The following true stresses produce the corresponding true plastic strains for a brass alloy: True Stress (psi) True Strain 48400 0.11 60400 0.19 What true stress is necessary to produce a true plastic strain of 0.26

Answer:

For the Top Side

- Strain ε  = 0.00021739

- Elongation is 0.00260868 cm

For The Right side

- Strain ε  = 0.00021739

-Elongation is 0.00347826 cm

Explanation:

Given the data in the question;

Length of the squared titanium plate = 12 cm by 12 cm = 0.12 m by 0.12 m

Thickness = 5 mm = 0.005 m

Force to the Top F[tex]_t[/tex] = 15 kN = 15000 Newton

Force to the right F[tex]_r[/tex] = 20 kN = 20000 Newton

elastic modulus, E = 115 GPa = 115 × 10⁹ pascal

Now, For the Top Side;

- Strain = σ/E = F[tex]_t[/tex]  / AE

we substitute

= 15000 / ( 0.12 × 0.005 × (115 × 10⁹) )

= 15000 / 69000000

Strain ε  = 0.00021739

- Elongation

Δl = ε × l

we substitute

Δl = 0.00021739 ×  12 cm

Δl = 0.00260868 cm

Hence, Elongation is 0.00260868 cm

For The Right side

- Strain = σ/E = F[tex]_r[/tex]  / AE

we substitute

Strain = 20000 / ( 0.12 × 0.005 × (115 × 10⁹) )

= 20000 / 69000000

Strain ε = 0.000289855

- Elongation

Δl = ε × l

we substitute

Δl = 0.000289855×  12 cm

Δl = 0.00347826 cm

Hence, Elongation is 0.00347826 cm

Answer:

ICC

Explanation:

The International Building Code (IBC) is a model building code developed by the International Code Council (ICC). It has been adopted for use as a base code standard by most jurisdictions in the United States.

Answer: Hello your question is poorly written attached below is the complete question

answer :

: max value to avoid failure = 59 MPa

; max value to avoid failure = 34.064 Mpa

Explanation:

Attached below is the detailed solution of the given problem

For Tresca criterion : max value to avoid failure = 59 MPa

For Von-Nissen criterion ; max value to avoid failure = 34.064 Mpa

Answer:

13.9357 horse power

Explanation:

Annealed copper

Given :

Width, b = 9 inches

Thickness, [tex]$h_0=2.2$[/tex] inches

K= 90,000 Psi

μ = 0.2, R = 14 inches, N = 150 rpm

For the maximum possible draft in one pass,

[tex]$\Delta h = H_0-h_f=\mu^2R$[/tex]

     [tex]$=0.2^2 \times 14 = 0.56$[/tex] inches

[tex]$h_f = 2.2 - 0.56$[/tex]

     = 1.64 inches

Roll strip contact length (L) = [tex]$\sqrt{R(h_0-h_f)}$[/tex]

                                             [tex]$=\sqrt{14 \times 0.56}$[/tex]

                                             = 2.8 inches

Absolute value of true strain, [tex]$\epsilon_T$[/tex]

[tex]$\epsilon_T=\ln \left(\frac{2.2}{1.64}\right) = 0.2937$[/tex]

Average true stress, [tex]$\overline{\gamma}=\frac{K\sum_f}{1+n}= 31305.56$[/tex] Psi

Roll force, [tex]$L \times b \times \overline{\gamma} = 2.8 \times 9 \times 31305.56$[/tex]

                                 = 788,900 lb

For SI units,

Power = [tex]$\frac{2 \pi FLN}{60}$[/tex]  

           [tex]$=\frac{2 \pi 788900\times 2.8\times 150}{60\times 44.25\times 12}$[/tex]

           = 10399.81168 W

Horse power = 13.9357

Answer:

$Monitor

Explanation:

The command that would be used when running a test bench to monitor variables or signals ( i.e. changes in the values of specific variables and signa)

is the $Monitor command

This command is also used to monitor the varying values of signals during simulation.