OPEN COILED HELICAL SPRING CLASS NOTES FOR ENGINEERING

 

OPEN COILED HELICAL SPRING

CLASS NOTES FOR ENGINEERING

Spring is a very useful item. Open coiled

helical spring finds use in many applications.

It absorbs shock and store energy. It has

significant deformation within elastic

limits. Use spring steel for its manufacture.

  OPEN COILED HELICAL SPRING UNDER AXIAL LOAD

Spring will be under both bending and torsion. Thus it is under complex stresses. Find maximum principal stress and maximum shear stresses.

Let α be the angle of the helix

Then the length of n turns of spring L =πD n/cos α

Consider a section of the spring[as shown in Fig. passing through the axis of the helix and perpendicular to the fig shown.]

Consider a small length of the spring as CD

When an axial load W acts, the spring twists and have bending (Coil radius will change).

Resolve Torque =T =  W R  into two components

T cos α=W R cosα will produce twist on all sections of the spring wire

Component T sin α=W R Sinα will cause bending of the spring wire. Hence produce bending moment on the wire. As a result of this bending, the coil diameter tends to reduce.

(i) Determination of axial deflection δ

J = (π/32)d⁴ and I =(π/64)d⁴

½ W δ = Work done under torsion + work done under bending

½ W δ =(1/2) (WR Cos α)ϴ + (1/2) (W R Sin α) Ф

δ =(R Cos α)ϴ+ (R Sin α) Ф =R [Cos α)ϴ+ (Sin α) Ф]

Put the value of θ andφ, we get

    =R [Cos α)(TL/JG)+ (Sin α) (ML/EI)]

    =R [Cos α)( WR Cos α L/JG)+ (Sin α) (W R Sin α L/EI)]

=W R² L[Cos² α/JG)+ (Sin² α/EI)]

=(W R² 2πRn/Cos α)[Cos² α/JG)+ (Sin² α/EI)]

=(2 πW R³ n/Cos α)[Cos² α/JG)+ (Sin² α/EI)]

=(64W R³ n/d⁴Cos α)[Cos² α/G)+ (Sin² α/E)]

δ =(8W D³ n/d⁴Cos α)[Cos² α/G)+ (Sin² α/E)]

(ii) RESILIENCE OPEN COILED HELICAL SPRING

U = ½ W δ= (4W² D³ n/d⁴Cosα)[Cos²α/G)+ (Sin²α/E)]

u =U/V=(4W² D³ n/d⁴Cos α)[Cos² α/G)+ (Sin² α/E)]/Volume of wire

(iii) Shear stress-open coiled helical spring

The component WD Cosα/2 will cause the shear stress in the spring wire.

ԏ =T d/2J = (WD Cosα/2)(d/2)/[(π/32) d⁴]

ԏ  = 8WDCosα/ πd³

 (iv) Bending stress σ_{b, open coiled helical spring}

The component (WD Sin α)/2 will cause the bending stress in the spring wire.

σ_b  =(M/I)y= (M/I)d/2 =[(WD Sin α)/2]x(d/2) / [ (π/64) d⁴]

σ_b  =16WD Sin α /πd³

(v) Maximum Principal Stresses in open coiled helical spring

σ₁ = (1/2) σ_b + (1/2) (σ_b²  + 4 ԏ²)^0.5

 (vi) Maximum shear stress ԏ_max

ԏ_max =(1/2) (σ_b²  + 4 ԏ²)^0.5

(vii)Angle of twist ϴ and angle of end rotation Φ

 WD cosα/2 and WD Sinα/2 both contribute angle of twist

EFFECT OF WD cosα/2

Let WD cosα/2  contributes dϴ₁ in a small length dl

Therefore dϴ^’ = [WD cosα/2] dl/JG

=[WD cosα/2] dl/[(π/32)d⁴G]

dϴ^’= 16WD Cosα] dl/[πd⁴G]

AND for length ‘L’

ϴ^’= 16WD Cosα] dl/[πd⁴G]

ϴ^’ =16WD Cosα] dl/[πd⁴G] L

 Component about X’X’ 

=16WD Cosα] L/[πd⁴G] Cos α

= 16WD Co²sαL/πd⁴G

Its about Y’Y’ will be

= 16WD Cosα] L/[πd⁴E] Sinα

=16WD L/[πd⁴E] Cosα Sinα

EFFECT OF WD Sinα/2

Let dФ be the rotation contributed by WD Sinα/2 in a small length dl

Therefore dФ = [WD Sinα/2] dl/EI

=[WD Sinα/2] dl/[(π/64)d⁴E]

dФ = [32WD Sinα] dl/[πd⁴E]

AND for length ‘L’

Ф = 32WD Sinα] dl/[πd⁴E]

Ф =32WD Sinα] L/[πd⁴E]

Its component about y y axis will be

Ф Cos α

= 32WDL Sinα Cos α /πd⁴E

Component about XX axis

= 32WDLSin²α/πEd⁴

Total twist about the horizontal axis will be

=16WD Co²sαL/πd⁴G + 32WDLSin²α/πEd⁴

= 16WD L/πd⁴(Cos²α/G + 2Sin²α/E)

Total rotation of the free end  about the vertical axis will be

=16WD L/[πd⁴E] Cosα Sinα –32WDL Sinα Cos α /πd⁴E

= 16WD L/(πd⁴) Cosα Sinα[1/G—2/E)

OPEN COILED HELICAL SPRING – AXIAL COUPLE

M is the couple acting to the spring. The applied couple tends to open the spring more. Or decrease the number of turns in the spring.

Positive couple if increases the radius of curvature. Negative bending moment if decreases the radius of curvature. Couple is acting at the free end of the spring. It will have two components M Cos α about yy axis. It will cause bending effect. M Sin α about the XX axis. It will cause twist in the spring wire.

EFFECT OF M Sin α

Consider length dl.

dϴ’ = M Sinα/(π/32)d⁴G dl

dϴ’ = [32M Sinα/πd⁴G] dl

Total twist in length L

ϴ =[32M Sinα/πd⁴G] dl=32M L Sinα/πd⁴G

It will two components.

Component about the xx-axis will be

ϴ₁ =(32MLSinα/πd⁴G)Cosα

= (32 ML/ πd⁴G)Cosα Sinα

Component about yy axis will be

Ф₁ =32MLSin²α/πd⁴G

 EFFECT OF COMPONENT M Cosα

Component about vertical axis

Ф₂=64 M L Cos²α/πd⁴E

Total rotation of the free end will be

Ф₁ + Ф₂ = Ф= 32 M LSin²α/πd⁴G+64 M LCos²α/πd⁴E

Ф = (32 M L/πd⁴)[ Sin²α/G+ 2 Cos²α/E)

 Total angle of twist will be ϴ₁ – ϴ₂

ϴ₁ – ϴ₂  =ϴ

ϴ = (32 M L/ πd⁴G)Cosα Sinα– 64 M L Cosα Sin α/πd⁴E

 The axial deflection will be 

δ=Rϴ=[ 32 M L/ πd⁴G)Cosα Sinα- 64 M L Cosα Sin α/πd⁴E] D/2

δ = (16 M L D/ πd⁴) ( 1/G—2/E)

 STRAIN ENERGY

U = (1/2) MФ= (16M²L/πd⁴)[ Sin²α/G+ 2 Cos²α/E)

 SHEAR STRESS

ԏ  = (MSinα/πd⁴) (d/2)=16MSinα/πd³

ԏ  =16MSinα/πd³

 BENDING STRESS

σ_b = (MCosα /πd⁴(d/2) = 32 M Cosα / πd³

σ_b = 32 M Cosα / πd³

Put α=0, we get the results of close closed helical springs from open coiled helical springs.

https://www.mesubjects.net/wp-admin/post.php?post=7618&action=edit      Close coiled under axial force

https://www.mesubjects.net/wp-admin/post.php?post=2273&action=edit           Q. ANS. Springs