Determine modulus of resilience, Mechanical Engineering

Determine modulus of resilience:

A steel specimen of 10 mm diameter & 50 mm gauge length was tested in tension and following observations were recorded.

Load at upper yield point = 20600 N

 Load at lower yield point = 19650 N

Maximum load = 35550 N

Gauge length after fracture = 62.43 N

Determine modulus of resilience and modulus of toughness. Also compute % elongation. E = 210 × 103  N/mm2 .

Solution

Area of cross-section of specimen,

A0  = (π/4 ) d 2

A 0 = (π /4 )(10) 2  = 78.57 mm2

∴          Yield strength,

σ Y = 19650 /78.57 = 250 N/mm2

Ultimate tensile strength, σ u  = 35550 /78.57 = 452.5 N/mm2

            Strain at fracture or % elongation =ε f = 62.43 - 50 = 0.25 or 25%

∴          Modulus of resilience  =

2383_Determine modulus of resilience.png

=          (250)2  /2 × 210 × 103 N-mm/mm3

= 148.8 × 10- 3  N-mm/mm3

 

(i)                  (Compare with Example 1 to note that for higher yield strength modulus of resilience is larger).

Modulus of toughness =       (su  +sY /2)   ε f

                                      = (452.5 + 250 /2 )× 0.25 N-mm/mm3

                                     = 87.8 N-mm/mm3

% elongation = 25%

Posted Date: 1/23/2013 2:39:42 AM | Location : United States







Related Discussions:- Determine modulus of resilience, Assignment Help, Ask Question on Determine modulus of resilience, Get Answer, Expert's Help, Determine modulus of resilience Discussions

Write discussion on Determine modulus of resilience
Your posts are moderated
Related Questions
m=4.9Kg, K=3694 N/m, x=0.013m. I have this much details, now I want to find the damping constant ''b''. I hv tried so much, but this question needs damping ratio ''zeta'' or damped

its definition, principle, explanation, application, and a diagram

A motor cycle with rider, weight of 250 Kg., the centre of Gravity of the machine and rider combined being 60 cm above from ground level when the machine is standing upright. Each

Enthalpy: Sol: The enthalpy is total energy of gaseous system. It takes into consideration, internal energy and pressure, volume effect. Thus, it can be defined as follows:

what are the common application of statics in industry



STATE - Thermodynamics: The State of system is its condition or configuration described in detail. State is the condition of the system which is identified by thermodynamic

EXPLOSIVE WELDING I CLADDING In this process, two pieces of metal are impacted together at an extremely high velocity of impact achieved by the detonation of an explosive cha

Heat Treatment of Cast Iron: Castings in iron are frequently heat treated for enhancing mechanical properties and microstructure. The treatments known to cast iron are explai