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elena-14-01-66 [18.8K]
2 years ago
9

Determine the critical load if the bottom is fixed and the top is pinned. ewew = 1. 6 ×(10)3ksi×(10)3ksi ,σyσy = 5 ksiksi

Engineering
1 answer:
Katen [24]2 years ago
4 0
<h3>What is a Critical Load?</h3>

Critical load Fcr or buckling load is the value of load that causes the phenomenon of change from stable to unstable equilibrium state.

With that beign said, first it is neessary to calculate the moment of inercia about the x-axis:

Ix= \frac{db^3}{12}\\ Ix = \frac{2.(4)^3}{12} = 10.667in

Then it is necessary to calculate the moment of inercia about the y-axis:

Iy = \frac{db^3}{12}\\ Iy = \frac{4.(2)^3}{12} = 2.662in

Comparing both moments of inercia it is possible to assume that the minimun moment of inercia is the y-axis, so the minimun moment of inercia is 2662in.

And so, it is possible to calculate the critical load:

Pc\gamma = \frac{2046\pi ^2E.I}{L^2} \\Pc\gamma= \frac{2046.\pi ^2.(1,6.10^3.10^3).2662}{(10.12)^2} \\Pc\gamma= 5983,9db

See more about critical load at: brainly.com/question/22020642

#SPJ1

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Answer:

True

Explanation:

8 0
3 years ago
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Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 280
stiks02 [169]

Answer:

Minimum Allowable diameter for each of the 5 bolts = 0.0394 m = 3.94 cm

Explanation:

Maximum Working Stress = Ultimate Shear Stress/factor of safety

Maximum Working Stress = 280 MPa/3.8 = 73.68 MPa

Working Stress = Applied load/Minimum allowable Area = L/A

Minimum Allowable Area = Applied Load/Maximum Working Stress

A = 450000/73680000 = 0.00611 m²

This area is supplied by 5 bolts, so each bolt supplies A/5 = 0.0061/5 = 0.00122 m²

Cross sectional Area of bolts = πD²/4

0.00122 = πD²/4

D² = 4 × 0.00122/π = 0.00155

D = √0.00155 = 0.0394 m = 3.94 cm

Each of the five bolt can have a minimum diameter of 3.94 cm

Hope this Helps!!!

8 0
3 years ago
Find E[x] when x is sum of two fair dice?
Ksenya-84 [330]

Answer:

When two fair dice are rolled, 6×6=36 observations are obtained.

P(X=2)=P(1,1)=

36

1

​

P(X=3)=P(1,2)+P(2,1)=

36

2

​

=

18

1

​

P(X=4)=P(1,3)+P(2,2)+P(3,1)=

36

3

​

=

12

1

​

P(X=5)=P(1,4)+P(2,3)+P(3,2)+P(4,1)=

36

4

​

=

9

1

​

P(X=6)=P(1,5)+P(2,4)+P(3,3)+P(4,2)+P(5,1)=

36

5

​

P(X=7)=P(1,6)+P(2,5)+P(3,4)+P(4,3)+P(5,2)+P(6,1)=

36

6

​

=

6

1

​

P(X=8)=P(2,6)+P(3,5)+P(4,4)+P(5,3)+P(6,2)=

36

5

​

P(X=9)=P(3,6)+P(4,5)+P(5,4)+P(6,3)=

36

4

​

=

9

1

​

P(X=10)=P(4,6)+P(5,5)+P(6,4)=

36

3

​

=

12

1

​

P(X=11)=P(5,6)+P(6,5)=

36

2

​

=

18

1

​

P(X=12)=P(6,6)=

36

1

​

Therefore, the required probability distribution is as follows.

Then, E(X)=∑X

i

​

⋅P(X

i

​

)

=2×

36

1

​

+3×

18

1

​

+4×

12

1

​

+5×

9

1

​

+6×

36

5

​

+7×

6

1

​

+8×

36

5

​

+9×

9

1

​

+10×

12

1

​

+11×

18

1

​

+12×

36

1

​

=

18

1

​

+

6

1

​

+

3

1

​

+

9

5

​

+

6

5

​

+

6

7

​

+

9

10

​

+1+

6

5

​

+

18

11

​

+

3

1

​

=7

E(X

2

)=∑X

i

2

​

⋅P(X

i

​

)

=4×

36

1

​

+9×

18

1

​

+16×

12

1

​

+25×

9

1

​

+36×

36

5

​

+49×

6

1

​

+64×

36

5

​

+81×

9

1

​

+100×

12

1

​

+121×

18

1

​

+144×

36

1

​

=

9

1

​

+

2

1

​

+

3

4

​

+

9

25

​

+5+

6

49

​

+

9

80

​

+9+

3

25

​

+

18

121

​

+4

=

18

987

​

=

6

329

​

=54.833

Then, Var(X)=E(X

2

)−[E(X)]

2

=54.833−(7)

2

=54.833−49

=5.833

∴ Standard deviation =

Var(X)

​

=

5.833

​

=2.415

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Explanation:

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