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geniusboy [140]
3 years ago
8

A square-thread power screw is used to raise or lower the basketball board in a gym, the weight of which is W = 100kg. See the f

ollowing two figures. For power screws, assume a major diameter d = 20mm, and a pitch of p = 2mm with double threads. The friction coefficient of steel parts is f = 0.1. Gravity constant is g = 10????/kg. (a) (7 points) If the purchased Cordless Electric Power Screwdrivers rotates at 300rpm, determine the power you need to raise to the basket board. (b) (2 points) Check if the self-locking condition of the power screw is satisfied or not.
Engineering
1 answer:
KIM [24]3 years ago
8 0

Answer:

power = 49.95 W

and it is self locking screw

Explanation:

given data

weight W = 100 kg = 1000 N

diameter d = 20mm

pitch p = 2mm

friction coefficient of steel f = 0.1

Gravity constant is g = 10 N/kg

solution

we know T is

T = w tan(α + φ ) \frac{dm}{2}     ...................1

here dm is = do - 0.5 P

dm = 20 - 1

dm = 19 mm

and

tan(α) = \frac{L}{\pi dm}      ...............2

here lead L = n × p

so tan(α) = \frac{2\times 2}{\pi 19}

α = 3.83°  

and

f = 0.1

so tanφ = 0.1

so that φ = 5.71°

and  now we will put all value in equation 1 we get

T = 1000 × tan(3.83 + 5.71 ) \frac{19\times 10^{-3}}{2}  

T = 1.59 Nm

so

power = \frac{2\pi N \ T }{60}     .................3

put here value

power = \frac{2\pi \times 300\times 1.59}{60}

power = 49.95 W

and

as φ > α

so it is self locking screw

 

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A plate (A-C) is connected to steelflat bars by pinsat A and B. Member A-E consists of two 6mm by 25mm parallel flat bars. At C,
juin [17]

Answer:

stress_ac = 5.333 MPa

shear stress_c = 1.763 MPa

Explanation:

Given:

- The missing figure is in the attachment.

- The dimensions of member AC = ( 6 x 25 ) mm x 2

- The diameter of the pin d = 19 mm

- Load at point A is P = 2 kN

Find:

-  Find the axial stress in AE and the shear stress in pin C.

Solution:

- The stress in member AE can be calculated using component of force P along the member AE  as follows:

                                    stress_ac = P*cos(Q) / A_ae

Where, Angle Q: A_E_B   and A_ac: cross sectional area of member AE.

                                    cos(Q) = 4 / 5   ..... From figure ( trigonometry )

                                    A_ae = 0.006*0.025*2 = 3*10^-4 m^2

Hence,

                                    stress_ae = 2*(4/5) / 3*10^-4

                                    stress_ae = 5.333 MPa

- The force at pin C can be evaluated by taking moments about C equal zero:

                                   (M)_c = P*6 - F_eb*3

                                      0 = P*6 - F_eb*3

                                      F_eb = 0.5*P

- Sum of horizontal forces for member AC is zero:

                                      P - F_eb - F_c = 0

                                      F_c = 0.5*P

- The shear stress of double shear bolt is given by an expression:

                                     shear stress = shear force / 2*A_pin

Where, The area of the pin C is:

                                     A_pin = pi*d^2 / 4

                                     A_pin = pi*0.019^2 / 4 = 2.8353*10^-4 m^2

Hence,

                                     shear stress = 0.5*P / 2*A_pin

                                     shear stress = 0.5*2 / 2*2.8353*10^-4

                                    shear stress = 1.763 MPa

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