
Area ≥
σy
There is less warning for buckling, so FOS=3
2
1
Fdemand ≤ π ∗E∗I
∗ F OS
L22
π ∗E∗I
σdemand ≤ A∗L
2 ∗F OS
q
I
Radius of gyration: r = A
L
Slenderness ratio: r
Slenderness ratio must be less than 200 in Canada
Design allowable stress=min(σsquash , σbuckle )
Check for area (Use tension and compression force): Area ≥
2
F OS∗Fdemand
σy
∗L
Check for I: I ≥ F OS∗Fπdemand
2 ∗E
L
Check for slenderness ratio: r < 200
NOTE: Remember to try to minimize self-weight
W indF orce
Wind load: Wwind = .5 ∗ ρ ∗ v 2 ∗ Cd = ExternalP
arts
kg
ρ is air density, equal to 1.2 m3
v is speed of wind in m
s
Cd is the coefficient of drag (depends on shape)
Sphere=.75
Race car=.2
Box=1.5
For trusses: Wwind =2.0 kPa
Joint force for wind (part of truss without handrails): Fwind = Area ∗ Wwind
For part of truss with handrails: Fwind = L ∗ (handrail + bottomHSS) ∗ Wwind
Dynamic Effector
Damped free vibrators:
p
x(t) = A ∗ exp(−β ∗ ωn ∗ T ) ∗ sin(ωn ∗ 1 − β 2 ∗ t + ϕ) + ∆o
β is usually less than 1, and for bridges, β might be .02
Natural frequency:
√
- For a single load on a bridge: fn = 15.76
∆o
√
- For a uniform load on a bridge: fn = 17.76
∆o
Dynamic amplification factor:
1
DAF = q
f 2 2
2∗β∗f 2
(1−( fn ) ) +(
fn
)
f=loading frequency
fn =natural frequency
β=natural constant=.05
Fmax = DAF ∗ Fo + m ∗ g
BMD and SFD
´
M (x) = V (x)dx
Moment´ is the area under the SFD
S(x) = w(x)dx
Note: To decide whether bending moments is positive or negative, a happy smile
(U) is positive (sagging), and a sad face (∩) is negative (hogging)
Shear is the accumulated applied loads along the beam
D
Curvature diagram: ϕ = BM
E∗I
m∗y
Mavier’s Equation: σ = I
Note: Choose max y (from centroid to top or bottom of the cross section)
Moment Area Theorems
MAT 1:
´B
∆θAB = A ϕ(x) = θA − θB
Note: To find any angle, I must know at least 1 other angle.
MAT 2:
δAB means the distance from the displaced beam at point A vertically down to
the tangent line taken at point B.
´B
δAB = A ϕ(x)dx ∗ x
where x is the lever arm distance taken from point A.
AB
Similar triangle ”trick”: L
LAC ∗ δCA = ∆B + δBA
Shear Stress
τxy = τyx which means on a very tiny section of a beam, the vertical shear is
equal to the horizontal shear.
∗Q
Shear Force: τ = VI∗B
Q = A ∗ d where d is the distance from the global centroid to the centroid of the
section you want to find, and A is the area of that section.
Things to note: shear stress is max at centroid, and 0 at stress free surfaces.
Tensile and compressive stresses on the other hand is 0 at the centroid and max
on the top and bottom.
Wood
Anisotropy: Different properties in different directions
Large members have different failure stresses than smaller ones
FOS=1.5
E05 for failure calculations
E50 for deflection calculations
Buckling Members
Poisson’s Ratio = µ
Different
for each material
ϵy
ϵx = −µ
ϵtotal = ϵstress + ϵpoisson
E
σy = 1−µ
2 ∗ (ϵy + µ ∗ ϵx )
E
σx = 1−µ
2 ∗ (ϵx + µ ∗ ϵy )
E
For 2D stress, the equivalent of E is 1−µ
2
Thin plate buckling:
General formula of local buckling stress:
k∗π 2 ∗E
t 2
σcr = 12∗(1−µ
2) ∗ ( b )
where t is the thickness, b is the width that is perpendicular to the
direction of applied stresses, µ is poisson’s ratio and k is a constant.
Case 1) Uniform compression, both sides restrained: k=4
4∗π 2 ∗E
t 2
σcr = 12∗(1−µ
2) ∗ ( b )
Case 2) Uniform compression, 1 side restrained: k=.425
2
t 2
∗E
σcr = .425∗π
12∗(1−µ2 ) ∗ ( b )
Case 3) Non-uniform compression, both sides restrained: k=6
6∗π 2 ∗E
t 2
σcr = 12∗(1−µ
2) ∗ ( b )
Case 4) Shear buckling:
t 2
t 2
5∗π 2 ∗E
σcr = 12∗(1−µ
2 ) ∗ (( b ) + ( h ) )
Stone
Stone is linearly inelastic
Durability is good
Heavy
Good compressive strength, bad tensile strength
When designing as an arch, design it such that there is no tensile-strength.
What is the optimal shape of an arch?
Hooke’s theory of arches states that ”as hangs a flexible cable, so, inverted, should
the touching points of an arch stand.”
If the thrust line ever exits the shape of the arch, the arch will fail.
P is always applied at thrust line location.
Remember to consider self weight.
σn =stress at base caused by axial load
γ=unit density of material
N =axial force
Vo =Original volume of tower
y=distance from the side to the center.
−Vo ∗γ
σn = −N
A =
A
For the side that’s in compression:
M ∗y
σtotal = −N
A − I
For the side that’s in tension:
M ∗y
σtotal = −N
A + I
If cross-section is constant:
σ = VoA∗γ = h ∗ γ
Concrete
σc′ = fc′ = strength
of control cylinders tested in compression
p
σt′ = ft′ = .33 σc′
p
Ec = 4500 σc′
As = Area of rebar
Av = Area of stirrups
Ac = Area of concrete = b ∗ h
Vc = Shear Force of Concrete
Vs = Shear Force of Reinforcement
d=distance from centroid of the rebar to the other side (compression zone)
ϵconcrete = ϵsteel = ϕ ∗ d ∗ (1 − k)
We want steel to carry tension and concrete to carry compression.
Max Force: Fmax = Ac ∗ σc′ + Ay ∗ σy
Ψ = distance from max compression of concrete to the tension of the steel.
Ts =Tension of steel=fs ∗ As = Es ∗ ϵs ∗ As
Cc =Compression of concrete
M = Ts ∗ Ψ = Cc ∗ Ψ
s
Modular ratio: n = E
Ec
As
Reinforcement percentage: ρ = b∗d
where
pb is the width of the section without rebar.
k = (n ∗ ρ)2 + 2 ∗ n ∗ ρ − n ∗ ρ
j = 1 − k3
M = Cc ∗ j ∗ d = Ts ∗ j ∗ d
If steel is still linear elastic:
M
ϕ = As ∗Es ∗j∗d
2 ∗(1−k)
σc = fc = Stress in concrete
σs = fs = Stress in reinforcement
k
σc = 1−k
∗ n∗AM
s ∗j∗d
M
σs = As ∗j∗d
Max moment carried by member (with stirrups):
Myield = As ∗ σy ∗ j ∗ d
jd=flexural lever arm
bw =total width of all webs
dv =effective depth=.9 ∗ d
v=shear stresses
V =shear force
fy =strength of stirrups
M = Ts ∗ j ∗ d = Cc ∗ j ∗ d
s ∗j∗d
V = ∆T∆x
v ∗ bw ∗ ∆X = ∆Ts
V
v = bw ∗j∗d
vmax = .25 ∗ σc′
Shear Strength of a Member: Vr = .5Vc + .5Vs ≤ .5Vmax
Vult = Vc + Vs = Vstrength ≤ Vmax
c
s
Vsaf e = FVOS
+ FVOS
≤ VFmax
OS
FOS = 2
Vmax = .25 ∗ fc′ ∗ bw ∗ dv
Concrete Continued
No stirrups:
√
√
230∗ fc′
230∗ fc′
Vc = 1000+dv ∗ bw ∗ dv = 1000+dv ∗ bw ∗ j ∗ d
√
√
230∗ f ′
230∗ f ′
σc = 1000+dvc = 1000+dvc
With stirrups:
p
Vc = .18 ∗ fc′ ∗ bw ∗ dv
A ∗f ∗d
Vs = v Sy v ∗ cot(35)
p
A ∗f
Minimum stirrup quantity: bvw ∗sy > .06 fc′
Shear design of reinforced concrete
1) Solve for BMD and SFD: v=max shear
2) Check Vmax equation
vmax = .25 ∗ σc′ ∗ bw ∗ dv
v ≤ Vmax
2
If false, then cross section too small.
3) Determine if shear reinforcement is required
230∗f ′
Vc = .5 1000+dcv ∗ bw ∗ dv
If v ≤ Vc , then you are done.
Else, then you need stirrups.
4) You need stirrups:
Av ∗fy
√ ′
S=
.06∗
fc ∗bw
RecalculatepVc
1
Vc = .18 ∗ fc′ ∗ bw ∗ dv ∗ F OS
A
∗f
∗d
1
Vs = F OS
∗ v Sy v ∗ cot(35)
If v ≤ Vc + Vs , you are done.
Else, solve for required spacing.
5) Solve for required spacing:
.5∗Av ∗fy ∗dv ∗cot(35)
√ ′
S=
v−.5∗.18∗
fc ∗bw ∗dv
6) Sketch resulting design
Evaluating shear strength of reinforced concrete
- No FOS
1) SFD, BMD
2) Check if it has minimum stirrups
√
p
230∗ f ′
A ∗f
If bvw ∗sy ≤ .06 ∗ fc′ , Vc = 1000+dvc ∗ bw ∗ dv
p
Else, Vc = .18 ∗ fc′ ∗ bw ∗ dv
3) If you have any shear reinforcements
A ∗f ∗d ∗cot(35)
Vs = v y Sv
4) Check Vmax
Vmax = .25 ∗ σc′ ∗ bw ∗ dv
5) Find strength
Strength=min(Vmax ,Vc + Vs )