CIV102 formulae
Height required to drop weight down rod to break rod (L.09, Sept. 27):
Suspension Bridge (L.05, Sept. 19):
∆=
wL 2 Hh = 8
maximum loss of PE= 0.5* Pfail * Δ
Young’s Modulus: σ E= E=Young’s Modulus, σ=stress, ε=strain ε
height above initial position: h = x − ∆
Horsepower (L.07): 1 HP = 746 W
max strain energy without damage PEmax = area under P-Δ curve before yield plateau (triangle)
mass suspended by massless spring (L.08, Sept. 26): 1 f= 2π
k 15 .8 = m ∆0
f=natural frequency of spring, k=spring constant, m=mass, Δ0= initial deflection caused by mass alone (in mm)
Pfail ⋅ l EA
extension below original position: x =
Max height to drop object (L.09, Sept. 27):
x=
PE max weight
h = x −∆ = x −
Pyield ⋅ l EA
Safe stress (L.09, Sept. 27): Definitions: Resilience, Toughness: Resilience – ability of object to absorb energy without damage Toughness – ability of object to absorb energy without breaking Failure load of statically loading a rod: Pfail = Aσbreak
Pfail=failure load, A=cross-sectional area of rod, σbreak=breaking stress of rod
PE max weight
safe stress ≈ 60% yield stress
Design of spherical pressure vessel (L.10, Sept. 28) σ=
d ⋅p 4t
σ=stress, d=diameter, t=thickness, p=pressure differential
Design of cylindrical pressure vessel (L.10, Sept. 28) σL =
d ⋅p 4t
σL =longitudinal stress, d=diameter, t=thickness, p=pressure differential
σC =
d ⋅p 2t
σC =circumferential stress, d=diameter, t=thickness, p=pressure differential
A horizontal rip always occurs first, because σC = 2 σL Moment and Inertia (L.11, Oct.3): M=Fd A couple (two parallel forces) have same moment relative to any point α=
M Im
α=angular acceleration, M=turning moment, Im=mass moment of inertia=2mr2
2
I m = ∫ y mdA = A
h 2
2 ∫ y mbdy = m
−
h 2
I = Inertia = ∫ y 2 dA = A
bh 3 12
bh 3 12 (for rectangular cross-section)
Equlibrium of a 2-D object with 3 external forces (L.14, Oct.10): All forces must be parallel OR intersect at a point Poisson’s Ratio (tut, Oct. 6): ratio of longitudinal elongation to transverse contraction Buckling of a Rope (L.16, Oct.12): M = EI Φ E=Young’s Modulus, Φ=curvature=1/R
∆ mid =
L2 ⋅Φ 8
Δmid=deflection at middle, L=length of stick, Φ=curvature
Overturning Moment=P Δmid Righting Moment = E I Φmid PE =
π 2 EI L2
PE=Euler load (buckling load)
Truss Design (L.17, Oct. 17): weight of truss approx. 10% of load it resists HSS member calculation – compression (L.18, Oct.18): First find P, the load a particular member must resist. Then: Aσ y from P < , find minimum value of A.
Virtual Work – to find deflection at a point (L.19,Oct.19): First, solve the truss by finding the forces on each member. Then, solve the truss with only a 1kN force at the point we want to find deflection. Table: Member
P Force
A Area
σ=P/A Stress
ε=σ/E Strain
L Length
Δ=εL Deform
P* Dummy Force
P* Δ Work
2
A=cross-sectional area of member, σy=yielding stress=350 MPa (steel), Aσy=crushing load, 2=safety factor
π 2 EI from P < , find minimum value of I. 3L2 I=moment, L=length of member, 3=safety factor
HSS member calculation – tension (L.18, Oct.18): First find P, the load a particular member must resist. Then: Aσ y from P < , find minimum value of A. 2
(tension +ve. compression –ve.) Total Internal Work=Total External Work=1kN x Δ
if truss is loaded at only one point, and we need to find deflection at that particular point:
A=cross-sectional area of member, σy=yielding stress=350 MPa (steel), Aσy=crushing load, 2=safety factor
internal work for one member= P ′∆ = P ′
L L from r > , find minimum value of r.( < 200 ) 200 r
let P ′ = P (rather than P ′ =1 kN). Then:
Use HSS table: find appropriate values of A and r. if r not supplied (such as for a cylindrical rod): r=
I A
r=radius of gyration, I=moment, A=area
find Σ of work
Pi 2 Li total internal work= ∑ Ai E i total external work = P ′∆ = P∆
PL AE
( P ′ = dummy force) work=
P2L AE
Frequency (L.20, Oct. 24):
Bending of Beams (L.23, Oct. 31):
Wtotal ∆ = max Wself ∆0
ε = Φy
For a point load: f =
M = E ΦI
∆0
Φ=
M EI
Flexural Stress: σ=
15 .8
or:
My (Navier’s Equation) I
f=natural frequency of bridge, Δ0=deflection of bridge under point load
Finding I of a cross-section (L.24, Nov.1) table method
For a distributed load [kN/m]:
Uniformly distributed load (L.25, Nov.2): equation for moment with only Uniformly Distributed Load on beam:
f ≈
18 ∆0
f=natural frequency of bridge, Δ0=deflection of bridge from self-weight at midspan Bracing for buckling of truss bridge (L.22, Oct. 26): R ≥ 0.02 P
R=force to prevent buckling, P=force causing buckling thin steel rods: Aσ y from P < , find minimum value of A. σy =350 MPa (steel). 2
HSS members: π 2 EI from P < , find minimum value of I. 3L2
M max =
wL 2 w in [kN/m], L=length 8
Moment-Area Theorems (L.26, Nov.7): First Moment-Area Theorem: (for finding slope) The change of angle between any two points on a beam is equal to the area under the Φ (curvature) diagram between these two points. Second Moment-Area Theorem: (for finding deflection) For any two points D and T along the length of a deflected beam, the deviation of point D from the tangent drawn at point T equals the area under the Φ (curvature) diagram between D and T multiplied by the distance from the centroid of the diagram to point D. D Alternate saying: δDT = ∫T Φxdx
UDL on Beam Fixed to Wall (L.28, Nov.9): wL 8 EI
end deflection: ∆ =
σ critical =
6π 2 E t 12 b
2
shear stress:
Shear Stress (L.29, Nov.14): V ⋅Q τ= I ⋅b
side view:
4
(Jourawski’s Equation)
t 2 t 2 + a h a = distance between stiffeners. If no stiffeners, second term disappears.
τ critical =
5π 2 E 12
V = shear force, I = second moment of area, Q = first moment of area = area times distance from its centroid to centroidal axis b = length for which we are finding shear stress of, length of section that would fail
Thrust Line for an Arch (L.36, Nov.29):
for a rectangular cross-section:
e = eccentricity = distance between thrust line and central axis
3V τ= 2bh
M=Pe
Box-Girder Bridges (L.32, Nov.): note: need factor of safety of 3 for critical stresses Two sides held in place: σ critical =
π 2E t
2
12 L
Four sides held in place: σ critical = σ critical
4π 2 E t 12 L
4π 2 E t = 12 b
2
(L
2
(L>b)
Three sides held in place: σ critical
0.45π 2 E t = 12 b
2
(L>2b)
L=length, b=width, t=thickness
Box-Girder Bridge Designs – cont’d (L.33, Nov.22):
max eccentricity is 1/6 of thickness