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 Q3Shafts Subjected to Combined Twisting Moment and Bending MomentWhen the shaft is subjected to combined twisting moment and bending moment, then the shaftmust be designed on the basis of the two moments simultaneously. Various theories have beensuggested to account for the elastic failure of the materials when they are subjected to varioustypes 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 mildsteel.2. Maximum normal stress theory or Rankine's theory. It is used for brittle materials such ascast iron.1 = Shear stress induced due to twisting moment, ando, = Bending stress (tensile or compressive) induced due to bendingLetmoment.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 have132м16T16+ 42ndnd³ndxd3 = JM? +or16The expression JM?+T² is known as equivalent twisting moment and is denoted by T Theequivalent twisting moment may be defined as that twisting moment, which when acting alone, producesthe 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 asT = JM² +7? = " xtxdFrom this expression, diameter of the shaft ( d') may be evaluated..(i)16Now according to maximum normal stress theory, the maximum normal stress in the shaft,Ozmax)Os + V(o,)? + 4.(ii)232M32M= -X2 Id16T+ 4nd11+nd32(M + JM² +T²)X06 (max)32+ VM² +r³].(iv)or The expression ((M+ JM² +r²)] is known as equivalent dending moment and is denotedby M, The equivalent bending moment may be defined as that moment which when acting aloneproduces the same tensile or compressive stress (0,) as the actual bending moment. By limitingthe maximum normal stress [0mal equal to the allowable bending stress (0,), then the equation (v)may be written asM, = LM + VM² + T²] = ",x 0, x d'From this expression, diameter of the shaft ( d') may be evaluated...-(v)32Notes: 1. In case of a hollow shaft, the equations (ii) and (v) may be written asT, = M² +T° = xt (d.)' (1 – k*)16M, = }(M + /M² + T³ )-×o, (d,³ (1 – x*)and322. It is suggested that diameter of the shaft may be Obtained by using both the theories and the larger of thetwo values is adopted.

Question

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 Q3Shafts Subjected to Combined Twisting Moment and Bending MomentWhen the shaft is subjected to combined twisting moment and bending moment, then the shaftmust be designed on the basis of the two moments simultaneously. Various theories have beensuggested to account for the elastic failure of the materials when they are subjected to varioustypes 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 mildsteel.2. Maximum normal stress theory or Rankine's theory. It is used for brittle materials such ascast iron.1 = Shear stress induced due to twisting moment, ando, = Bending stress (tensile or compressive) induced due to bendingLetmoment.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 have132м16T16+ 42ndnd³ndxd3 = JM? +or16The expression JM?+T² is known as equivalent twisting moment and is denoted by T Theequivalent twisting moment may be defined as that twisting moment, which when acting alone, producesthe 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 asT = JM² +7? = &#34; xtxdFrom this expression, diameter of the shaft ( d') may be evaluated..(i)16Now according to maximum normal stress theory, the maximum normal stress in the shaft,Ozmax)Os + V(o,)? + 4.(ii)232M32M= -X2 Id16T+ 4nd11+nd32(M + JM² +T²)X06 (max)32+ VM² +r³].(iv)or The expression ((M+ JM² +r²)] is known as equivalent dending moment and is denotedby M, The equivalent bending moment may be defined as that moment which when acting aloneproduces the same tensile or compressive stress (0,) as the actual bending moment. By limitingthe maximum normal stress [0mal equal to the allowable bending stress (0,), then the equation (v)may be written asM, = LM + VM² + T²] = &#34;,x 0, x d'From this expression, diameter of the shaft ( d') may be evaluated...-(v)32Notes: 1. In case of a hollow shaft, the equations (ii) and (v) may be written asT, = M² +T° = xt (d.)' (1 – k*)16M, = }(M + /M² + T³ )-×o, (d,³ (1 – x*)and322. It is suggested that diameter of the shaft may be Obtained by using both the theories and the larger of thetwo values is adopted.

Accepted Answer

Given
Bending moment = 3000 Nm
Torque = 100000 Nm
Material 45C8
Maximum tensile stress = 700 MPa…

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,