Answer:
hello below is missing piece of the complete question
minimum size = 0.3 cm
answer : 0.247 N/mm2
Explanation:
Given data :
section span : 10.9 and 13.4 cm
minimum load applied evenly to the top of span : 13 N
maximum load for each member ; 4.5 N
lets take each member to be 4.2 cm
Determine the max value of P before truss fails
Taking average value of section span ≈ 12 cm
Given minimum load distributed evenly on top of section span = 13 N
we will calculate the value of by applying this formula
= 
= 1.56 * 10^-5
next we will consider section ; 4.2 cm * 0.3 cm
hence Z (section modulus ) = BD^2 / 6
= ( 0.042 * 0.003^2 ) / 6 = 6.3*10^-8
Finally the max value of P( stress ) before the truss fails
= M/Z = ( 1.56 * 10^-5 ) / ( 6.3*10^-8 )
= 0.247 N/mm2
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:
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Answer:
Explanation:
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