22 MECHANICAL ENGINEERING PRINCIPLESA10203040CBExtension /mmLoad /kN0 1.0 2.0 3.0 4.0 5.0PFigure 2.5(b) Gradient of straight line portion of graph is
given by:
BC
AC=25000 N
0. 35 × 10 −^3 m= 71. 43 × 106 N/mYoung’s modulus of elasticity=(gradient of graph)(
L
A)=( 71. 43 × 106 )(
120 × 10 −^3
100 × 10 −^6)= 85. 72 × 109 Pa= 85 .72 GPa(c) Ultimate tensile strength=maximum load
original cross-sectional area=38. 5 × 103 N
100 × 10 −^6 m^2= 385 × 106 Pa=385 MPa(d) Percentage elongation
=extension at fracture point
original length× 100=5 .0mm
120 mm× 100 = 4 .17%(e) Strain ε =extensionx
original lengthlfrom which,extensionx=εl= 0. 01 × 120= 1 .20 mm.From the graph, the load corresponding to an
extensionof1.20mmis36kN.Stress at a strain of 0.01 is given by:σ=force
area=36000 N
100 × 10 −^6 m^2= 360 × 106 Pa=360 MPa(f) When the stress is 200 MPa, thenforce=area×stress=( 100 × 10 −^6 )( 200 × 106 )=20 kNFrom the graph, the corresponding extension is
0.30 mm.Problem 5. A mild steel specimen of
cross-sectional area 250 mm^2 and gauge
length 100 mm is subjected to a tensile test
and the following data is obtained:
within the limit of proportionality, a load of
75 kN produced an extension of 0.143 mm,
load at yield point = 80 kN, maximum load
on specimen = 120 kN, final cross-sectional
area of waist at fracture = 90 mm^2 ,andthe
gauge length had increased to 135 mm at
fracture.
Determine for the specimen: (a) Young’s
modulus of elasticity, (b) the yield stress,
(c) the tensile strength, (d) the percentage
elongation, and (e) the percentage reduction
in area.(a) ForceF =75 kN=75000 N, gauge length
L=100 mm= 0 .1 m, cross-sectional area
A=250 mm^2 = 250 × 10 −^6 m^2 , and extension
x= 0 .143 mm= 0. 143 × 10 −^3 m.Young’s modulus of elasticity,E=stress
strain=F/A
x/L=FL
Ax=( 75000 )( 0. 1 )
( 250 × 10 −^6 )( 0. 143 × 10 −^3 )= 210 × 109 Pa=210 GPa