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vekshin1
3 years ago
12

For beam design, the cross section dimensions of the beam are determined: A. Based on allowable normal stress, but the allowable

shear stress should always be checked to be sure it is not exceeded. B. Based on the allowable shear stress, but the allowable normal stress should always be checked to be sure it is not exceeded. C. Such that both the allowable normal stress and the allowable shear stress are reached at the same time. D. Based on the allowable normal stress. E. The allowable shear stress is not an important design consideration.
Engineering
1 answer:
noname [10]3 years ago
8 0

Answer:

B. Based on the allowable shear stress, but the allowable normal stress should always be checked to be sure it is not exceeded

Explanation:

Shear stress is analyzed to determine the shear forces along the lenght of the beam. This is represented in a shear force diagram. The beam cross sectional design is determined in such a way as to minimize the shear stress. Allowable normal stress should always be checked in a structure if failure is to be prevented.

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A circular section of material is tested. The original specimen is 200 mm long and has a diameter of 13 mm. When loaded to its p
n200080 [17]

Answer:

modulus of elasticity = 100.45 Gpa,

proportional limit = 150.68 N/mm^2.

Explanation:

We are given the following parameters or data in the question as;

=> "The original specimen = 200 mm long and has a diameter of 13 mm."

=> "When loaded to its proportional limit, the specimen elongates by 0.3 mm."

=> " The total axial load is 20 kN"

Step one: Calculate the area

Area = π/ 4 × c^2.

Area = π/ 4 × 13^2 = 132.73 mm^2.

Step two: determine the stress induced.

stress induced = load/ area= 20 × 1000/132.73 = 150.68 N/mm^2.

Step three: determine the strain rate:

The strain rate = change in length/original length = 0.3/ 200 = 0.0015.

Step four: determine the modulus of elasticity.

modulus of elasticity = stress/strain = 150.68/0.0015 = 100453.33 N/mm^2 = 100.45 Gpa.

Step five: determine the proportional limit.

proportional limit = 20 × 1000/132.73 = 150.68 N/mm^2.

7 0
3 years ago
Read 2 more answers
A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 78 MPa (70.98 ksi). If the plate is
Reika [66]

Answer:

minimum length of a surface crack is 15.043 mm

Explanation:

given data

strain fracture toughness K = 78 MPa

tensile stress = 345 MPa

Y = 1.04

to find out

minimum length of a surface crack

solution

we find here length of critical interior flaw from formula that is

α  =  \frac{1}{\pi} (\frac{K}{\sigma Y})^2     ....................1

put here value we get

α  =  \frac{1}{\pi} (\frac{78*\sqrt{10^3} }{345*1.04})^2

α  = 15.043 mm

so minimum length of a surface crack is 15.043 mm

7 0
3 years ago
A reciprocating compressor takes a compresses it to 5 bar. Assuming that the compression is reversible and has an index, k, of 1
Gelneren [198K]

Answer:

final temperature is 424.8 K

so correct option is e 424.8 K

Explanation:

given data

pressure p1 = 1 bar

pressure p2 = 5 bar

index k = 1.3

temperature t1 = 20°C = 293 k

to find out

final temperature  t2

solution

we have given compression is reversible and has an index k

so we can say temperature is

\frac{t2}{t1}= [\frac{p2}{p1}]^{\frac{k-1}{k} }  ...........1

put here all these value and we get t2

\frac{t2}{293}= [\frac{5}{1}]^{\frac{1.3-1}{1.3} }

t2 = 424.8

final temperature is 424.8 K

so correct option is e

5 0
3 years ago
A pumping test was made in pervious gravels and sands extending to a depth of 50 ft. ,where a bed of clay was encountered. The n
Vikki [24]

Answer:per minute from the pumping well, a steady state was attained in about 24 hr. The draw-down at a distance of 10 ft. was 5.5 ft. and at 25 ft. was 1.21 ft.

Explanation:

6 0
2 years ago
Maggie discovered that a pipe in her basement has sprung a leak. She calls a plumber but in the meantime she grabs a roll of duc
dem82 [27]

Answer:

The answer is "Option a".

Explanation:

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3 0
3 years ago
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