Assignment #3 Metal Casting IE 337 Manufacturing Material - Classes

Therefore, D = H = (1.273) 0.333
= 1.084 dm
Cylinder area A = 2πD 2
/4 + πDL = 2π(1.084) 2
/4 + π(1.084)(1.084) = 5.536 dm 2
V/A = 1.0/5.536 = 0.1806 dm
Chvorinov’s Rule T = (0.1806)
TS 2
C = 0.0326C
m m
(3) Cube: V = L 3
=1.0 dm 3
. L = 1.0 dm
Cube area = 6L 2
= 6(1) 2
= 6.0 dm 2
V/A = 1.0/6.0 = 0.1667 dm
Chvorinov’s Rule T = (0.1667)
TS 2
C = 0.02778C
m m
(b) Sphere would be the best riser, since V/A ratio is greatest.
(c) Given that C m = 3.5 min/cm 2
= 350 min/dm 3
Sphere: T = 0.0428(350) = 14.98 min
TS
Cylinder: T = 0.0326(350) = 11.41 min
TS
Cube: T = 0.02778(350) = 9.72 min
TS
10.24 A cylindrical riser is to be designed for a sand casting mold. The length of the cylinder is to be 1.25
times its diameter. The casting is a square plate, each side = 10 in and thickness = 0.75 inch. If the metal
is cast iron, and the mold constant = 16.0 min/in 2
in Chvorinov's Rule, determine the dimensions of the
riser so that it will take 30% longer for the riser to solidify.
Solution: Casting volume V = tL 2
= 0.75(10.0) 2
= 75 in 3
Casting area A = 2L 2
+ 4Lt = 2(10.0) 2
+ 4(10.0)(0.75) = 230.0 in 2
V/A = 75/230 = 0.3261
Casting T = 16(0.3261)
TS 2
= 1.70 min
Riser T = 1.30(1.70) = 2.21 min
TS
Riser volume V = πD 2
H/4 = 0.25πD 2
(1.25D) = 0.3125πD 3
Riser area A = 2πD 2
/4 + πDH = 0.5πD 2
+ 1.25πD 2
= 1.75πD 2
V/A = 0.3125πD 3
/1.75πD 2
= 0.1786D
Riser T = 16.0(0.1786D)
TS 2
= 16.0(0.03189)D 2
= 0.5102D 2
= 2.21 min
D 2
= 2.21/0.5102 = 4.3316
D = (4.3316) 0.5
= 2.081 in
H = 1.25(2.081) = 2.602 in.