Final Year Test Flashcards
1
Q
Converting distributed load to concentrated load
A
Total force = area of load (width * height) (height is the force)
Point of action: centre point of area
2
Q
Pin vs roller vs fixed support (reaction forces)
A
Pin support has x and y forces
Roller support only has y force
Fixed has x, y and moment forces
3
Q
Finding shear force and bending moments in any section
A
- Draw free body diagram
- Find reactions
- Find SF’s and BM’s-select sections before and after each applied load and reaction
- Calculate SF and BM for section using equilibrium equations
- Plot SF and BM diagrams
4
Q
Youngs modulus, stress strain
A
YM=stress/strain
Stress=Force/Area
Strain=Difference in length/length
5
Q
T/J = Gθ/L = τ/r
A
τ = shear stress (force/area) G = shear modulus L = length
6
Q
Macaulays method for working out deflection
A
Use the equation (d^2v)/(dx^2) = M/EI
- Do normal bending moment but cut off just before the end
- Get an equation for M(x)
- Sub in for M in equation above
- Intergrate twice to find V. You can just square the X brackets and divide by 2. You’ll end up with two constants
- Look back at your original diagram. Under each support V = 0. Set x to the support values to find the two constants
- Sub in x for any point to find the deflection there
7
Q
Izz formula
A
Izz = (b*h^3)/12 + b*h*(ȳ - y1)^2 b = width h = height ȳ = centroid y1 = centre of block 1
Izz = Izz1 + Izz2
8
Q
Finding shear force and bending moments when there’s a distributed load
A
Q(x): Draw into it a bit Do F(x-x1) where F is the force and x1 is the point at which it starts to act M(x): F(x-x1) * (x-x1)/2