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3 years ago

I need help with this question please

Engineering

1 answer:

solniwko [45]3 years ago

6 0

Answer:

The resultant moment is 477.84 N·m

Explanation:

We note that the resultant moment is given by the moment about a given point

The length of the sides of the formed triangles are;

l = sin(40°) × 4/sin(110°) ≈ 2.736

Taking the moment about the lower left hand corner of the figure, with the convention that clockwise moments are positive, we have;

The resultant moment, ∑m, is given as follow;

∑M = 250 N × 4 m + 400 N × cos(40°) × 4 m - 400 N × cos(40°) × 2 m + 400 N × sin(40°) × 2 m × tan(40°) - 600 N × cos(40°) × 2 m - 600 N× sin(40°) × 2 m × tan(40°) = 477.837084 N·m

Therefore, the resultant moment, ∑m ≈ 477.84 N·m clockwise.

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3 years ago

What are the controlling LRFD load combinations for dead and floor live load?

yuradex [85]

Answer:

1) 1.4(D + F)

2) 1.2(D + F + T) + 1.6(L + H) + 0.5(Lr or S or R)

3) 1.2D + 1.6(Lr or S or R) + ((0.5 or 1.0)*L or 0.8W)

4) 1.2D + 1.6W + (0.5 or 1.0)*L + 0.5(Lr or S or R)

5) 1.2D + 1.0E + (0.5 or 1.0)*L + 0.2S

6) 0.9D + 1.6W + 1.6H

7) 0.9D + 1.0E + 1.6H

Explanation:

Load and Resistance Factor Design

there are 7 basic load combination of LRFD that is

1) 1.4(D + F)

2) 1.2(D + F + T) + 1.6(L + H) + 0.5(Lr or S or R)

3) 1.2D + 1.6(Lr or S or R) + ((0.5 or 1.0)*L or 0.8W)

4) 1.2D + 1.6W + (0.5 or 1.0)*L + 0.5(Lr or S or R)

5) 1.2D + 1.0E + (0.5 or 1.0)*L + 0.2S

6) 0.9D + 1.6W + 1.6H

7) 0.9D + 1.0E + 1.6H

and

here load factor for L given ( * ) mean it is permitted = 0.5 for occupancies when live load is less than or equal to 100 psf

here

D is dead load and L is live load

E is earth quake load and S is snow load

W is wind load and R is rain load

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3 0

3 years ago

Steam enters a two-stage adiabatic turbine at 8 MPa and 5008C. It expands in the first stage to a state of 2 MPa and 3508C. Stea

Nataly [62]

Answer:

1) The exergy of destruction is approximately 456.93 kW

2) The reversible power output is approximately 5456.93 kW

Explanation:

1) The given parameters are;

P₁ = 8 MPa

T₁ = 500°C

From which we have;

s₁ = 6.727 kJ/(kg·K)

h₁ = 3399 kJ/kg

P₂ = 2 MPa

T₂ = 350°C

From which we have;

s₂ = 6.958 kJ/(kg·K)

h₂ = 3138 kJ/kg

P₃ = 2 MPa

T₃ = 500°C

From which we have;

s₃ = 7.434 kJ/(kg·K)

h₃ = 3468 kJ/kg

P₄ = 30 KPa

T₄ = 69.09 C (saturation temperature)

From which we have;

h₄ = $h_{f4}$ + x₄× $h_{fg}$ = 289.229 + 0.97*2335.32 = 2554.49 kJ/kg

s₄ = $s_{f4}$ + x₄× $s_{fg}$ = 0.94394 + 0.97*6.8235 ≈ 7.563 kJ/(kg·K)

The exergy of destruction, $\dot X_{dest}$ , is given as follows;

$\dot X_{dest}$ = T₀ × $\dot S_{gen}$ = T₀ × $\dot m$ × (s₄ + s₂ - s₁ - s₃)

$\dot X_{dest}$ = T₀ × $\dot W$ ×(s₄ + s₂ - s₁ - s₃)/(h₁ + h₃ - h₂ - h₄)

∴ $\dot X_{dest}$ = 298.15 × 5000 × (7.563 + 6.958 - 6.727 - 7.434)/(3399 + 3468 - 3138 - 2554.49) ≈ 456.93 kW

The exergy of destruction ≈ 456.93 kW

2) The reversible power output, $\dot W_{rev}$ = $\dot W_{}$ + $\dot X_{dest}$ ≈ 5000 + 456.93 kW = 5456.93 kW

The reversible power output ≈ 5456.93 kW.

6 0

3 years ago