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# CIV102 Formula Sheets

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University of Toronto St. George

Civil Engineering

CIV102H1

M.P.Collins

Fall

Description

CIV102 formulae Height required to drop weight down rod to break rod (L.09, Sept. 27):
Δ = P faill
Suspension Bridge (L.05, Sept. 19): EA
wL 2 maximum loss of PE= 0.5* P * Δ
Hh = fail
8 PE max
extension below original position:x = weight
Young’s Modulus:
height above initial position:= x − Δ
E = σ E=Young’s Modulus, σ=stress, ε=strain
ε Max height to drop object (L.09, Sept. 27):
Horsepower (L.07): max strain energy without damage
1 HP = 746 W PEmax= area under P-Δ curve before yield plateau (triangle)
mass suspended by massless spring (L.08, Sept. 26):
PE max
1 k 15.8 x = weight
f = =
2π m Δ0 P yield
f=natural frequency of spring, k=spring constant, m=mass, h = x − Δ = x −
Δ0= initial deflection caused by mass alone (in mm) EA
Safe stress (L.09, Sept. 27):
Definitions: Resilience, Toughness:
Resilience – ability of object to absorb energy without damage safe stress 60% yield stress
Toughness – ability of object to absorb energy without breaking
Failure load of statically loading a rod:
P = Aσ
fail break
Pfaililure load, A=cross-sectional area breakreaking stress of rod Design of spherical pressure vessel (L.10, Sept. 28) Equlibrium of a 2-D object with 3 external forces (L.14, Oct.10):
d All forces must be parallel OR intersect at a point
σ = ⋅ p
4t
σ=stress, d=diameter, t=thickness, p=pressure differential Poisson’s Ratio (tut, Oct. 6):
ratio of longitudinal elongation to transverse contraction
Design of cylindrical pressure vessel (L.10, Sept. 28)
d Buckling of a Rope (L.16, Oct.12):
σ L ⋅ p M = EIΦ
4t E=Young’s Modulus, Φ=curvature=1/R
σL=longitudinal stress, d=diameter, t=thickness, p=pressure differential
d
σ C ⋅ p L 2
2t Δ mid= ⋅Φ
σC=circumferential stress, d=diameter, t=thickness, p=pressure differential 8
Δmideflection at middle, L=length of stick, Φ=curvature
A horizontal rip always occurs first, because σC= 2 σ L
Overturning Moment=P Δ mid
Moment and Inertia (L.11, Oct.3): Righting Moment = E I Φ mid
M=Fd 2
A couple (two parallel forces) have same moment relative to any point π EI
PE= 2
L
α = M
I PE=Euler load (buckling load)
m 2
α=angular acceleration, M=turning momenm, I =mass moment of inertia=2mr
h
2 3
I = y mdA = y mbdy = m bh
m ∫A ∫ 12
−h
2
2 bh 3
I = Inertia = ∫ y dA =
A 12 (for rectangular cross-section) Truss Design (L.17, Oct. 17): Virtual Work – to find deflection at a point (L.19,Oct.19):
weight of truss approx. 10% of load it resists 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.
HSS member calculation – compression (L.18, Oct.18): Table:
First find P, the load a particular member must resist. Then: P A σ=P/A ε=σ/E L Δ=εL P* P* Δ
Member Force Area Stress Strain Length Deform Dummy Work
Aσ y
from P < , find minimum value of A. Force
2
A=cross-sectional area of membey, σ =yielding stress=350 MPa (steel),
Aσ ycrushing load, 2=safety factor (tension +ve. compression –ve.) find Σ of work
π EI
from P < 2 , find minimum value of I.
3L Total Internal Work=Total External Work=1kN x Δ
I=moment, L=length of member, 3=safety factor
HSS member calculation – tension (L.18, Oct.18):
if truss is loaded at only one point, and we need to find deflection at that
First find P, the load a particular member must resist. Then: particular point:
Aσ y
from P <

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