Engineer drawing:
How can i draw this? Any simple way?

Answer:

50421.6 m³

Explanation:

The river has an average rate of water flow of 59.6 m³/s.

Tributary B accounts for 47% of the rate of water flow. Therefore the rate of water flow through tributary B is:

Flow rate of water through tributary B = 47% of 59.6 m³/s = 0.47 * 59.6 m³/s = 28.012 m³/s

The volume of water that has been discharged through tributary B = Flow rate of water through tributary B * time taken

time = 30 minutes = 30 minutes * 60 seconds / minute = 1800 seconds

The volume of water that has been discharged through tributary B in 30 seconds = 28.012 m³/s * 1800 seconds = 50421.6 m³

Answer:

a) the maximum stress that can be applied without plastic deformation is 398.72 N/mm²  

b) length of the specimen is 66.97 mm

Explanation:

Given the data in the question;

a) Determine the maximum stress that can be applied without plastic deformation

when know that; maximum stress σ[tex]_{max}[/tex]  = F / A

where F is the force in the rod ( 39872 N )

A is the cross-sectional area of the rod ( 100 mm² )

so we substitute;

σ[tex]_{max}[/tex]  = 39872 N / 100 mm²

σ[tex]_{max}[/tex]  = 398.72 N/mm²

Therefore, the maximum stress that can be applied without plastic deformation is 398.72 N/mm²  

b)  

strain in the members can be calculated using the expression

ε = σ / E

where σ is the stress in the rod

E is the module of elasticity (  110 GPa = 110000 N/mm² )

(Sl-L) / L = σ/E

where Sl-L is the change in length of the member

L is the original length of the specimen

so we substitute

(67.21 - L) / L = 398.72 / 110000

110000( 67.21 - L) = 398.72L

7393100 - 110000L = 398.72L

7393100 = 398.72L+ 110000L

7393100 = 110398.72L

L = 7393100 / 110398.72

L = 66.97 mm

Therefore; length of the specimen is 66.97 mm