Question

A cylindrical shaft made of steel of yield strength 700 mpa is subjected to static loads consisting of bending moment 10 kn-m and a torsional moment 30 kn-m. Determine the diameter of the shaft using: 1. Maximum shear stress theory, 2. Maximum strain energy theory, and assuming a factor of safety of 2. Take e = 210 gpa and poisson's ratio = 0.25.

Answer 1

Answer:

cylindrical shaft made of steel of yield strength 700 MPa is subjected to static loads consisting of a bending moment of 10 kN-m and a torsional moment of 30 kN-m. Determine the diameter of the shaft using two different theories of failure and assuming a factor of safety of 2. [Ans. 100 mm 6. Fig. 14.17 shows a shaft from a hand-operated machine. The frictional torque in the journal bearings at A and B is 15 N-m each. Find the diameter ( d) of the shaft (on which the pulley is mounted) using maximum distortion energy criterion. The shaft material is 40 C 8 steel for which the yield stress in tension is 380 MPa and the factor of safety is 1.5.cylindrical shaft made of steel of yield strength 700 MPa is subjected to static loads consisting of bending moment 10 kN-m and a torsional moment 30 kN-m. Determine the diameter of the shaft using two different theories of failure, and assuming a factor of safety of 2. Take E = 210 GPa and poisson's ratio = 0.25.FROM OBSERVATION, THE ANGLE OF TWIST OF THE SHAFT IS PROPORTIONAL TO THE APPLIED TORQUE AND TO THE SHAFT LENGTH. WHEN SUBJECTED TO TORSION, EVERY CROSS-SECTION OF A CIRCULAR SHAFT REMAINS PLANE AND UNDISTORTED. ... CROSS-SECTIONS OF NONCIRCULAR (NON- AXISYMMETRIC) SHAFTS ARE DISTORTED WHEN SUBJECTED TO TORSION.

Explanation:

Shaft Made Of Mild Steel Is Required To Transmit 100 KW At 300 R.p.m. The Supported Length Of The Shaft Is 3 Metres. It Carries Two Pulleys Each Weighing 1500 N Supported At A Distance Of 1 Metre From The Ends Respectively. Assuming The Safe Value Of Stress, Determine The Diameter Of The Shaft.metal bar of 10mm dia when subjected to a pull of 23.55KN gave and elongation of 0.3mm on a gauge length of 200mm. In a torsion test maximum shear stress of 40.71N/mm2 was measured on a bar of 50mm dia. The angle of twist measured over a length of 300mm being 0°21’. Deter poisson’s.

 

Solution:                                                                       

Given data:                                                                   

diad            =                 10mm                  

pull P                   =                 23.55KN             

elongation SL =    0.3mm                                    

Gauge length        =       200mm                

Torsion Test                                                                           

(or)             =                 40.71 N/mm2                

Dia              =                 500mm                

Twist angle           =                 0o21’          

length                   =                 300 mm               

Asked                                                        

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