age 2Figure Q2 (b)Given : The allowable shear stress i:kPa80Q2 c) 0 The tor sion formula :=JI (0.019)A (0.019)^4Therefore, the max torque T =Nm80x102.Q2 c) (i)For point lying on the outside suurface of the shaft:x10° = (0.019)I (0.019*-0.0125 *): Therefore, the max torque TNm80Q2 c) (iii) The shear stress at point located at inner surface of shaft,kPaT'cA (c,* -c,T'cT= -=.)(0.0125) =7 (0.019*.0.0125)

age 2 Figure Q2 (b) Given : The allowable shear stress i: c) () s0 kPa Ic The tor sion formula :r = J I (0.019) A (0.019)^4 so x10': ; Therefore, the max torque T = 2 For noint king on the outside s uface of the shaft:

age 2 Figure Q2 (b) Given : The allowable shear stress i: c) () s0 kPa Ic The tor sion formula :r = J I (0.019) A (0.019)^4 so x10': ; Therefore, the max torque T = 2 For noint king on the outside s uface of the shaft:

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Combined Loading

Any functioning component, such as machine parts, structural members, pressure vessels, and so on, are all under the influence of more than one force. When anybody is under the influence of more than one force such as shear forces, bending moments, axial forces, torsion, and so on, then such body is said to be under combined loading. A very practical example of combined loading is the crankshaft of an internal combustion engine (IC engine). The crankshaft of the IC engine is under high rpm, which is caused due to the torsional forces. Due to the presence of piston, counterbalances, and internal imbalances, the crankshaft experiences lateral forces and due to the rise in temperature, the crankshaft undergoes thermal expansion, which pulls the crankshaft along the axial direction and lateral direction. The presence of unbalanced loads along the periphery of the crankshaft also induces a bending moment.

Question

age 2
Figure Q2 (b)
Given : The allowable shear stress i:
kPa
80
Q2 c) 0 The tor sion formula :=
J
I (0.019)
A (0.019)^4
Therefore, the max torque T =
Nm
80
x10
2.
Q2 c) (i)
For point lying on the outside suurface of the shaft:
x10° = (0.019)
I (0.019*-0.0125 *)
: Therefore, the max torque T
Nm
80
Q2 c) (iii) The shear stress at point located at inner surface of shaft,
kPa
T'c
A (c,* -c,
T'c
T= -=.
)(0.0125) =
7 (0.019*.0.0125)

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