Ask question

Ask question

All categories

3 years ago

12

After you check coolant freeze point... a) replace cap immediately after inspecting the cooling system b) call out “Coolant syst

em looks great ,Bay 1” C) inform guest of the results D) both a and c

1 answer:

Oxana [17]3 years ago

3 0

Hello I think the answer is D

You might be interested in

For a bolted assembly with eight bolts, the stiffness of each bolt is kb = 1.0 MN/mm and the stiffness of the members is km = 2.

rjkz [21]

Answer:

a) 0.978

b) 0.9191

c) 1.056

d) 0.849

Explanation:

Given data :

Stiffness of each bolt = 1.0 MN/mm

Stiffness of the members = 2.6 MN/mm per bolt

Bolts are preloaded to 75% of proof strength

The bolts are M6 × 1 class 5.8 with rolled threads

Pmax =60 kN, Pmin = 20kN

<u>a) Determine the yielding factor of safety</u>

$n_{p} = \frac{S_{p}A_{t} }{CP_{max}+ F_{i} }$ ------ ( 1 )

Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N

Input the given values into the equation above

equation 1 becomes ( np ) = $\frac{380*20.1}{0.277*7500*5728.5}$ = 0.978

note : values above are derived values whose solution are not basically part of the required solution hence they are not included

<u>b) Determine the overload factor of safety</u>

$n_{L} = \frac{S_{p}A_{t}-F_{i} }{C(P_{max} )}$ ------- ( 2 )

Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N

input values into equation 2 above

hence : $n_{L}$ = 0.9191 $n_{L} = 0.9191$

<u>C) Determine the factor of safety based on joint separation</u>

$n_{0} = \frac{F_{i} }{P_{max}(1 - C ) }$

Fi = 5728.5 N, Pmax = 7500 N, C = 0.277,

input values into equation above

Hence $n_{0}$ = 1.056

<u>D) Determine the fatigue factor of safety using the Goodman criterion.</u>

nf = 0.849

attached below is the detailed solution .

4 0

3 years ago

HELP ME PLEASE RN

IRISSAK [1]

Answer:

information

Explanation:

see picture

8 0

3 years ago

Why is tubing sometimes coiled when installed in a car or vechile

andriy [413]

Answer:

Coiled tubing is often used to carry out operations similar to wire lining.

8 0

3 years ago

Find the Rectangular form of the following phasors?

almond37 [142]

Answer:

The angles are missing in the question.

The angles are :

45, 30, 60, 90, -34, -56, 20, -42, -65, -15

P=10, P=5, P=25, P=54, P=65, P=95, P=250, P=8, P=35, P=150

Explanation:

1. P = 10, θ = 45° rectangular coordinates

x = r cosθ , y = r sinθ

So, rectangular form is x + iy

x = P cosθ = 10 cos 45°

= 7.07

y =P sinθ = 10 sin 45°

= 7.07

Therefore, rectangular form

x + iy = 7.07 + i (7.07)

2. P = 5 , θ = 30°

x = 5 cos 30° = 4.33

y = 5 sin 30° = 2.5

So, (x+iy) = 4.33 + i (2.5)

3. P = 25 , θ = 60°

x = 25 cos 60° = 12.5

y = 25 sin 60° = 21.65

So, (x+iy) = 12.5 + i (21.65)

4. P = 54 , θ = 90°

x = 54 cos 90° = 0

y = 54 sin 90° = 54

So, (x+iy) = 0+ i (54)

5. P = 65 , θ = -34°

x = 65 cos (-34°) = 53.88

y = 65 sin (-34°) = -36.34

So, (x+iy) = 53.88 - i (36.34)

6. P = 95 , θ = -56°

x = 95 cos (-56)° = 53.12

y = 95 sin (-56)° = -78.75

So, (x+iy) = 53.12 - i (78.75)

7. P = 250 , θ = 20°

x = 250 cos 20° = 234.92

y = 250 sin 20° = 85.5

So, (x+iy) = 234.92 + i (85.5)

8. P = 8 , θ = (-42)°

x = 8 cos (-42)° = 5.94

y = 8 sin (-42)° = -5.353

So, (x+iy) = 5.94 - i (5.353)

9. P = 35 , θ = (-65)°

x = 35 cos (-65)° = 14.79

y = 35 sin (-65)° = -31.72

So, (x+iy) = 14.79 - i (31.72)

10. P = 150 , θ = (-15)°

x = 150 cos (-15)° = 144.88

y = 150 sin (-15)° = -38.82

So, (x+iy) = 144.88 - i (38.82)

6 0

3 years ago

The state of plane strain on an element is:

balu736 [363]

Answer:

a. ε₁=-0.000317

ε₂=0.000017

θ₁= -13.28° and θ₂=76.72°

b. maximum in-plane shear strain =3.335 *10^-4

Associated average normal strain ε(avg) =150 *10^-6

θ = 31.71 or -58.29

Explanation:

$\epsilon _{1,2} =\frac{\epsilon_x + \epsilon_y}{2} \pm \sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\epsilon _{1,2} =\frac{-300 \times 10^{-6} + 0}{2} \pm \sqrt{(\frac{-300 \times 10^{-6}+ 0}{2}) ^2 + (\frac{150 \times 10^-6}{2})^2}\\\\\epsilon _{1,2} = -150 \times 10^{-6} \pm 1.67 \times 10^{-4}$

ε₁=-0.000317

ε₂=0.000017

To determine the orientation of ε₁ and ε₂

$tan 2 \theta_p = \frac{\gamma_xy}{\epsilon_x - \epsilon_y} \\\\tan 2 \theta_p = \frac{150 \times 10^{-6}}{-300 \times 10^{-6}-\ 0}\\\\tan 2 \theta_p = -0.5$

θ= -13.28° and 76.72°

To determine the direction of ε₁ and ε₂

$\epsilon _{x' }=\frac{\epsilon_x + \epsilon_y}{2} + \frac{\epsilon_x -\epsilon_y}{2} cos2\theta + \frac{\gamma_xy}{2}sin2\theta \\\\\epsilon _{x'} =\frac{-300 \times 10^{-6}+ \ 0}{2} + \frac{-300 \times 10^{-6} -\ 0}{2} cos(-26.56) + \frac{150 \times 10^{-6}}{2}sin(-26.56)\\\\$

=-0.000284 -0.0000335 = -0.000317 =ε₁

Therefore θ₁= -13.28° and θ₂=76.72°

b. maximum in-plane shear strain

$\gamma_{max \ in \ plane} =2\sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\gamma_{max \ in \ plane} = 2\sqrt{(\frac{-300 *10^{-6} + 0}{2} )^2 + (\frac{150 *10^{-6}}{2})^2}$

=3.335 *10^-4

$\epsilon_{avg} =(\frac{\epsilon_x + \epsilon_y}{2} )$

ε(avg) =150 *10^-6

orientation of γmax

$tan 2 \theta_s = \frac{-(\epsilon_x - \epsilon_y)}{\gamma_xy} \\\\tan 2 \theta_s = \frac{-(-300*10^{-6} - 0)}{150*10^{-6}}$

θ = 31.71 or -58.29

To determine the direction of γmax

$\gamma _{x'y' }= - \frac{\epsilon_x -\epsilon_y}{2} sin2\theta + \frac{\gamma_xy}{2}cos2\theta \\\\\gamma _{x'y' }= - \frac{-300*10^{-6} - \ 0}{2} sin(63.42) + \frac{150*10^{-6}}{2}cos(63.42)$

= 1.67 *10^-4

4 0

4 years ago