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grin007 [14]
3 years ago
8

B1) 20 pts. The thickness of each of the two sheets to be resistance spot welded is 3.5 mm. It is desired to form a weld nugget

that is 5.5 mm in diameter and 5.0 mm thick after 0.3 sec welding time. The unit melting energy for a certain sheet metal is 9.5 J/mm3 . The electrical resistance between the surfaces is 140 micro ohms, and only one third of the electrical energy generated will be used to form the weld nugget (the rest being dissipated), determine the minimum current level required.
Engineering
1 answer:
kap26 [50]3 years ago
4 0

Answer:

minimum current level required =  8975.95 amperes

Explanation:

Given data:

diameter = 5.5 mm

length = 5.0 mm

T = 0.3

unit melting energy = 9.5 j/mm^3

electrical resistance = 140 micro ohms

thickness of each of the two sheets = 3.5mm

Determine the minimum current level required

first we calculate the volume of the weld nugget

v = \frac{\pi }{4} * D^2 * l = \frac{\pi }{4} * 5.5^2 * 5 = 118.73 mm^3

next calculate the required melting energy

= volume of weld nugget * unit melting energy

= 118.73 * 9.5 = 1127.94 joules

next find the actual required electric energy

= required melting energy / efficiency

= 1127 .94 / ( 1/3 )  = 3383.84 J

TO DETERMINE THE CURRENT LEVEL REQUIRED  use the relation below

electrical energy =  I^2 * R * T

3383.84 / R*T = I^2

3383.84 / (( 140 * 10^-6 ) * 0.3 ) = I^2

therefore  8975.95 = I ( current )

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A silicon carbide plate fractured in bending when a blunt load was applied to the plate center. The distance between the fractur
anyanavicka [17]

Answer:

hello your question has some missing values below are the missing values

Mirror Radius (mm) Bending Failure Stress (MPa)

.603                                         225

.203                                         368

.162                                          442

answer : 191 mPa

Explanation:

<u>Determine the stress present at the time of fracture for the original plate</u>

Bending stress ∝ 1 / ( mirror radius )^n ------ ( 1 )

at 0.603  bending stress = 225

at 0.203  bending stress = 368

at 0.162  bending stress = 442

<u>applying equation 1   determine the value of n for several combinations</u>

 ( 225 / 368 ) = ( 0.203 / 0.603 )^n

hence : n = 0.452

also

 ( 368/442 ) = ( 0.162 / 0.203 ) ^n

hence : n = 0.821

also

( 225 / 442 ) = ( 0.162 / 0.603 ) ^n

hence : n = 0.514

Next determine the average value of n

n ( mean value ) =  ( 0.452 + 0.821 + 0.514 ) / 3 = 0.596

Calculate estimated stress present at the time of fracture for the original plate

= bending stress at x =  0.796 / bending stress at x = 0.603

= x / 225 = ( 0.603 / 0.796 ) ^ 0.596

therefore X ( stress present at the time of fracture of original plate )

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     <em>=  191 mPa </em>

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Answer:

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