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kupik [55]
4 years ago
7

To use Allowable Stress Design to calculate required dimensions using a specified factor of safety. A structural element that ca

rries a load must be designed so that it can support the load safely. There are several reasons why an element may fail at loads that are less than the theoretical limit. For example, material properties may not exactly equal the reference values used in the design. The actual loading may differ from the design loading. The exact dimensions of the member may be different from the nominal values. These scenarios, and others, make it important to design structural members so that the expected load is less than the expected load that would make the member fail. One method of doing this uses a factor of safety. The allowed load Fallow can be related to the load that causes failure, Ffail, using a constant called the factor of safety, F.S.=FfailFallow, which should be larger than 1. For preliminary analysis, the stresses that are developed are assumed to be constant, so that the load and the stress are related by N=σA or V=τA, where A is the area subjected to the load.
An anchor rod with a circular head supports a load Fallow = 11 kN by bearing on the surface of a plate and passing through a hole with diameter h = 2.6 cm. One way the anchor could break is by the rod failing in tension.
What is the minimum required diameter of the rod if the factor of safety for tension failure is F.S. = 1.6, given that the material fails in tension at σfail = 60 MPa? Assume a uniform stress distribution.
Engineering
1 answer:
Kamila [148]4 years ago
6 0

Answer:

Explanation:

For preliminary analysis, the stresses that are developed are assumed to be constant, so that the load and the stress are related by N=σA or V=τA, where A is the area subjected to the load.

 

a.) An anchor rod with a circular head supports a load  Pallow = 11 kN by bearing on the surface of a plate and passing through a hole with diameter h = 2.6 cm as shown (Figure 1). One way the anchor could break is by the rod failing in tension, as shown (Figure 2). What is the minimum required diameter of the rod if the factor of safety for tension failure is F.S. = 1.6, given that the material fails in tension at σfail = 60 MPa ? Assume a uniform stress distribution.

b.)The anchor from Part A can also fail in shear in the circular head, as shown (Figure 3). What is the minimum thickness t required for the head to support the allowed load Pallow = 11 kN if the material fails in shear at τfail = 30 MPa ? Use a factor of safety F.S. = 2.2.

c.)The anchor in Part A could also fail if the surface of the support plate is crushed by the bearing load (Figure 4). What is the required minimum head diameter using a factor of safety F.S. = 1.4? The maximum bearing stress is σ fail = 60 MPa .

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aleksley [76]

Answer:

Uair = 0.0749 KW/k = 74.9 W/k

Explanation:

The natural air change per hour is given by the formula:

Natural Air Change per Hour = ACPH = 60*Volume Flow/Volume

where,

ACPH = 0.4

Volume Flow = ? in ft³/min

Volume = 19456 ft³

Therefore,

0.4 = (60 min)(Volume Flow)/(19456 ft³)

Volume Flow = (0.4)(19456 ft³)/(60 min) = (129.7 ft³/min)(1 min/60 s)

Volume Flow =  (2.16 ft³/s)(0.3048 m/1 ft)³ = 0.061 m³/s

Now, we find heat loss coefficient:

Uair = Volumetric Flow*Density of air*Specific Heat Capacity of air

Uair = (0.061 m³/s)(1.225 kg/m³)(1 KJ/kg.k)

<u>Uair = 0.0749 KW/k = 74.9 W/k</u>

6 0
3 years ago
How many ( 1/8") in one inch?
Anna35 [415]

Answer:

1 inch

Explanation:

There for there are 8 1/8 inch that make up 1 inch so the answer is 1 inch

4 0
3 years ago
Read 2 more answers
The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter
Korvikt [17]

Answer:

the ratio of the etched to the original crack tip radius is 30.24

Explanation:

Given the data in the question;

we determine the initial fracture stress using the following expression;

(σf)₁ = 2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2 ----- let this be equation 1

where; (σ₀)₁ is the initial fracture strength

(p_t)₁ is the original crack tip radius

α₁ is the original crack length.

first, we determine the final crack length;

α₂ = α₁ - 16% of α₁

α₂ = α₁ - ( 0.16 × α₁)

α₂ = α₁ - 0.16α₁

α₂ = 0.84α₁

next, we calculate the final fracture stress;

the fracture strength is increased by a factor of 6;

(σ₀)₂ = 6( σ₀ )₁

Now, expression for the final fracture stress

(σf)₂ = 2(σ₀)₂ [ α₂/(p_t)₂ ]^{1/2 ------- let this be equation 2

where (p_t)₂ is the etched crack tip radius

value of fracture stress of glass is constant

Now, we substitute 2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2 from equation for (σf)₂  in equation 2.

0.84α₁ for α₂.

6( σ₀ )₁ for (σ₀)₂.

∴

2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2  = 2(6( σ₀ )₁) [ 0.84α₁/(p_t)₂ ]^{1/2  

divide both sides by 2(σ₀)₁

[ α₁/(p_t)₁ ]^{1/2  =  6 [ 0.84α₁/(p_t)₂ ]^{1/2

[ 1/(p_t)₁ ]^{1/2  =  6 [ 0.84/(p_t)₂ ]^{1/2

[ 1/(p_t)₁ ]  =  36 [ 0.84/(p_t)₂ ]

1 / (p_t)₁ = 30.24 / (p_t)₂

(p_t)₂ = 30.24(p_t)₁

(p_t)₂/(p_t)₁ = 30.24

Therefore, the ratio of the etched to the original crack tip radius is 30.24

6 0
3 years ago
A room in the lower level of a cruise ship has a 30 cm diameter circular window If the midpoint of the window is 4 m below the w
AleksAgata [21]

Answer:

F = 2840.3 N

Explanation:

Given:

- Diameter of window D = 0.3 m

- Midpoint of window from sea level h = 4 m

- Specific gravity of sea water S.G = 1.024

- Density of water p = 1000 kg/m^3

Find:

The hydro-static force F_r acting on the mid-point of the window.

Solution:

- The average pressure P acting on the midpoint of the window:

                               P = S.G p*g*h

                               P = 1.024*1000*9.81*4

                               P = 40181.76

- The hydro-static force F_r acting on the mid-point of the window:

                               F = P*A = P*pi*D^2 / 4

                               F = 40181.76*pi*0.3^2 / 4

                               F = 2840.3 N

7 0
3 years ago
If your accelerator pedal gets stuck, what is the first thing you should do?
Anna35 [415]

If your accelerator gets stuck down, do the following: Shift to neutral. Apply the brakes. Keep your eyes on the road and look for a way out.If your accelerator gets stuck down, do the following:

Shift to neutral.

Apply the brakes.

Keep your eyes on the road and look for a way out.

Warn other drivers by blinking and flashing your hazard lights.

Try to drive the car safely off the road.

Turn off the ignition when you no longer need to change direction.

8 0
3 years ago
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