CIV102 Formula Sheets

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University of Toronto St. George
Civil Engineering

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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