# 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)d^{4} and I =(π/64)d^{4}

#### 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 R^{2} L[Cos^{2} α/JG)+ (Sin^{2} α/EI)]

#### =(W R^{2} 2πRn/Cos α)[Cos^{2} α/JG)+ (Sin^{2} α/EI)]

#### =(2 πW R^{3} n/Cos α)[Cos^{2} α/JG)+ (Sin^{2} α/EI)]

#### =(64W R^{3} n/d^{4}Cos α)[Cos^{2} α/G)+ (Sin^{2} α/E)]

#### δ =(8W D^{3} n/d^{4}Cos α)[Cos^{2} α/G)+ (Sin^{2} α/E)]

**(ii) RESILIENCE OPEN COILED HELICAL SPRING**

#### U = ½ W δ= (4W^{2} D^{3} n/d^{4}Cosα)[Cos^{2}α/G)+ (Sin^{2}α/E)]

#### u =U/V=(4W^{2} D^{3} n/d^{4}Cos α)[Cos^{2} α/G)+ (Sin^{2} 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) d^{4}]

#### ԏ = 8WDCosα/ πd^{3}

** **(iv) **Bending stress σ**_{b, open coiled helical spring}

_{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) d^{4}]

#### σ_{b} =16WD Sin α /πd^{3}

**(v) Maximum Principal Stresses in open coiled helical spring**

#### σ_{1} = (1/2) σ_{b} + (1/2) (σ_{b}^{2} + 4 ԏ^{2})^{0.5}

** ****(vi) Maximum shear stress ԏ**_{max}

_{max}

#### ԏ_{max }=(1/2) (σ_{b}^{2} + 4 ԏ^{2})^{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 dϴ_{1 }in a small length dl

#### Therefore dϴ^{’} = [WD cosα/2] dl/JG

^{’}

#### =[WD cosα/2] dl/[(π/32)d^{4}G]

#### dϴ^{’}= 16WD Cosα] dl/[πd^{4}G]

^{’}

#### AND for length ‘L’

#### ϴ^{’}= 16WD Cosα] dl/[πd^{4}G]

^{’}

#### ϴ^{’} =16WD Cosα] dl/[πd^{4}G] L

^{’}

** Component about X’X’ **

#### =16WD Cosα] L/[πd^{4}G] Cos α

**=** 16WD Co^{2}sαL/πd^{4}G

#### Its about Y’Y’ will be

#### = 16WD Cosα] L/[πd^{4}E] Sinα

#### =16WD L/[πd^{4}E] 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)d^{4}E]

#### dФ = [32WD Sinα] dl/[πd^{4}E]

#### AND for length ‘L’

#### Ф = 32WD Sinα] dl/[πd^{4}E]

#### Ф =32WD Sinα] L/[πd^{4}E]

#### Its component about y y axis will be

#### Ф Cos α

#### = 32WDL Sinα Cos α /πd^{4}E

**Component about XX axis**

#### = 32WDLSin^{2}α/πEd^{4}

#### Total twist about the horizontal axis will be

#### =16WD Co^{2}sαL/πd^{4}G + 32WDLSin^{2}α/πEd^{4}

#### = 16WD L/πd^{4}(Cos^{2}α/G + 2Sin^{2}α/E)

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

#### =16WD L/[πd^{4}E] Cosα Sinα –32WDL Sinα Cos α /πd^{4}E

#### = 16WD L/(πd^{4}) 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)d^{4}G dl

#### dϴ’ = [32M Sinα/πd^{4}G] dl

#### Total twist in length L

#### ϴ =[32M Sinα/πd^{4}G] dl=32M L Sinα/πd^{4}G

**It will two components.**

**Component about the xx-axis will be **

#### ϴ_{1} =(32MLSinα/πd^{4}G)Cosα

#### = (32 ML/ πd^{4}G)Cosα Sinα

#### Component about yy axis will be

#### Ф_{1} =32MLSin^{2}α/πd^{4}G

** ****EFFECT OF COMPONENT M Cosα**

#### Component about vertical axis

#### Ф_{2}=64 M L Cos^{2}α/πd^{4}E

#### Total rotation of the free end will be

#### Ф_{1} + Ф_{2} = Ф= 32 M LSin^{2}α/πd^{4}G+64 M LCos^{2}α/πd^{4}E

#### Ф = (32 M L/πd^{4})[ Sin^{2}α/G+ 2 Cos^{2}α/E)

** ****Total angle of twist will be ϴ**_{1} – ϴ_{2}

_{1}– ϴ

_{2}

#### ϴ_{1} – ϴ_{2 } =ϴ

#### ϴ = (32 M L/ πd^{4}G)Cosα Sinα– 64 M L Cosα Sin α/πd^{4}E

** ****The axial deflection will be **

#### δ=Rϴ=[ 32 M L/ πd^{4}G)Cosα Sinα- 64 M L Cosα Sin α/πd^{4}E] D/2

#### δ = (16 M L D/ πd^{4}) ( 1/G—2/E)

** ****STRAIN ENERGY **

#### U = (1/2) MФ= (16M^{2}L/πd^{4})[ Sin^{2}α/G+ 2 Cos^{2}α/E)

** ****SHEAR STRESS**

#### ԏ = (MSinα/πd^{4}) (d/2)=16MSinα/πd^{3}

#### ԏ =16MSinα/πd^{3}

** ****BENDING STRESS **

#### σ_{b} = (MCosα /πd^{4}(d/2) = 32 M Cosα / πd^{3}

#### σ_{b} = 32 M Cosα / πd^{3}

#### 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