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kumpel [21]
3 years ago
14

A solid square rod is cantilevered at one end, as shown below. The rod is 0.6 m long and supports a completely reversing transve

rse load at the other end of \pm2 kN. The material is AISI 1080 hot-rolled steel with Sut = 770 Mpa and Sy = 420 MPa. If the rod must support this load for 104 cycles with a design factor of 1.5, what dimension should the square cross section have? Neglect any stress concentration at the support end. Use surface factor ka = 0.488, size factor kb = 0.85, and all other factors equal to one.

Engineering
1 answer:
jeyben [28]3 years ago
3 0

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

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The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upp
Aliun [14]

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1

The Lower Surface Cp is given as

Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1

The difference of the Cp over the airfoil is given as

\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32

Now the Lift Coefficient is given as

C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192

Now the coefficient of moment about the leading edge is given as

C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306

So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

5 0
3 years ago
A new approval process is being adapted by Ursa Major Solar. After an opportunity has been approved, the contract is sent to the
BigorU [14]
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4 0
3 years ago
You are hired as the investigators to identify the root cause and describe what should have occurred based on the following info
creativ13 [48]

Answer:

The mass of fuel added, which is 10,166.2 kg is less than 22,300 kg which is the mass of fuel required to travel from Toronto to Edmonton, the plane therefore crashed.

Explanation:

Since density ρ = m/v where m = mass of fuel and v = volume of fuel, we need to find the mass of each volume of fuel.

So, m = ρv now ρ = specific gravity × density of water = 0.803 × 1000 kg/m³ = 803 kg/m³.

To find the mass of the 7,682 L of fuel, its volume is 7,682 dm³ = 7,682 dm³ × 1 m³/1000 dm³ = 7.682 m³.

It's mass, m = 803 kg/m³ × 7.682 m³ = 6168.646 kg

To find the mass of the extra 4,916 L of fuel added, we have

m' = ρv' where v' = 4,916 L = 4,916 dm³ = 4916 dm³ × 1 m³/1000 dm³ = 4.916 m³

m' =  803 kg/m³ × 4.916 m³ = 3947.548 kg

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<u>Since this mass of fuel added, which is 10,166.2 kg is less than 22,300 kg which is the mass of fuel required to travel from Toronto to Edmonton, the plane therefore crashed.</u>

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vlabodo [156]

Answer:

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Explanation:

5 0
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Define waves as it applies to electromagnetic fields
julsineya [31]

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5 0
2 years ago
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