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Unformatted text preview: ©Prep101 Civ 100 Equation Sheet
Equilibrium
Particle
Σ Fx = 0, Σ Fy = 0, Σ Fz = 0
Rigid BodyTwo Dimensions
Σ Fx = 0, Σ Fy = 0, Σ M0 = 0
Rigid BodyThree 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(x2)
M = 10025xNm 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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 Spring '10

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