Question

The shaft is made of A992 steel. It has a diameter of 1 in. and is supported by bearings at A and D, which allows free rotation. For A992 steel, G = 11 × 103 ksi. (1) Determine the angle of twist of B with respect to D.(2) Determine the angle of twist of C with respect to D.Answer unit: degree or radians, two decimal places

Answer 1

the angle of twist of B with respect to D is -1.15°

the angle of twist of C with respect to D is 1.15°

Step-by-step explanation:

The missing diagram that is supposed to be added to this image is attached in the file below.

From the given information:

The shaft is made of A992 steel. It has a diameter of 1 in and is supported by bearing at A and D.

For the Modulus of Rigidity G = 11 × 10³ Ksi = 11 × 10⁶ lb/in²

The objective are :

1) To determine the angle of twist of B with respect to D

Considering the Polar moment of Inertia at the shaft
`J\tau`

shaft
`J\tau` =
`(\pi)/(2)r^4`

where ;

r = 1 in /2

r = 0.5 in

shaft
`J \tau` =
`(\pi)/(2) * 0.5^4`

shaft
`J\tau` = 0.098218

Now; the angle of twist at B with respect to D is calculated by using the expression

`\phi_(B/D) = \sum (TL)/(JG)`

`\phi_(B/D) = (T_(CD)L_(CD))/(JG)+(T_(BC)L_(BC))/(JG)`

where;

`T_(CD) \ \ and \ \ L_(CD)` are the torques at segments CD and length at segments CD

`{T_(BC) \ \ and \ \ L_(BC)}` are the torques at segments BC and length at segments BC

Also ; from the diagram; the following values where obtained:

`L_(BC)}` = 2.5 in

`J\tau` = 0.098218

G = 11 × 10⁶ lb/in²

`T_{BC` = -60 lb.ft

`T_{CD` = 0 lb.ft

`L_{CD` = 5.5 in

`\phi_(B/D) = 0+ ([(-60 * 12 )] (2.5 * 12 ))/( (0.9818)(11 * 10^6))`

`\phi_(B/D) = ([(-720 )] (30 ))/(1079980)`

`\phi_(B/D) = (-21600)/(1079980)`

`\phi_(B/D) =` − 0.02 rad

To degree; we have

`\phi_(B/D) = -0.02 * (180)/(\pi)`

`\mathbf{\phi_(B/D) = -1.15^0}`

Since we have a negative sign; that typically illustrates that the angle of twist is in an anti- clockwise direction

Thus; the angle of twist of B with respect to D is 1.15°

(2) Determine the angle of twist of C with respect to D.Answer unit: degree or radians, two decimal places

For the angle of twist of C with respect to D; we have:

`\phi_(C/D) = (T_(CD)L_(CD))/(JG)+(T_(BC)L_(BC))/(JG)`

`\phi_(C/D) = 0+(T_(BC)L_(BC))/(JG)`

`\phi_(B/D) = 0+ ([(60 * 12 )] (2.5 * 12 ))/( (0.9818)(11 * 10^6))`

`\phi_(C/D) = (21600)/(1079980)`

`\phi_(C/D) =` 0.02 rad

To degree; we have

`\phi_(C/D) = 0.02 * (180)/(\pi)`

`\mathbf{\phi_(C/D) = 1.15^0}`