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Lilit [14]
3 years ago
15

A rectangular concrete beam has dimensions b=16 in. and h=30 in. The location of the Gr. 60 reinforcing bars, which are placed a

ll in one row in the bottom of the section, is d=27 in. Using a specified compressive strength of concrete of 5,000 psi and the assumptions and simplifications of the ACI 318-14 Code, find the nominal flexural strength Mn for positive bending if the reinforcing bars are: a. Three No. 9 bars b. Four No. 9 bars c. Three No. 8 + two No. 10 bars d. None (i.e., the beam is unreinforced)

Engineering
1 answer:
klio [65]3 years ago
5 0

Answer:

attached below

Explanation:

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Spring-loaded rack guide yokes are made of ______ and have a spring that pushes on the back side of the rack to help reduce the
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3 years ago
Air in a large tank at 300C and 400kPa, flows through a converging diverging nozzle with throat diameter 2cm. It exits smoothly
-Dominant- [34]

Answer:

The answer is "3.74 \ cm\ \ and \ \ 0.186 \frac{kg}{s}"

Explanation:

Given data:  

Initial temperature of tank T_1 = 300^{\circ}\ C= 573 K

Initial pressure of tank P_1= 400 \ kPa

Diameter of throat d* = 2 \ cm

Mach number at exit M = 2.8

In point a:

calculating the throat area:

A*=\frac{\pi}{4} \times d^2

      =\frac{\pi}{4} \times 2^2\\\\=\frac{\pi}{4} \times 4\\\\=3.14 \ cm^2

Since, the Mach number at throat is approximately half the Mach number at exit.  

Calculate the Mach number at throat.  

M*=\frac{M}{2}\\\\=\frac{2.8}{2}\\\\=1.4

Calculate the exit area using isentropic flow equation.

\frac{A}{A*}= (\frac{\gamma -1}{2})^{\frac{\gamma +1}{2(\gamma -1)}}  (\frac{1+\frac{\gamma -1}{2} M*^2}{M*})^{\frac{\gamma +1}{2(\gamma -1)}}

Here: \gamma is the specific heat ratio. Substitute the values in above equation.

\frac{A}{3.14}= (\frac{1.4-1}{2})^{-\frac{1.4+1}{2(1.4 -1)}}  (\frac{1+\frac{1.4-1}{2} (1.4)^2}{1.4})^{\frac{1.4+1}{2(1.4-1)}} \\\\A=\frac{\pi}{4}d^2 \\\\10.99=\frac{\pi}{4}d^2 \\\\d = 3.74 \ cm

exit diameter is 3.74 cm

In point b:

Calculate the temperature at throat.

\frac{T*}{T}=(1+\frac{\Gamma-1}{2} M*^2)^{-1}\\\\\frac{T*}{573}=(1+\frac{1.4-1}{2} (1.4)^2)^{-1}\\\\T*=411.41 \ K

Calculate the velocity at exit.  

V*=M*\sqrt{ \gamma R T*}

Here: R is the gas constant.  

V*=1.4 \times \sqrt{1.4 \times 287 \times 411.41}\\\\=569.21 \ \frac{m}{s}

Calculate the density of air at inlet

\rho_1 =\frac{P_1}{RT_1}\\\\=\frac{400}{ 0.287 \times 573}\\\\=2.43\  \frac{kg}{m^3}

Calculate the density of air at throat using isentropic flow equation.  

\frac{\rho}{\rho_1}=(1+\frac{\Gamma -1}{2} M*^2)^{-\frac{1}{\Gamma -1}} \\\\\frac{\rho *}{2.43}=(1+\frac{1.4-1}{2} (1.4)*^2)^{-\frac{1}{1.4-1}} \\\\\rho*= 1.045 \ \frac{kg}{m^3}

Calculate the mass flow rate.  

m= \rho* \times A* \times V*\\\\= 1.045 \times 3.14 times 10^{-4} \times 569.21\\\\= 0.186 \frac{kg}{s}

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