Screws and Threaded Fasteners a) b) c)portal.ku.edu.tr/.../MDesign/problems_exams/Chapter14_prob01.pdf · 4 a) If the members of the joint have the same modulus of elasticity and

1

Screws and Threaded Fasteners

1) A single square-thread power screw has an input power of 3 kW at a speed of 1 rev/s. The screw has a major diameter of 36 mm and a pitch of 6 mm. The frictional coefficients are 0.14 for the threads and 0.09 for the collar, with a collar friction radius of 45 mm. a) Find the axial resisting load F and the combined efficiency of the screw and collar. b) Indicate whether the screw is self-locking or not. c) Find the most critical section and determine safety factor assuming that the load is static. Nut and power screw are made of AISI 1040 HR steel. (Do not consider buckling)

a) Root diameter, dr = d-p = 36-6 dr = 30 mm Mean diameter, dm = d-p/2 =36-6/2 dm =33 mm Since screw is single threaded lead is equal to pitch, l = p = 6 mm

Input torque is T = P/n T =

revrad

srev

kW

π21

3

⋅ T = 477N.m

The relation between torque and axial resisting force (F) is, Frld

dlFdT cc

m

mm µµπ

πµ+

−+

=2

The first part of this equation is the torque required to resist axial force and thread friction and the second part is the torque required for collar friction.

FmmmmmmFmN ⋅⋅⋅+

⋅⋅−⋅⋅⋅⋅⋅+⋅⋅= −

−−

−−−3

33

333

104509.010614,01033103314,0106

21033.477

ππ

F = 65kN

The overall efficiency is: inputWork

outputWorke =

13,02.477

106652

3

=⋅

⋅⋅==−

ππ mNmkN

TFle or %13

b) Self-locking is obtained for square threads when λµ tg> where md

ltg⋅

= −

πλ 1

In this problem 06,033

614.0 =⋅

>π

so screw is self-locking alone.

2

c) Compressive stress between collar and nut: 23

3

2 )1030(106544

mN

dF

AF

rrc −⋅⋅

⋅⋅===ππ

σ

MPac 92=σ

Shear stress due to collar torque between collar and nut: 3

)(16

r

ccc

dFr

JcT

⋅⋅⋅

=⋅

=πµτ

MPammN 6.4930

).10263(163

3

≅⋅

⋅=π

τ

Combined stress between collar and nut: 22

22

max 6.492

922

+

=+

= τστ c

MPa6.67max =τ Shear stress between the motor and the nut: Tinput = 477 N.m

3

3

3 30.104771616

⋅⋅⋅==

ππτ mmN

dT

r

MPa90=τ *

Average thread shear stress: MPaN

pHpd

FAF

rs

23

26030

1065

2

3

≅⋅⋅

⋅===ππ

τ

Average bearing stress in the screw threads: ( ) ( )

6603036

106522

3

22 −

⋅=−

==ππ

σ N

pHdd

FAF

rb

b

MPab 2.5=σ The most critical section is at the left side of the nut. Here τ due to torsion is 90 MPa.

The safety factor is ττ

2/ysy SSn == Sy =290 MPa from Table A-20

MPaMPan

902290⋅

= = 1.6

2) A permanent bolted joint is designed with a M20 coarse-pitch (5.8 grade, rolled-threads) bolt. The bolt is tightened by 30 kN preload. Then an external load, P, varying between 10 and 36 kN is applied to the joint. a) Does separation occur between parts when external load is maximum? If not, at what maximum external load separation occurs? b) Find the factor of safety of the bolt against fatigue failure. (Take km/kb = 3, R=%99) a) At separation resultant load on members is zero, Fm = 0

0=−+

= isepmb

mm FP

kkk

F

sep

m

bi P

kkF

1

1

+= kNFP isep 30

34

34 ⋅== Psep = 40 kN

When a 40 kN external force is applied, separation occurs between parts. In this problem the maximum external load is 36 kN so separation does not occur. b) The portion of external load (P = 10….36 kN) taken by bolt is,

3

361

1maxmax ⋅

+=

+=

b

mmb

bb

kk

Pkk

kP kNPb 9max = , 5.2min =bP

So, the maximum bolt load is kNFPF ibb 39maxmax =+= And the minimum bolt load is kNFPF ibb 5.32minmin =+= Hence the mean and the alternating bolt loads are:

kNFFF bbbm 75.35

2minmax =

+= kNFFF bb

ba 25.32

minmax =−

=

Tensile strength area of M20 coarse-pitch bolt is At = 245 mm2 (Table 14-2).

MPammkN

AKf*F

t

baa 245

15.72 =29.2==σ MPa

mmNk

AF

t

bmm 92.145

245.75.35

2 ===σ

From Table 14-7, Sut =520MPa Kf = 2.2 (Table14-8, grade 5.8, rolled threads, Cload = 0.7, Csurf ≅ 0.86 (Table14-7), remaining factors

are unity (note that reliability factor = 0.814).

( ) MPaMPaSe .41275205.00 81486.00.7 =⋅⋅⋅⋅⋅⋅=

Plot the modified Goodman diagram and the loading line as shown in Fig. 13-17

n = 2.8 (Using Eq. 14-16 or from geometry) 3) The head of a 300 mm diameter cylinder vessel will be designed. The internal pressure in the cylinder is Pin = 1MPa. The bolts have initial tightening load of 12 kN. a) Find the stiffness of the bolt and members. b) Determine the number of bolts (N) for a safety factor of 2 for fatique loading.

1MPa

D = 300 mm

20

20

cast ironM 10*1.5 grade 5.8

Eb = 200 GPa Eci = 90 GPa

MPammkN

AKfm*F

t

ii 245

301*2 =122.4==σ

4

a) If the members of the joint have the same modulus of elasticity and assuming α = 45°, we can find the stiffness with the formula below.

++⋅

=

dldl

dEk cı

m

5.25.05ln2

π lgrip = 20+20=40 mm 1140

105.240105.0405ln2

10101090 39

=

⋅+⋅+

⋅⋅⋅⋅=−π

mk kN/mm

The bolt stiffness is,

lEd

lEAk bbb

b 4

2π== ( )

mmkNkb 393

10404102001010

3

923

=⋅⋅

⋅⋅⋅= −

−π

Stiffness of the members can also be calculated by other approaches.(see Section 14-8)

b) External force per bolt: kNNNN

APP in 68.704

3.010 26

=⋅

⋅⋅=⋅

= π

Resultant bolt load: imb

bb FP

kkk

F ++

= kNNN

Fb 1214.181268.701140393

393 +=+⋅+

=

Tensile stress area for M10×1.5 is At =58 mm2 (Table14-2) Sy =520 MPa (Table14-7)

t

yy A

F=σ tyy AF ⋅= σ

nS y

all =σ ty

all An

SF ⋅=

MPammFall 252058 2 ⋅= kN

NkNFall 1214.1808.15 +== N = 5.89 so, use 6 bolts.

Preload must be checked against separation.(Fm = 0)

0=−⋅+

= ibm

mm F

NP

kkk

F kNFi 1276.8668.70

39311401140 <=⋅

+= Preload is enough.