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CIV102H1 (4)
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# CIV102 Formula Sheets

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School
University of Toronto St. George
Department
Civil Engineering
Course
CIV102H1
Professor
M.P.Collins
Semester
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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