Answer:
Upper bounds 22.07 GPa
Lower bounds 17.59 GPa
Explanation:
Calculation to estimate the upper and lower bounds of the modulus of this composite.
First step is to calculate the maximum modulus for the combined material using this formula
Modulus of Elasticity for mixture
E= EcuVcu+EwVw
Let pug in the formula
E =( 110 x 0.40)+ (407 x 0.60)
E=44+244.2 GPa
E=288.2GPa
Second step is to calculate the combined specific gravity using this formula
p= pcuVcu+pwTw
Let plug in the formula
p = (19.3 x 0.40) + (8.9 x 0.60)
p=7.72+5.34
p=13.06
Now let calculate the UPPER BOUNDS and the LOWER BOUNDS of the Specific stiffness
UPPER BOUNDS
Using this formula
Upper bounds=E/p
Let plug in the formula
Upper bounds=288.2/13.06
Upper bounds=22.07 GPa
LOWER BOUNDS
Using this formula
Lower bounds=EcuVcu/pcu+EwVw/pw
Let plug in the formula
Lower bounds =( 110 x 0.40)/8.9+ (407 x 0.60)/19.3
Lower bounds=(44/8.9)+(244.2/19.3)
Lower bounds=4.94+12.65
Lower bounds=17.59 GPa
Therefore the Estimated upper and lower bounds of the modulus of this composite will be:
Upper bounds 22.07 GPa
Lower bounds 17.59 GPa
Answer:
Manufacturer’s Recommendations means the instructions, procedures, and recommendations which are issued by the manufacturer of any equipment used at the Facility relating to the operation, maintenance, or repair of such equipment, and any revisions or updates thereto from time to time issued by the manufacturer.
Manufacturer’s Recommendations means the instructions, procedures and recommendations which are issued by any manufacturer of the Equipment relating to the operation, maintenance and repair of the Equipment and any revisions to such instructions, procedures and recommendations agreed to by any manufacturer of the Equipment and which are valid at the time such operation, repair and maintenance is being carried out.
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Explanation:
Answer:
material remove in 3 min is 16790.4 mm³/s
Explanation:
given data
length L = 80 cm = 800 mm
width W = 30 cm
height H = 15 cm
make grove length = 80 cm
width = 8 cm
depth = 10 cm
mill toll diameter = 4 mm
axial cutting depth = 20 mm
to find out
How much material removed in 3 minutes
solution
first we find time taken for length of advance that is
time = 
here advance is given as 0.001166 mts / sec
so time = 
time = 686.106 seconds
now we find material remove rate that is
remove rate = mill toll rate × axial cutting depth × advance
remove rate = 4 × 20×0.001166 ×1000
remove rate = 93.28 mm³/s
so
material remove in 3 minute = 3 × 60 = 180 sec
so material remove in 3 min = 180 × 93.28
material remove in 3 min is 16790.4 mm³/s
The answer is true because if the effect is neglected, the saturation id region is considered true