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makvit [3.9K]
3 years ago
13

Construct a link mechanism of crank oa 30mm rotating clockwise rod ab 100mm and bc 50mm

Engineering
1 answer:
m_a_m_a [10]3 years ago
5 0

Answer:

the answer is 180mm

Explanation:

the answer is 180mm.

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Is it a problem that the stress intensity factor, K, from Irwin's near-tip approximation approaches infinity as you get close to
valina [46]

Answer:

No it is not a problem

Explanation:

It is not a problem because the stress intensity factor K would approach infinity as you get close to a crack tip and the intensity factor would approach Zero as you get too far away from the crack tip and this is simply because a crack is a notch with zero tip radius .

and The application of stress intensity factor k in respect to present fatigue crack tip is termed " linear elastic fracture mechanics "

3 0
3 years ago
La liberacion de energia se facilita por
lakkis [162]

Answer:

Enzimas

Explanation:

6 0
3 years ago
An object is supported by a crane through a steel cable of 0.02m diameter. If the natural swinging of the equivalent pendulum is
devlian [24]

Answer:

22.90 × 10⁸ kg

Explanation:

Given:

Diameter, d = 0.02 m

ωₙ = 0.95 rad/sec

Time period, T = 0.35 sec

Now, we know

T= 2\pi\sqrt{\frac{L}{g}}

where, L is the length of the steel cable

g is the acceleration due to gravity

0.35= 2\pi\sqrt{\frac{L}{9.81}}

or

L = 0.0304 m

Now,

The stiffness, K is given as:

K = \frac{\textup{AE}}{\textup{L}}

Where, A is the area

E is the elastic modulus of the steel = 2 × 10¹¹ N/m²

or

K = \frac{\frac{\pi}{4}d^2\times2\times10^11}{0.0304}

or

K = 20.66 × 10⁸ N

Also,

Natural frequency, ωₙ = \sqrt{\frac{K}{m}}

or

mass, m = \sqrt{\frac{K}{\omega_n^2}}

or

mass, m = \sqrt{\frac{20.66\times10^8}{0.95^2}}

mass, m = 22.90 × 10⁸ kg

4 0
3 years ago
I NEED HELP NOW!!!!!!!!!!!!!!
Greeley [361]

Answer:

S or the 4th option

Explanation:

THIS IS SO EASY

8 0
3 years ago
Read 2 more answers
Problem
anzhelika [568]

Answer:

In the clevis shown in Fig. 1-11b, find the minimum bolt diameter and the minimum thickness of each yoke that will support a load P = 14 kips without exceeding a shearing stress of 12 ksi and a bearing stress of 20 ksi.

127-clevis-double-shear-bolt.gif

Solution 127

Hide Click here to show or hide the solution

127-fbd-clevis-double-shear-bolt.gifFor shearing of rivets (double shear)

P=τA

14=12[2(14πd2)]

d=0.8618in → diameter of bolt answer

For bearing of yoke:

P=σbAb

14=20[2(0.8618t)]

t=0.4061in → thickness of yoke answer

3 0
3 years ago
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