Manufacturing Processes- Dr Simin Nasseri 3.20

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29 Νοε 2013 (πριν από 3 χρόνια και 10 μήνες)

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Manufacturing Processes
-

Dr Simin Nasseri

Groover’s Chapter 3 Problems:


3.20

-

A metal alloy has been tested in a tensile test with the following results for the flow curve
parameters: strength coefficient = 620.5 MPa and strain
-
hardening exponent = 0.26.

The same metal
is now tested in a compression test in which the starting height of the specimen = 62.5 mm and its
diameter = 25 mm. Assuming that the cross section increases uniformly, determine the load
required to compress the specimen to a height of (a
) 50 mm and (b) 37.5 mm.


Solution
: Starting volume of test specimen
V
= π
hD
o
2
/4 = 62.5π(25)
2
/4 = 30679.6 mm
3
.

(a) At
h
= 50 mm, ε = ln(62.5/50) = ln(1.25) = 0.223

Y
f
= 620.5(.223)
.26
= 420.1 MPa

A
=
V
/
L
= 30679.6/50 = 613.6 mm
2

F
= 420.1(613.6) =
257,7
70 N

(b) At
h
= 37.5 mm, ε = ln(62.5/37.5) = ln(1.667) = 0.511

Y
f
= 620.5(0.511)
.26
= 521.1 MPa

A
= V/L = 30679.6 /37.5 = 818.1 mm
2

F
= 521.1(818.1) =
426,312 N




3.23

A bend test is used for a certain hard material. If the transverse rupture strengt
h of the material
is

known to be 1000 MPa, what is the anticipated load at which the specimen is likely to fail, given
that

its dimensions are: width of cross section = 15 mm, thickness of cross section = 10 mm, and
length =

60 mm?



Solution
:
F
= (
TRS
)(
bt
2
)/1.5
L
= 1000(15 x 10
2
)/(1.5 x 60) =
16,667 N.



3.30

In a Brinell hardness test, a 1500 kg load is pressed into a specimen using a 10 mm diameter

hardened steel ball. The resulting indentation has a diameter = 3.2 mm. Determine the Brinell

hardness number for the metal.



Solution
:
HB
= 2(1500)/(10π(10
-

(10
2
-

3.2
2
)
.5
) = 3000/(10π x 0.5258) =
182 BHN