A solid circular shaft is subjected to a bending moment of 2800 Nm and a torque of 10,000 Nm. The shaft is made of 45 C 8 steel having ultimate tensile stress of 700 MPa and ultimate shear stress of 500 MPa. Assuming a factor of safety as 5, determine,
A solid circular shaft is subjected to a bending moment of 2800 Nm and a torque of 10,000 Nm. The shaft is made of 45 C 8 steel having ultimate tensile stress of 700 MPa and ultimate shear stress of 500 MPa. Assuming a factor of safety as 5, determine,
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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Consider the case of shafts subjected to combined twisting moment and bending moment.
A solid circular shaft is subjected to a bending moment of 2800 Nm and a torque of 10,000 Nm. The shaft is made of 45 C 8 steel having ultimate tensile stress of 700 MPa and ultimate shear stress of 500 MPa.
Assuming a factor of safety as 5, determine,
- the diameter of the solid circular shaft.
- if a hollow circular shaft is to be used in place of the solid shaft, find the inside and outside diameter when the ratio of inside to outside diameter is 0.5.
![Attachment Q3
Shafts Subjected to Combined Twisting Moment and Bending Moment
When the shaft is subjected to combined twisting moment and bending moment, then the shaft
must be designed on the basis of the two moments simultaneously. Various theories have been
suggested to account for the elastic failure of the materials when they are subjected to various
types of combined stresses.
The following two theories are important from the subject point of view:
1. Maximum shear stress theory or Guest's theory. It is used for ductile materials such as mild
steel.
2. Maximum normal stress theory or Rankine's theory. It is used for brittle materials such as
cast iron.
1 = Shear stress induced due to twisting moment, and
o, = Bending stress (tensile or compressive) induced due to bending
Let
moment.
According to maximum shear stress theory, the maximum shear stress in the shaft,
Fmar = V(o,)? + 4?
Substituting the values of t and o, from Art. 14.9 and Art. 14.10, we have
1
32м
16T
16
+ 4
2
nd
nd³
nd
xd3 = JM? +
or
16
The expression JM?+T² is known as equivalent twisting moment and is denoted by T The
equivalent twisting moment may be defined as that twisting moment, which when acting alone, produces
the same shear stress (1) as the actual twisting moment. By limiting the maximum shear stress (T)
equal to the allowable shear stress (t) for the material, the equation (i) may be written as
T = JM² +7? = " xtxd
From this expression, diameter of the shaft ( d') may be evaluated.
.(i)
16
Now according to maximum normal stress theory, the maximum normal stress in the shaft,
Ozmax)
Os + V(o,)? + 4
.(ii)
2
32M
32M
= -X
2 Id
16T
+ 4
nd
1
1
+
nd
32
(M + JM² +T²)
X06 (max)
32
+ VM² +r³]
.(iv)
or](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F685d5111-7372-4863-a6da-fda0d5313222%2Ffbf69f78-3dd9-4bdd-bdcb-447aea8f4276%2Fwgym9c7_processed.png&w=3840&q=75)
Transcribed Image Text:Attachment Q3
Shafts Subjected to Combined Twisting Moment and Bending Moment
When the shaft is subjected to combined twisting moment and bending moment, then the shaft
must be designed on the basis of the two moments simultaneously. Various theories have been
suggested to account for the elastic failure of the materials when they are subjected to various
types of combined stresses.
The following two theories are important from the subject point of view:
1. Maximum shear stress theory or Guest's theory. It is used for ductile materials such as mild
steel.
2. Maximum normal stress theory or Rankine's theory. It is used for brittle materials such as
cast iron.
1 = Shear stress induced due to twisting moment, and
o, = Bending stress (tensile or compressive) induced due to bending
Let
moment.
According to maximum shear stress theory, the maximum shear stress in the shaft,
Fmar = V(o,)? + 4?
Substituting the values of t and o, from Art. 14.9 and Art. 14.10, we have
1
32м
16T
16
+ 4
2
nd
nd³
nd
xd3 = JM? +
or
16
The expression JM?+T² is known as equivalent twisting moment and is denoted by T The
equivalent twisting moment may be defined as that twisting moment, which when acting alone, produces
the same shear stress (1) as the actual twisting moment. By limiting the maximum shear stress (T)
equal to the allowable shear stress (t) for the material, the equation (i) may be written as
T = JM² +7? = " xtxd
From this expression, diameter of the shaft ( d') may be evaluated.
.(i)
16
Now according to maximum normal stress theory, the maximum normal stress in the shaft,
Ozmax)
Os + V(o,)? + 4
.(ii)
2
32M
32M
= -X
2 Id
16T
+ 4
nd
1
1
+
nd
32
(M + JM² +T²)
X06 (max)
32
+ VM² +r³]
.(iv)
or
![The expression ((M+ JM² +r²)] is known as equivalent dending moment and is denoted
by M, The equivalent bending moment may be defined as that moment which when acting alone
produces the same tensile or compressive stress (0,) as the actual bending moment. By limiting
the maximum normal stress [0mal equal to the allowable bending stress (0,), then the equation (v)
may be written as
M, = LM + VM² + T²] = ",x 0, x d'
From this expression, diameter of the shaft ( d') may be evaluated.
..-(v)
32
Notes: 1. In case of a hollow shaft, the equations (ii) and (v) may be written as
T, = M² +T° = xt (d.)' (1 – k*)
16
M, = }(M + /M² + T³ )-×o, (d,³ (1 – x*)
and
32
2. It is suggested that diameter of the shaft may be Obtained by using both the theories and the larger of the
two values is adopted.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F685d5111-7372-4863-a6da-fda0d5313222%2Ffbf69f78-3dd9-4bdd-bdcb-447aea8f4276%2Fo7apac9_processed.png&w=3840&q=75)
Transcribed Image Text:The expression ((M+ JM² +r²)] is known as equivalent dending moment and is denoted
by M, The equivalent bending moment may be defined as that moment which when acting alone
produces the same tensile or compressive stress (0,) as the actual bending moment. By limiting
the maximum normal stress [0mal equal to the allowable bending stress (0,), then the equation (v)
may be written as
M, = LM + VM² + T²] = ",x 0, x d'
From this expression, diameter of the shaft ( d') may be evaluated.
..-(v)
32
Notes: 1. In case of a hollow shaft, the equations (ii) and (v) may be written as
T, = M² +T° = xt (d.)' (1 – k*)
16
M, = }(M + /M² + T³ )-×o, (d,³ (1 – x*)
and
32
2. It is suggested that diameter of the shaft may be Obtained by using both the theories and the larger of the
two values is adopted.
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