All categories

2 years ago

Whats the best used for Arch bridge

Engineering

1 answer:

SOVA2 [1]2 years ago

6 0

Answer:

China has also been constructing arch bridges for many years. They built the Zhaozhou Bridge in 605 AD, and it is still around today! Modern arch bridges use materials such as concrete, steel and iron.

Explanation:

You might be interested in

State the mathematical expression to define the availability of equipment over a specified time and operational availability?

Gelneren [198K]

Answer:

$Availability=\dfrac{Up\ time}{Down\ time+Up\ time}$

Explanation:

Availability:

It define as the probability of system which perform desired task before showing any failure .

The availability can be define as follows

$Availability=\dfrac{Up\ time}{Down\ time+Up\ time}$

Or we can say that

$Availability=\dfrac{Up\ time}{total\ time}$

Availability can also be express as

$Availability=\dfrac{MTBF}{MTBF+MTTR}$

Where MTBF is the mean time between two failure.

MTTR is the mean time to repair.

7 0

2 years ago

Is EPA is the organization that was formed to ensure safe and health working conditions for workers by setting and enforcing sta

mr_godi [17]

Let me search this up

3 0

2 years ago

It is desired to produce and aligned carbon fiber-epoxy matrix composite having a longitudinal tensile strength of 630 MPa. Calc

ratelena [41]

Answer:

The answer is below

Explanation:

Given that:

Diameter (D) = 0.03 mm = 0.00003 m, length (L) = 2.4 mm = 0.0024 m, longitudinal tensile strength $(\sigma_{cd})=630\ MPa = 630*10^6\ Pa$ , Fracture strength

$(\sigma_f)=5100\ MPa=5100*10^6\ Pa,fiber-matrix\ stres(\sigma_m)=17.5\ MPa=17.5*10^6\ Pa,matrix\ strength=\tau_c=17\ MPa=17 *10^6\ Pa$

a) The critical length ( $L_c$ ) is given by:

$L_c=\sigma_f*(\frac{D}{2*\tau_c} )=5100*10^6*\frac{0.00003}{2*17*10^6}=0.0045\ m=4.5\ mm$

The critical length (4.5 mm) is greater than the given length, hence th composite can be produced.

b) The volume fraction (Vf) is gotten from the formula:

$\sigma_{cd}=\frac{L*\tau_c}{D}*V_f+\sigma_m(1-V_f)\\\\V_f=\frac{\sigma_{cd}-\sigma_{m}}{\frac{L*\tau_c}{D}-\sigma_{m}} \\\\Substituting:\\\\V_f=\frac{630*10^6-17.5*10^6}{\frac{0.0024*17*10^6}{0.00003} -17.5*10^6} \\\\V_f=0.456$

6 0

2 years ago

A spacecraft is fueled using hydrazine (N2H4; molecular weight of 32 grams per mole [g/mol]) and carries 1 comma 630 kilogram

Varvara68 [4.7K]

Answer:

attached below

Explanation:

7 0

3 years ago

If the surface energy of a magnesium oxide - nickel oxide (MgO-NiO) solid solution is 1.05 J/m2 and its elastic modulus is 198 G

Nastasia [14]

Answer:

The maximum length is 3.897×10^-5 mm

Explanation:

Extension = surface energy/elastic modulus

surface energy = 1.05 J/m^2

elastic modulus = 198 GPa = 198×10^9 Pa

Extension = 1.05/198×10^9 = 5.3×10^-12 m

Strain = stress/elastic modulus = 27×10^6/198×10^9 = 1.36×10^-4

Length = extension/strain = 5.3×10^-12/1.36×10^-4 = 3.897×10^-8 m = 3.897×10^-8 × 1000 = 3.897×10^-5 mm

7 0

3 years ago