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svlad2 [7]
3 years ago
11

Question #4

Engineering
1 answer:
inn [45]3 years ago
8 0

Answer:

Deconstruction

Explanation:

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Gray cast iron, with an ultimate tensile strength of 31 ksi and an ultimate compressive strength of 109 ksi, has the following s
suter [353]

Using an appropriate failure theory, find the factor of safety in each case. State the name of the theory that you are using the theory is max stress theory.

<h3>Wat is the max stress theory?</h3>

The most shear strain concept states that the failure or yielding of a ductile fabric will arise whilst the most shear strain of the fabric equals or exceeds the shear strain fee at yield factor withinside the uniaxial tensile test.”

Stress states at various critical locations are f= 2.662.

Read more about strain:

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3 0
2 years ago
A standard penetration test has been conducted on a coarse sand at a depth of 16 ft below the ground surface. The blow counts ob
scoray [572]

Solution :

Given :

The number of blows is given as :

0 - 6 inch = 4 blows

6 - 12 inch = 6 blows

12 - 18 inch = 6 blows

The vertical effective stress $=1500 \ lb/ft^2$

                                              $= 71.82 \ kN/m^2$

                                             $ \sim 72 \ kN/m^2 $

Now,

$N_1=N_0 \left(\frac{350}{\bar{\sigma}+70} \right)$

$N_1 = $ corrected N - value of overburden

$\bar{\sigma}=$ effective stress at level of test

0 - 6 inch, $N_1=4 \left(\frac{350}{72+70} \right)$

                      = 9.86

6 - 12 inch, $N_1=6 \left(\frac{350}{72+70} \right) $

                        = 14.8

12 - 18 inch, $N_1=6 \left(\frac{350}{72+70} \right) $

                         = 14.8

$N_{avg}=\frac{9.86+14.8+14.8}{3}$

       = 13.14

       = 13

8 0
3 years ago
A gas tank is known to have a thickness of 0.5 inches and an internal pressure of 2.2 ksi. Assuming that the maximum allowable s
sergiy2304 [10]

Answer:

D_o=11.9inch

Explanation:

From the question we are told that:

Thickness T=0.5

Internal PressureP=2.2Ksi

Shear stress \sigma=12ksi

Elastic modulus \gamma= 35000

Generally the equation for shear stress is mathematically given by

 \sigma=\frac{P*r_1}{2*t}

Where

r_i=internal Radius

Therefore

 12=\frac{2.2*r_1}{2*0.5}

 r_i=5.45

Generally

 r_o=r_1+t

 r_o=5.45+0.5

 r_o=5.95

Generally the equation for outer diameter is mathematically given by

 D_o=2r_o

 D_o=11.9inch

Therefore

Assuming that the thin cylinder is subjected to integral Pressure

Outer Diameter is

 D_o=11.9inch

7 0
3 years ago
8. Describe and correct the error in stating the domain. Xf * (x) = 4x ^ (1/2) + 2 and g(x) = - 4x ^ (1/2) The domain of (f + g)
konstantin123 [22]

Answer:

hi

Explanation:

4 0
3 years ago
The sliders A and B are connected by a light rigid bar of length l = 20 in. and move with negligible friction in the slots, both
DedPeter [7]

Answer:

Explanation:

Given:

- The Length of the rigid bar L = 20 in

- The position of slider a, x_a = 16 in

- The position of slider b, y_b

- The velocity of slider a, v_a = 3 ft /s

- The velocity of slider b, v_b

- The acceleration of slider a, a_a

- The acceleration of slider b, a_b

Find:

-Determine the acceleration of each slider and the force in the bar at this instant.

Solution:

- The relationship between the length L of the rod and the positions x_a and x_b of sliders A & B is as follows:

                               L^2 = x_a^2 + y_b^2   ....... 1

                               y_b = sqrt( 20^2 - 16^2 )

                               y_b = 12

- The velocity expression can derived by taking a derivation of Eq 1 with respect to time t:

                               0 = 2*x_a*v_a + 2*y_b*v_b

                               0 = x_a*v_a + y_b*v_b   ..... 2

                               0 = 16*36 + 12*v_b

                               v_b = - 48 in /s = -4 ft/s

- Similarly, the acceleration expression can be derived by taking a derivative of Eq 2 with respect to time t:

                               0 = v_a^2 + x_a*a_a + v_b^2 + y_b*a_b

                               0 = 9 + 4*a_a/3 + 16 + a_b

                               4*a_a/3 + a_b = -25

                               4*a_a + 3*a_b = -75  .... 3

- Use dynamics on each slider. For Slider A, Apply Newton's second law of motion in x direction:

                               F_x = m_a*a_a

                               P - R_r*16/20 = m_a*a_a

                               

- For Slider B, Apply Newton's second law of motion in y direction:

                               F_y = m_b*a_b

                               - R_r*12/20 = m_b*a_b

- Combine the two dynamic equations:

                               P - 4*m_b*a_b / 3 = m_a*a_a

                               3P = 3*m_a*a_a + 4*m_b*a_b  ... 4

- Where,                  P = Is the force acting on slider A

                               P , m_a and m_b are known quantities but not given in question. We are to solve Eq 3 and Eq 4 simultaneously for a_a and a_b.                    

5 0
3 years ago
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