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Torsion of Circular Section 5-62 The two shafts are made of A-36 steel. Each has a diameter of 25 mm, and they are supported by bearings at A, B and C, which allow free rotation. If the support at D is fixed, Determine the angel of twist of ends B and A. G = 75 GPa (p. 211) T [Nm] L -60 Nm A F G B A G F B 60 Nm D H C E 120 Nm 1. Internal Torque: T E = 90 Nm T D = 30 Nm T [Nm] L -90 Nm D C H E 30 Nm F E F EF F’ FE F EF = F’ EF T F /T E = R F /R E H G E F

Week 6 - 1 slides

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Page 1: Week 6 - 1 slides

Torsion of Circular Section

5-62 The two shafts are made of A-36 steel. Each has a diameter of 25 mm, and they are supported by bearings at A, B and C, which allow free rotation. If the support at D is fixed, Determine the angel of twist of ends B and A. G = 75 GPa (p. 211)

T [Nm]

L

-60 Nm

A F G

B

A G F B

60 Nm

D H C E

120 Nm

1. Internal Torque:

TE = 90 Nm TD = 30 Nm

T [Nm]

L

-90 Nm

D C H

E 30 Nm

F

E

FEF

F’FE

FEF = F’EF

TF /TE = RF/RE

H

G E

F

Page 2: Week 6 - 1 slides

Torsion of Circular Section

5-62 The two shafts are made of A-36 steel. Each has a diameter of 25 mm, and they are supported by bearings at A, B and C, which allow free rotation. If the support at D is fixed, Determine the angel of twist of ends B and A. G = 75 GPa (p. 211)

T [Nm]

L

-60 Nm

A F G

B

A G F B

60 Nm

D H C E

120 Nm

1. Internal Torque:

TE = - 90 Nm TD = 30 Nm

T [Nm]

L

-90 Nm

D C H

E 30 Nm

H

G E

F

Segment DH: TDH = 30 Nm Shaft DE

Segment HE: THE = -90 Nm

Segments AG and FB: T = 0 Shaft AB

Segment GF: TGF = -60 Nm

Page 3: Week 6 - 1 slides

Torsion of Circular Section

1. Internal Torque: Segment DH: TDH = 30 Nm Shaft DE

Segment HE: THE = -90 Nm

Segments AG and FB: T = 0 Shaft AB

Segment GF: TGF = -60 Nm 2. Angle of twist:

= - 0.02086

Ans.

Ans.

5-62 The two shafts are made of A-36 steel. Each has a diameter of 25 mm, and they are supported by bearings at A, B and C, which allow free rotation. If the support at D is fixed, Determine the angel of twist of ends B and A. G = 75 GPa (p. 211)

H

G E

F

F

E

ϕE

RE

Page 4: Week 6 - 1 slides

Torsion

External Loading Types

Deformation

Strain

Shear Strain

τ = Gγ

τmax ≤ τallow

(Compatibility)

Internal Loading Diagram x

T

0

γ = π/2 – θ’

Page 5: Week 6 - 1 slides

Axial Load

External Loading Types

Deformation

Strain

Average Normal Strain

σ = Eε

σ ≤ σallow

(Compatibility)

Internal Loading Diagram

N

x

F

0

Page 6: Week 6 - 1 slides

Bending

Page 7: Week 6 - 1 slides

Bending Moments and Shear Forces

BEAM SIGN CONVENTION

Where distributed load acts downward on the beam; internal shear force causes a clockwise rotation of the beam segment on which it acts; and the internal moment causes compression in the top fibers of the segment, or to bend the segment so that it holds water.

Page 8: Week 6 - 1 slides

Bending Moments and Shear Forces

Page 9: Week 6 - 1 slides

1. Determine the ground reactions:

2. Plotting the shear and moment diagrams:

Graphical Method for Constructing Shear and Moments Diagrams

; MA = 0 ; MB = 0

A B

A B

Page 10: Week 6 - 1 slides

A B

Graphical Method for Constructing Shear and Moments Diagrams

1. Determine the ground reactions:

; MA = 0 ; MB = 0

2. Plotting the shear and moment diagrams:

A B

A B

Page 11: Week 6 - 1 slides

A B

Graphical Method for Constructing Shear and Moments Diagrams

6-19 Draw the shear and moment diagrams for the beam. (p. 295)

30 kN/m 45 kN m

1.5 m 1.5 m 1.5 m