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AnnZ [28]
2 years ago
12

A center-point bending test was performed on a 2 in. x d in. wood lumber according to ASTM D198 procedure with a span of 4 ft an

d the 4 in. side is positioned vertically. If the maximum load was 240 kips and the modulus of rupture was 940.3 ksi, what is the value of d
Engineering
1 answer:
Zigmanuir [339]2 years ago
8 0

Answer:

3.03 INCHES

Explanation:

According to ASTM D198 ;

Modulus of rupture = ( M / I ) * y  ----- ( 1 )

M ( bending moment ) = R * length of span / 2

                                     = (120 * 10^3 ) * 48 / 2 = 288 * 10^4 Ib-in

I ( moment of inertia ) = bd^3 / 12

                                    = ( 2 )*( d )^3  / 12 =  2d^3 / 12

b = 2 in ,  d = ?

length of span = 4 * 12 = 48 inches

R = P  / 2 =  240 * 10^3 / 2 =   120 * 10^3 Ib

y ( centroid distance ) = d / 2  inches

back to equation ( 1 )

( M / I ) * y

940.3 ksi = ( 288 * 10^4 / 2d^3 / 12 ) * d / 2

                = ( 288 * 10^4 * 12 ) / 2d^3 )  * d / 2

940300  = 34560000* d / 4d^3

4d^3 ( 940300 ) = 34560000 d  ( divide both sides with d )

4d^2 = 34560000 / 940300

d^2 = 9.188   ∴ Value of d ≈ 3.03 in

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3 years ago
Explain what a margin of safety is in driving as well as how it can help minimize risk.
Yakvenalex [24]

Answer:

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2 years ago
Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 10 MPa, 450°C, and 80 m/s, and the exit
8090 [49]

Answer:

a) The change in Kinetic energy, KE = -1.95 kJ

b) Power output, W = 10221.72 kW

c) Turbine inlet area, A_1 = 0.0044 m^2

Explanation:

a) Change in Kinetic Energy

For an adiabatic steady state flow of steam:

KE = \frac{V_2^2 - V_1^2}{2} \\.........(1)

Where Inlet velocity,  V₁ = 80 m/s

Outlet velocity, V₂ = 50 m/s

Substitute these values into equation (1)

KE = \frac{50^2 - 80^2}{2} \\

KE = -1950 m²/s²

To convert this to kJ/kg, divide by 1000

KE = -1950/1000

KE = -1.95 kJ/kg

b) The power output, w

The equation below is used to represent a  steady state flow.

q - w = h_2 - h_1 + KE + g(z_2 - z_1)

For an adiabatic process, the rate of heat transfer, q = 0

z₂ = z₁

The equation thus reduces to :

w = h₁ - h₂ - KE...........(2)

Where Power output, W = \dot{m}w..........(3)

Mass flow rate, \dot{m} = 12 kg/s

To get the specific enthalpy at the inlet, h₁

At P₁ = 10 MPa, T₁ = 450°C,

h₁ = 3242.4 kJ/kg,

Specific volume, v₁ = 0.029782 m³/kg

At P₂ = 10 kPa, h_f = 191.81 kJ/kg, h_{fg} = 2392.1 kJ/kg, x₂ = 0.92

specific enthalpy at the outlet, h₂ = h_1 + x_2 h_{fg}

h₂ = 3242.4 + 0.92(2392.1)

h₂ = 2392.54 kJ/kg

Substitute these values into equation (2)

w = 3242.4 - 2392.54 - (-1.95)

w = 851.81 kJ/kg

To get the power output, put the value of w into equation (3)

W = 12 * 851.81

W = 10221.72 kW

c) The turbine inlet area

A_1V_1 = \dot{m}v_1\\\\A_1 * 80 = 12 * 0.029782\\\\80A_1 = 0.357\\\\A_1 = 0.357/80\\\\A_1 = 0.0044 m^2

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Answer:

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