Civ100 ES edited - ©Prep101 Civ 100 Equation Sheet...

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Unformatted text preview: ©Prep101 Civ 100 Equation Sheet Equilibrium Particle Σ Fx = 0, Σ Fy = 0, Σ Fz = 0 Rigid Body-Two Dimensions Σ Fx = 0, Σ Fy = 0, Σ M0 = 0 Rigid Body-Three Dimensions Σ Fx = 0, Σ Fy = 0, Σ Fz = 0 Σ Mx = 0, Σ My = 0, Σ Mz = 0 Moments Moment of a Force Mo = Fd i j k M O = r × F = rx ry rz Fx Fy Fz Moment of a Force About a Specified Axis ux uy uz M a = u • r × F = rx ry rz Fx Fy Fz Simplification of a Force and Couple System FR = ΣF (MR)O = ΣM O Lots more free study aids at http://www.prep101.com/freestuff ©Prep101 Distributed Loading Reduction of Distributed Loading wL w = L/2 L wL/2 w = 2L/3 L w2 w1 w1L (w2 – w1)(L/2) = L/2 L/3 L Lots more free study aids at http://www.prep101.com/freestuff ©Prep101 Structural Analysis Method of Joints Determine the external reaction forces using ΣFx = 0 and ΣFy = 0 for the entire structure as a whole. Draw the free body diagram of a joint with at least one known value and no more than two unknowns. Resolve all forces into x and y components. A member in compression pushes on the joint, and a member in tension pulls on the joint. Apply ΣFx = 0 and ΣFy = 0 to solve for unknown forces. Continue for remaining joints. 500N α 1 500N α 2 F1 ΣFx F2 =0 = 500 - F2sinα F2 = 500 / sinα ΣFy =0 = F2 cosα - F1 F1 = F2 cosα F1 = 500 cosα / sinα Lots more free study aids at http://www.prep101.com/freestuff ©Prep101 Structural Analysis Method of Sections Determine the external reaction forces using ΣFx = 0 and ΣFy = 0 for the entire structure as a whole. Cut the structure through no more than three members. Draw the free body diagram of the remaining structure with cut sections labeled with unknown forces. Resolve all forces into x and y components. A member in compression pushes on the joint, and a member in tension pulls on the joint. Apply ΣFx = 0, ΣFy = 0 and ΣM = 0 to solve for unknown forces. 1 α F1 2 α F2 F3 2m A 3 1000N ΣFy 2m 1000N =0 = -1000 + F2cosα F2 = 1000 / cosα ΣMA =0 = 1000(2) - F1 (2) F1 = 1000 ΣFx =0 = F1 - F3 + F2sinα F3 = 1000+ 1000 sinα/ cosα Lots more free study aids at http://www.prep101.com/freestuff ©Prep101 Shear and Bending Moment Diagrams Shear and Bending Moment Diagrams Determine the external reactions. Resolve all forces into x and y components. Specify separate x coordinates from the beam’s left end, to sections of the beam between changes in loading. Section the beam at each x coordinate and draw a free body diagram. Apply ΣFy = 0 and ΣM = 0 to solve for internal shear and moment. Plot the internal shear and moments on a diagram labeling important values. 50N 2m 2m 50 Reactions at the supports are 25N at each end. M M x x V V 25 25 Using ΣFy = 0 gives V = 25 N Using ΣM = 0 gives M = 25x Nm Using ΣFy = 0 gives V = -25 N Using ΣM = 0 gives M = 25x – 50(x-2) M = 100-25xNm V= 25 V= -25 M= 50 Lots more free study aids at http://www.prep101.com/freestuff ©Prep101 Fluid Pressure p = ρgz where p = fluid pressure r = mass density g = acceleration due to gravity z = distance below the fluid surface Typical distribution The direction of the pressure is always normal to the surface of the object. Lots more free study aids at http://www.prep101.com/freestuff ©Prep101 Design Equations Stress and Strain Axial (σ) or shear stress (τ) = F / A Axial strain ε = Δ / L Young’s Modulus E=σ/ε F/A=EΔ/L Elongation of an axially elongated member Δ = PL/EA The bending stress in a member σ = Mxy/Ix Axial stress without yielding σ=P/A Buckling P = π2 EI / L2 Lots more free study aids at http://www.prep101.com/freestuff ...
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