Chapter 8

Question 1

What are the sections that can be taken from a symmetrical beam?

Answer 1

Top Fibre: AB

Middle Fibre: GH

Bottom Fibre: CD

Random fibre: EF

Question 2

What happens to the top and bottom sections after a bending moment has been applied to the beam?

Answer 2

Top fibre: AB ⇒ A’B’

Bottom fibre: CD ⇒ C’D’

Question 3

What happens to the middle sections are a bending moment for a symmetrical beam?

Answer 3

The length of the middle fibre remains the same but curves

Question 4

How is the angle of bend calculated?

Answer 4

GH=G’H’ = Rdθ

GH/R = dθ

Question 5

For the random fibre, how is the strain of this fibre calculated using the change in the fibre length?

Answer 5

εx = (E’F’ - EF)/EF

Question 6

For the random fibre, how is the strain calculated from the angle of bend and lengths R and y? What is the chain of equations?

Answer 6

εx = (E’F’ - G’H’)/EF

εx = ((R+y)dθ - Rdθ)/Rdθ

εx = ydθ/Rdθ

εx = y/R

Question 7

How is R an indication of strain?

Answer 7

εx = 1/R

Large R means a small strain

Small R means a large strain

Question 8

Whats the equation for youngs modulus law?

Answer 8

E = σ/ε

Question 9

How can be stress of a fibre be related to the lengths R and y?

Answer 9

σx/E = y/R

σx = Ey/R

Question 10

What is the stress past the neutral plane and what is the trend?

Answer 10

Tension and increases linearly with distance from the neutral plane.

σx = Ey/R

E = Constant

R = Constant

Question 11

What is the stress before the neutral plane and what is the trend?

Answer 11

Compression and decreases linearly with distance up to the neutral plane.

σx = Ey/R

E = Constant

R = Constant

Question 12

What is the method to determine the centroid position for a non-uniform object

Answer 12

1. A reference plane is determined
2. The cross section is split into sections with a regular area which can be calculated
3. Measure the distance from the reference plane to the centre of the regular areas.
4. Use the equation

Question 13

What is the equation to determine the centroid position for a non-uniform object?

Answer 13

y = ∑yiAi / ∑Ai

For sub section i:

yi = centre of subsection

Ai = Area of subsection

Question 14

Prove that the centroid of a cross-section must be the neutral axis?

Answer 14

σ = dF/dA

∑F = 0

∫ dF = 0

∫ σdA = ∫ Ey/R dA = 0

(E/R) ∫ ydA = 0

E/R ≠ 0

so ∫ ydA = 0

y = 0

Therefor the centroid of a cross-section must be the neutral axis

Question 15

How is the bending moment calculated?

Answer 15

∑M = 0

∑Fy = 0

∫ y σ dA + M = 0

M = ∫ y σ dA

M = (E/R) ∫ y^2 dA

Question 16

How is the second moment of inertia calculated?

Answer 16

I = ∫^x_x (∫^y_y y² dy)dx

e.g. Rectangle section

I = ∫^b/2_-b/2 (∫^h/2_-h/2 y² dy)dx

I = ∫^b/2_-b/2 [(1/3)y³] dx

I = ∫^b/2_-b/2 [(1/3)(h/2)³-(1/3)(-h/2)³] dx

I = ∫^b/2_-b/2 [(1/12)h³] dx

I = [(1/12)h³​] × ∫^b/2_-b/2 dx

I = [(1/12)h³​] × [x]^b/2_-b/2

I = [(1/12)h³​] × [(b/2) + (b/2)]​

I = bh³/12

Question 17

What is the second moment of area for a rectangular section?

Answer 17

Izz = (bh^3)/12

b = Base

h = Height

Question 18

What is the second moment of area for a circular sections?

Answer 18

I = (πR^4)/4

R = Radius

Question 19

What is the second moment of area for a semicircular section?

Answer 19

I = 0.110R^4

R = Radius

Question 20

What is the second moment of area for a circular tube?

Answer 20

A = 2πR[m]t

R[m] = Radius from the centre to the middle of the tube section thickness

t = Thickness

This equation is only valid for t = R/10

Question 21

What is the second moment of area for a half circular tube?

Answer 21

I = 0.095π((R[m])^3)t

Question 22

What is the second moment of area for a triangular section?

Answer 22

I = (bh^3)/36

Question 23

How is the bending moment calculated with the distance R and the youngs modulus?

Answer 23

M = EI/R

Question 24

What is the bending relationship equations?

Answer 24

M / I = σ / y = E / R

Question 25

How can stress be calculated from the second moment of inertia?

Answer 25

σ = My/I

Question 26

How can the second moment of area be calculated when it is translated away from the neutral axis but is parallel?

Answer 26

Izz = Iz’z’ + A(y[=])^2

Izz = (bh^3)/12 + bh(y[=])^2

Question 27

For a non uniform cross-section, how is the second moment of area calculated?

Answer 27

The non-uniform shape is broken up into different uniform sections and then the second moment of areas are added together.

Izz1 = Iz’z’1 + Ay[=2]^2

Izz2 = Iz’z’2+ Ay[=2]^2

I = Izz1 + Izz2