HELP! It’s for an architecture class on PLATO
Select the correct answer.
Which association maintains the International Building Code?
A. NFPA
B. ICC
C. EPA
D. DOJ
Answers
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.
Related Questions
IV. An annealed copper strip 9 inches wide and 2.2 inches thick, is rolled to its maximum possible draft in one pass. The following properties of annealed copper are given: strength coefficient is 90,000 psi; true strain at the onset of non-uniform deformation is 0.45; and, engineering strain at yield is 0.11. The coefficient of friction between strip and roll is 0.2. The roll radius is 14inches and the rolls rotate at 150 rpm. Calculate the roll-strip contact length. Calculate the absolute value of thetrue strain that the strip undergoes in this operation. Determine the average true stress of the strip in theroll gap. Calculate the roll force. Calculate the horsepower required.
Answers
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
When you're running a test bench, you would like to include a printout to the screen to inform yourself on the status of the current simulation. Which command you could use to monitor specific variables or signals in a simulation every time one of the signals changes value?
Answers
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.
A square plate of titanium is 12cm along the top, 12cm on the right side, and 5mm thick. A normal tensile force of 15kN is applied to the top side of the plate. A normal tensile force of 20kN is applied to the right side of the plate. The elastic modulus, E, is 115 GPa for titanium. If the left and bottom edges of the plate are xed, calculate the normal strain and elongation of both the TOP and RIGHT side of the plate. Report your answer with proper units and signicant digits.
Answers
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