Final

Question 1

Steps for support reactions (2)

Answer 1

1) if any distributed load calculate the point load

2) use 3 equilibrium equations to solve supports

Question 2

Steps for method of joints (2)

Answer 2

1) solve support reactions

2) solve a joint that has support reactions, using fx and fy, should only have two unknowns

Negative answer is is compression

Postitive answer is tension

Question 3

Steps for method of sections (5)

Answer 3

1) highlight members wanted

2) make a cut through all members wanted

3) determine support reactions

4) pick side of cut you want to analyze and draw fbd

• draw all members not included in cut (include support reactions and any given forces)

-add tension forces to replace each member that was cut

5) solve members using 3 equilibrium equations

-take moment at point that has two unkown forces

-if stuck u can’t take moment at a point that isn’t on fbd

Question 4

Steps for internal forces (5)

Answer 4

1) determined ne point load from distributed load (area)

2) find support reactions

3) cut beam and draw new fbd with internal forces and all other forces

4) use similar slopes to find new area of distributed load (if triangle)

5) use 3 equilibrium equations to solve 3 unknowns

Question 5

Shear force diagram steps (5)

Answer 5

1) find support reactions (have to find point load first if given distributed load)

2) draw outline of graph

3) jump up if there’s a support reaction and down if point load

4) change in V= (-) area of distributed load

• if needing to draw a parabola: slope is negative of the value at that point

5) point loads/support reactions increase/decrease graph along that line

-moments don’t affect V

-no distributed loads/point loads = constant V

Question 6

Bending moment diagram steps (linear shear) (2)

Answer 6

1) jump up for clockwise moment and down for counterclockwise

2) change in M = area of shear force

-change in Vab = Vb - Va

• change in Vab = (-) area N/m • length solve for length
• change in Mab = (height from V diagram)(length) or half of that for triangle
• connect two points with parabola

-slope: equal sign to the value of the shear at the same point

Question 7

Bending moment diagram steps (parabolic shear) (2)

Answer 7

1) jump up for clockwise moment and down for counterclockwise

2) slope = (y2-y1)/(x2-x1) Numbers are referring to co ordinates of distributed load

• y-int = when distributed load touches y axis
• w(x) = mx + b just plug in numbers here
• V(x) = integral (-w(x) dx)
• M(x) = integral (V(x) dx)
• evaluate M(x value at that point)
• slope: equal sign to the value of the shear at that point

Question 8

Steps for centroids (continuous) (4)

Answer 8

1) take slice that doesn’t have weird gap

2) find dA = (length)(thickness of slice)

-put length in terms of thickness variable

3) find x squiggle: distance from y axis to centre of slice

4) limits of integration are limits of variable in differential element

Question 9

Steps for centroids discrete (1)

Answer 9

Make table

Question 10

Steps for centre of gravity (5)

Answer 10

1) determine which are 0

2) draw element

3) find x squiggle (distance from axis to centroid of element)

4) dV = volume of element = area * thickness

5) use limits of x bar to integrate

Question 11

Steps for moments of intertia continuous (3)

Answer 11

1) if Ix take hor slice if Iy take vert slice

2) find dA = length(thickness)

3) use limits of differentiable element to integrate