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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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There is a proposal in Brooklyn to construct a new mid-rise apartment building on a vacant lot at the intersection of Avenue A a
Soloha48 [4]

Answer:

a. Park should be adjacent to 48th Street, b. Difference in noise level = 707dBa, c. Yes

Explanation:

Data given for Avenue A

Cars = 496, Medium Truck = 52, Heavy Truck = 19, Buses = 10

Data given for 48th Street

Cars = 822, Medium Truck = 22, Heavy Truck = 8, Buses = 3

Consider the PCEs to be Cars = 1, Medium Truck = 13, Heavy Truck = 47, Buses = 18

a. Noise Level = Number of vehicles x PCE

For Avenue A

Noise level = (496 x 1) + (52 x 13) + (19 x 47) + (10 x 18) = 2245dBa

For 48th Street

Noise level = (822 x 1) + (22 x 13  + (8 x 47) + (3 x 18) = 1538dBa

The park should adjacent to 48th street as it is quieter than Avenue A

b. Let the Setback be 50ft. We know that the reduction of noise for 100ft = 5-8 dBa, hence

For Avenue A Noise Reduction due to 50 ft = (8/100) x 50 = 4dBa

Noise at Setback distance = 2245 - 4 = 2241dBa

Considering the same setback the noise at 48th street would be = 1538 - 4 = 1534 dBa

The difference is noise level between the two sides would be = 2241 - 1534 = 707 dBa

c. Yes the developer concerns are valid because there is a clear difference in noise levels of the two sites. This can be seen even after setting the same Setback. Locating the park next to Avenue A will cause serious noise problems.

8 0
4 years ago
WHAT IS THE EFFECT OF ICE ACCRETION ON THE LONGITUDINAL STABILITY OF AN AIRCRAFT?
soldier1979 [14.2K]

Answer:

The major effects of ice accretion on the aircraft is that it disturbs the flow of air and effects the aircraft's performance.

Explanation:

The ice accretion effects the longitudinal stability of an aircraft as:

1. The accumulation of ice on the tail of an aircraft results in the reduction the longitudinal stability and  the elevator's efficacy.

2. When the flap is deflected at 10^{\circ} with no power there is an increase in the longitudinal velocity.  

3. When the angle of attack is higher close to the stall where separation occurs in the early stages of flow, the effect of ice accretion are of importance.  

4. When the situation involves no flap  at reduced power setting results in the decrease in aircraft's longitudinal stability an increase in change in coefficient of pitching moment  with attack angle.

5 0
4 years ago
Choose the correct word or phrase to complete the sentence to explain human intervention in a machine system.
maksim [4K]

Answer:

Fully Automated

Periodic Maintenance Activities

6 0
3 years ago
Solar energy stored in large bodies of water, called solar ponds, is being used to generate electricity. If such a solar power p
fgiga [73]

Answer: 1.137*10^7 Btu/h.

Explanation:

Given data:

Efficiency of the plant = 4.5percent

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= 3333.333 *3412.152Btu/h.

= 11373840 Btu/h

= 1.137*10^7 Btu/h.

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3 years ago
Convert A'B'C'D' + A'B'C'D + A'B'CD' + A'BC'D + AB'C'D' + AB'C'D+ AB'CD' to SOP form
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Answer:

thats really hard how could you answerthis hhhhhhh

6 0
3 years ago
Read 2 more answers
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