8.4 Frictional Forces on Screws Screws used as fasteners Sometimes used to transmit power or motion...

8.4 Frictional Forces on Screws


Screws used as fastenersSometimes used to transmit power or

motion from one part of the machine to another

A square-ended screw is commonly used for the latter purpose, especially when large forces are applied along its axis

A screw is thought as an inclined plane or wedge wrapped around a cylinder



A nut initially at A on the screwwill move up to B when rotated 360° around the screw

This rotation is equivalent to translating the nut up an inclined plane of height l and length 2πr, where r is the mean radius of the head



The rise l for a single revolution is referred to as the lead of the screw, where the lead angle is given by θ = tan-

1 (l/2πr)



Frictional Analysis When a screw is subjected to

large axial loads, the frictional forces developed on the thread become important to determine the moment M* needed to turn the screw

Consider the square-threaded jack which supports vertical load W



Upward Screw Motion Relative forces of the jack to

this load are distributed over the circumference of the screw thread in contact with the screw hole in the jack, that is, within the region h

For simplicity, the thread can represented by a simple block resting on an inclined plane having screw’s lead angle θ



Upward Screw Motion The inclined plane represents the inside

supporting thread of the jack base 3 forces act on the block or screw Force W is the total axial load applied to

the screw Horizontal force S is caused by the applied

moment M such that by summing the moments about the axis of the screw, M = Sr where r is the screw’s mean radius



Upward Screw MotionAs a result of W and S,

the inclined plane exerts a resultant R on the block, which is shown to have components acting normal, N, and tangent F, to the contacting surfaces



Upward Screw Motion Provided M is large, the

screw and the block can either be brought to the verge of upward impending motion or motion can be occurring

R acts at an angle (θ + Φ) from the verticalΦ = tan-1 (F/N) = tan-1 (μN/N)

= tan-1μ



Upward Screw Motion Apply equilibrium equations

Solving

M is the moment needed to cause upward impending motion of the screw provided ss

y

x

WrM

WRF

RSF

1tan

)tan(

0)cos(;0

0)sin(;0



Upward Screw Motion If Φ is replaced by Φk = tan-1 μk (angle

of kinetic friction), a smaller value of M is needed to maintain uniform upwards motion of the screw



Downward Screw Motion If the surface of the screw is

very slippery, the screw may rotate downward if the magnitude of the moment is reduced to say M’ < M

This causes the effect of M’ to become S’

It requires the angle Φ to lie on the opposite side of the normal n to the plane supporting the block such as θ < Φ

M’ = Wr tan(θ – Φ)



Self Locking Screw If the moment M (or its effect S) is removed,

the screw will remain self-locking It will support the load W by friction forces

alone provided Φ ≥ θ To show this, consider the necessary

limiting case when Φ = θ Vertical equilibrium is maintained

since R is vertical and thus balances W



Downward Screw Motion When the surface of the screw

is very rough, the screw will not rotate downwards

Instead, the direction of the applied moment must be reversed in order to cause the motion

S’’ is caused by the applied (reverse) moment M’’

M’’ = Wr tan(Φ – θ)



Example 8.8 The turnbuckle has a square

thread with a mean radius of 5mm and a lead of 2mm. If the coefficient of static friction between the screw and the turnbuckle is μs = 0.25, determine the moment M that must be applied to draw the end screws closer together. Is the turnbuckle self-locking?



Solution Since friction at two screws must be overcome,

this requires

Solving

When the moment is removed, the turnbuckle will be self-locking

mNmmN

mmNM

mmmmr

mmrNW

WrM

ss

.37.6.7.6374

64.304.14tan520002

64.352/2tan2/tan

04.1425.0tantan,5,2000

tan2

11

11

View Free Body Diagram

8.5 Frictional Forces on Flat Belts


Whenever belt drives or hand brakes are designed, it is necessary to determine the frictional forces developed between the belt and its contacting surfaces

Consider the flat belt which passes over a fixed curved surface such that the total angle of belt to surface contact in radians is β and the coefficient of friction between the two surfaces is μ



Determine the tension T2 in the belt which is needed to pull the belt CCW over the surface and overcome both the frictional forces at the surface of contact and the known tension T1

Obviously T2 > T1



Frictional Analysis Consider FBD of the belt segment in contact

with the surface Normal force N and the frictional force F,

acting at different points on the belt, vary both in magnitude and direction

Due to this unknown force distribution, analysis the problem by studying the forces acting on a differentialelement of the belt



Frictional Analysis Consider FBD of an element having a length

ds Assuming either impending motion or

motion of the belt, the magnitude of the frictional force

dF = μ dN This force opposes the sliding

motion of the belt and thereby increases the magnitude of the tensile force acting in the belt by dT



Frictional Analysis Applying equilibrium equations

Since dθ is of infinitesimal size, sin(θ/2) and cos (θ/2) can be replaced by dθ/2 and 1 respectively

Product of the two infinitesimals dT and dθ/2 may be neglected when compared to infinitesimals of the first order

02

sin2

sin)(

;0

02

cos)(2

cos

;0

dT

ddTTdN

F

ddTTdN

dT

F

y

x



Frictional Analysis

Solving

eTT

T

TIn

dT

dT

TTTT

dT

dT

TddN

dTdN

T

T

12

1

2

0

21

2

1

,,0,



Frictional Analysis T2 is independent of the radius of the

drum and instead, it is a function of the angle of belt to surface contact, β

This equation is valid for flat belts placed on any shape of contacting surface

For application, it is valid only when impending motion or motion occurs



Example 8.9The maximum tension that can be developed In the cord is 500N. If the pulley at A is free to rotate and the coefficient of static friction at fixed drums B and C is μs = 0.25, determine the largest mass of cylinder that can be lifted by the cord. Assume that the force F applied at the end of the cord is directed vertically downward.



Solution Lifting the cylinder, which has a weight of

W = mg, causes the cord to move CCW over the drums at B and C, hence, the maximum tension T2 in the cord occur at D

Thus, T2 = 500N For section of the cord passing

over the drum at B 180° = π rad, angle of contact

between the drum and the cord β = (135°/180°)π = 3/4π rad

View Free Body Diagram



Solution

Since the pulley at A is free to rotate, equilibrium requires that the tension in the cord remains the same on both sides of the pulley

NN

e

NT

eTN

eTT s

4.27780.1

500500

500

;

4/325.01

4/325.01

12



Solution For section of the cord passing over the drum

at CW < 277.4N

kgsm

N

g

Wm

NW

Wen

eTT s

7.15/81.9

9.153

9.153

277

;

2

4/325.0

12

8.4 Frictional Forces on Screws Screws used as fasteners Sometimes used to transmit power or motion...

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Transcript of 8.4 Frictional Forces on Screws Screws used as fasteners Sometimes used to transmit power or motion...