4 static shear

Static Shear

(І) Direct shear

البرشام .1 مسامير صنع فى المستخدمة المعادن على يجرى

Single shear Double shear Punching shear

Smooth

Rough

= = =

1

τ P n A τ P

2 n Aτ P

πdt

PP PP/2

P/2

rτ

J L T G θ

E 2(1+υ)

Shear stress (τ) Shear strain (γ)

(ІІ) Torsion test

و -1 الموتور ادارة اعمدة مثل المنشات بعض فى االلتواء يحدثالطائرة مروحة عمود

و -2 دوران عنه ينشأ قص لقوى المقطع يتعرض االلتواء حالة فى. االخر المقطع على محوره حول المقطعين احد انزالق

= =

Where:τ : Shear stress T: Applied torque L: length r: Radius (D/2) θ: Angle of twist(rad) rad = deg x (π/180) J: (π/32)D4

G: Shear modulus of rigidity = =

Shear stress distribution:

Solid section

Hollow section

Failure shape:

Ductile materials (Steel) Brittle materials (Cast iron)

2

T

L

D θ

τmax

τmax

Tp.l x rJ

16 Tp.l π D3

J L Tp.l

G θp.l

Right sectionShear strength < Tensile strength

اجهادات المادة تتحمل والقص المستو فى الكسر عمودى ىيحدث

العينة محور على

Helicoidal sectionTensile strength < shear strength

المادة تتحمل الالشد ويحدثاجهادات

مستوى على الكسر مع 45º بزاوية يميل

عليه الذى المستوىاقصى

محور مع اى قص اجهادالعينة

القوانين

1- Shear stress at proportional limit (τp.l)

τp.l = =

2- Shear modulus of rigidity (G)

It is the property of a material to resist any type of deformation due to shear. It represents the stiffness of material in shear. For most materials E =2.5 G.

القص قوى بفعل للتشكل المادة مقاومة

=

3- Modulus of resilience

M.R =

4- Modulus of toughness

3

½ Tp.l x θp.l

Aо Lо

Lf - Lо

Lо

M.T =

5- Ductility

%Lf = Lо + (rθmax)2

6- Modulus of rupture in shear (τmax)

For solid sections For hollow sections

Ductile material Brittle material

t = thickness = (D-d)/2r = mean radius = D+d/4

القصفة المواد حالة فىللمادة Shear strengthقيمة الحقيقية القيمة تمثل ال االختبار من الناتجة

بسبب كسر لها يحدث العينة لقيمة Tensionالن الوصول max shearقبلstrength الحقيقية

7- Power (P)

P =

P: watt T : N/m2 n : rpm(rev/min)

خاصة حاالت4

⅔ Tmax x θmax

Aо Lо

2πnT 60

12 Tmax πD3

14 Tmax πD3

Tmax 2πtr2

D1 D3 D2

G J1

T1 L1

G J2

T2 L2

G JT L

JT x r

T1 = T2 = Tθtot = θ1 + θ2

T = T1 + T2 θ1 = θ2 = θ

Example:-The shaft bar shown is fixed at its two ends and subjected to twisting moment of 9 ton.m applied at point of thickness change. Calculate the maximum stress in the bar and the angle of twist at point O. take shear modulus G= 8 x 105 Kg/cm2

A O B

Solution

θ1 = θ2 =

J1 = π/32(20)4 J2 = π/32(10)4 L1 = 200 cm L1 = 100 cm

So T1 = 8 T2 T1 + T2 = 9 → T1 = 8 ton.m & T2 = 1 ton.m θ = τ =

5

D1D2

L1L2

( ) A torsion test was carried out on a solid mild steel rod specimen of 16 mm diameter and 160 mm length. The results of the test until rupture were as follows: - The torque at elastic limit is 2000 kg.cm - The maximum torque is 4500 kg.cm- The angle of twist at elastic limit is 20 degree- The maximum angle of twist is 600 degreeFind: a- The elastic shear strength b- The ultimate shear strengthc- The modulus of toughness d- The modulus of rigidity

Elastic limit Maximum Torque, T, kg.cm 2000 4500

Angle of twist, θ, deg 20 600

6

1.6 cm L = 16 cm J = π/32 D4 = π/32 (1.6)4 = 0.64 cm4 r = D/2 = 0.8 cm

r

e.l

J Te.l=

Elastic shear strength= l.e

J T l.er x = 8.0 x 0002

46.0= 0052 mc/gk 2

htgnerts raehs etamitlU= xam

D 3

T 21 xam = 0054 x 21 (6.1) 3

= 6914 mc/gk 2

(G) ytidigir fo suludoM= θ x J l.e

T l.eL x = 61 x 0002 46.0 x 02 x 081/

= 01 x 3.41 5 gk /mc 2

ssenhguot fo suludoM= ⅔ Txamθ x xam

A о L о= = gk 779 /mc 2⅔ 006 x 0054 x x 081/

π/4(6.1) 261 x

J T l.e

Lθ G l.e=

( ) A torsion test was carried out on a mild steel specimen of 3 cm diameter and 20 cm length. The following data was recordedTorque (Kg.cm) 0 1000 2000 2500 3000 3500 4000 4500θ, degree 0 10 20 30 50 100 250 600Draw Torque-angle of twist diagram and find the following:

i- Elastic shear strengthii- Modulus of rigidityiii- Operating stress in shear if the factor of safety is 2iv- Ultimate shear strengthv- Discuss the fracture shape of test specimenvi- Draw the shear stress distribution on the cross section

7

3 cm L = 20 cm J = π/32 D4 = π/32 (3)4 = 7.96 cm4 r = D/2 = 1.5 cm

0

1000

2000

3000

4000

5000

0 100 200 300 400 500 600 700

T

θ

r

e.l

J Te.l=

L

G θe.l=

Modulus of rigidity (G)= J x θe.l

Te.l x L = 2000 x 20

7.96 x 20 x /180= 1.44 x 104 kg/cm2

Failure shape is right section due to lower shear stress resistance

( ) A torsion test was carried out on a mild steel rod specimen of 200 mm length. The results of the test until rupture were as follows:T, KN.cm 0 10 20 25 30 35 40 45θ, in degree 0 10 20 30 50 100 250 600Draw T – θ diagram and find:

a- The diameter of the rod if the modulus of rigidity equals 90 KN/cm2 b- Elastic shear strength c- Modulus of toughnessd- Ultimate shear strength

8


J T l.er x = 5.1 x 0002

69.7= 773 mc/gk 2

sserts gnitarepO= po

S.F= 773

2= 5.881 mc/gk 2

l.e


D 3

T 21 xam = 0054 x 21 (3) 3

= 736 mc/gk 2

0

01

02

03

04

05

06

0 001 002 003 004 005 006 007

T

θhtgnerts raehs citsalE= l.e

J T l.er x = 7.1 x 02

21.31= 6.2mc/NK 2


D 3

T 21 xam = 54 x 21(4.3) 3 = 4.4 mc/NK 2


A о L о= =NK37.1 /mc 2⅔ 006 x 54 x x 081/

π/4(4.3) 202 x

rl.e

J T l.e=

Lθ G l.e=


T l.eL x =02 x 02

π/23D 4 x 02 x 081/= 09 NK /mc 2

mc Dmc 02 = L = J π/23D 4 = π/23(4.3) 4mc 21.31 = 4 mc 7.1 = 2/D = r

mc 4.3 = D

( ) A torsion test was carried out on a hollow mild steel rod specimen of 15 mm diameter and 150 mm length. The results of the test until rupture were as follows:

T, ton.cm 0 0.4 0.8 1 1.2 1.4 1.6 1.75 1.9 2θ, in degree 0 4 8 20 50 100 150 250 500 800

Draw T – θ diagram and find:i- Elastic shear strength ii- Modulus of toughnessiii- Modulus of rigidity iv- Modulus of Elasticity if poisson ratio is 0.3v- Ultimate shear strength vi- Discuss the fracture shape of test specimen

9

1.5 cm L = 15 cm J = π/32 D4= π/32 (1.5)4 = 0.5 cm4 r = D/2 = 0.75 cm

0

0.5

1

1.5

2

2.5

0 100 200 300 400 500 600 700 800 900

T

θ

Modulus of toughness= ⅔ Txamθ x xam

A о L о= =not 7.0 /mc 2⅔ 008 x 2 x x 081/

π/4(5.1) 251 x

htgnerts raehs citsalE= l.e

J T l.er x = 57.0 x 8.0

5.0= 2.1mc/not 2

rl.e

J T l.e=

Lθ G l.e=


T l.eL x =51 x 8.0

5.0 x 8 x 081/= not 271 /mc 2

(E) yticitsale fo suludoM= +1)2 x G υ ( 271 =(3.0 + 1) x 2 x = not 2.744 /mc 2

E +1)2 υ (

= G


D 3

T 21 xam = 2 x 21(5.1) 3 = not 3.2 mc/ 2

Failure shape is right section due to lower shear stress resistance( ) A torsion test was done on a mild steel rod specimen of diameter 30 mm and

length 20 cm. The relationship between the torque (kg.cm) and angle of twist in degree until rupture was as follows:Torque (Kg.cm) 0 1000 2000 2500 3000 3500 4000 4500θ, degree 0 10 20 30 50 100 250 600Draw the torque-angle of twist diagram and find:

i- Elastic shear strengthii- Operating stress in shear if the factor of safety is 3iii- Shear modulus or modulus of rigidity iv- Ultimate shear strengthv- Draw a shape of fracture of specimen and draw the shear stress

distribution on the cross – sectional area

10

0500

10001500

20002500

30003500

40004500

5000

0 100 200 300 400 500 600 700

T

θ

3 cm L = 20 cm J = π/32 D4= π/32 (3)4 = 8 cm4 r = D/2 = 1.5 cm

Failure shape is right section due to lower shear stress resistance

( ) A torsion test was carried out on a mild steel rod of 15 mm diameter and 15 cm length. The applied torque in kg.cm and the measured angle of twist in degree until rupture was as follows:T (Kg.cm) 0 800 1600 2000 2400 2800 3200 3500 3800 4000θ (degree) 0 8 16 40 100 200 300 500 1000 1600Draw the torque-angle of twist diagram and find:

a) Elastic shear strengthb) Operating stress in shear if the factor of safety is 3c) Modulus of rupture d) Ductilitye) Modulus of resilience f) Modulus of toughnessg) Draw the shear stress distribution on the rod cross – section and shape

of fracture

11


J T l.er x = 5.1 x 0002

8= gk 573 mc/ 2

rl.e

J T l.e=

Lθ G l.e=


S.F= 573

3= 521 mc/gk 2

l.e


T l.eL x =02 x 0002

8 x 02 x 081/= not 3.41 /mc 2


D 3

T 21 xam = 0054 x 21 (3) 3 = gk 6.636 mc/ 2

0

005

0001

0051

0002

0052

0003

0053

0004

0054

0 002 004 006 008 0001 0021 0041 0061 0081

T

θ

mc 5.1mc 51 = L = J π/23D 4 = π/23(5.1) 4mc 5.0 = 4 mc 57.0 = 2/D = r

( ) A torsion test was done on a metal rod specimen of diameter 15 mm, and its modulus of rigidity is 1.731 x 105 kg/cm2. The relationship between the torque (ton.cm) and angle of twist (θ) in degree until rupture was as follows:T (ton.cm) 0 0.4 0.8 1 1.2 1.4 1.6 1.75 1.9 2θ (degree) 0 4 8 20 50 100 150 250 500 800Draw the torque-angle of twist diagram and find:

Stress at proportional limit Length of the rod Modulus of toughness

12


J T l.er x =

57.0 x 0061 5.0

= gk 0042 mc/ 2

rl.e

J T l.e=

Lθ G l.e=


S.F=

0042 3= 008 mc/gk 2

l.e

erutpur fo suludoM= xam

D 3

T 21 xam = 0004 x 21 (5.1) 3 = gk 7254 mc/ 2


A о L о= =not 8.2 /mc 2

⅔ 0061 x 0004 x x 081/

π/4(5.1) 251 x

ecneiliser fo suludoM= T ½ l.eθ x l.e

A о L о= =gk 4.8 /mc 2 ½ 61 x 0061 x x 081/

π/4(5.1) 251 x

0

5.0

1

5.1

2

5.2

0 001 002 003 004 005 006 007 008 009

T

θ

mc 5.1mc ??? = L = J π/23D 4 = π/23(5.1) 4mc 5.0 = 4 mc 57.0 = 2/D = r

rl.e

J T l.e=

Lθ G l.e=


T l.eL x =L x 008

5.0 x 8 x 081/= 01x137.1 5mc/gk 2

mc 1.51 = L

timil lanoitroporp ta ssertS= l.e

J T l.er x =

57.0 x 008 5.0

= gk 0021 mc/ 2


A о L о= =gk 207 /mc 2

⅔ 008 x 0002 x x 081/

π/4(5.1) 251 x

( ) The given table shows the results of torsion test until rupture of mild steel hollow specimen of outside diameter 2 cm and inside diameter 1 cm and length 30 cm. Draw the torque – angle of twist (T- θ) curve considering 1 cm of torque axis representing 5 kg.m, and 1 cm of θ- axis representing 100°. Using torque results, find:Torque (Kg.cm) 0 10 20 25.8 30 35 40 45 50 53θ in degree 0 1.39 2.78 3.6 30 100 350 750 1120 1250(i) Shearing stress at proportional limit using torsion formula(ii) Operating stress in shear if the factor of safety is 3(iii) Shear modulus or modulus of rigidity (iv) Modulus of resilience in shear(v) Shearing modulus of rupture or shearing strength

13

L = 30 cm J = π/32 (D4 – d4) = π/32 (24 – 14) = 1.47 cm4 r = D/2 = 1 cm 21

T

051015202530354045505560

0 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400

θ

= 7.5 kg/cm2

( ) A steel hollow shaft of inner diameter equals half the outer diameter and equal 150cm length was subjected to twisting moment 500 ton.cm. If the modulus of Elasticity (E) of steel is 2.1 x 104 Kg/mm2 and poisson ratio (υ) is 0.3. Determine the external diameter:

(i) If the limiting shear stressing the steel is 9 kg/mm2

(ii) If the allowable angle of twist is 1º for 20cm length

Limiting shear stressing the steel is 9 kg/mm2

14

r

e.l

J Te.l=

L

G θe.l=

J Te.l x r =

25.8 x 1 1.47

Stress at proportional limit= l.e =gk 55.71 mc/ 2


S.F=

55.71 3= 58.5 mc/gk 2

l.e


T l.eL x =03 x 8.52

6.3 x 74.1 x 081/= mc/gk 0838 2

ecneiliser fo suludoM= T ½ l.eθ x l.e

A о L о= =gk 10.0 /mc 2 ½ 6.3 x 8.52 x x 081/

π/42) 21 – 203 x (

erutpur fo suludoM= xam =35

2(5.1) x 5.0 x 2

Txam rtπ2 2

suidar naem :r mc 5.0 = (1 – 2)½ = ssenkciht :t = 5.1 = (1 + 2)½

mc 051 = L = J π/23(D2)) 4(D) – 4 = ( π/23D 51) 4 ( D = 2/D2 = rD2

D01 x 5 = 01 x 0001 x 005 = T 6 mm .gK

01 x 1.2 = E 4mm/gK 23.0 = υ

(G) ytidigir fo suludoM= =01 x 1.2 4

(3.0 + 1)2= 7708 /gk mm 2 E

+1)2 υ (

rl.e

J T l.e=

Lθ G l.e=

J T l.e

Lθ G l.e=

rl.e

J T l.e=

D9

π/23D 51) 4( 01 x 5 6=

3.27 = Dmm 5.441 = D2 = retemaid retuO mm

01 x 02

x 1 x 7708 081/π/23D 51) 4( 01 x 5 6 =

58.64 = Dmm 7.39 = D2 = retemaid retuO mm

Allowable angle of twist is 1º for 20cm length

( ) A shaft of a material having an allowable shear stress of 60 N/mm2 transmits a torque of 10 KN.m. Assume linear elastic behavior, draw the shear stress distribution and get the shaft dimensions for the following cases:(i) Solid circular shaft

(ii) Hollow circular shaft with inner diameter equal 7/8 the outer diameter

Solid circular shaft

Hollow circular shaft with inner diameter equal 7/8 the outer diameter

15

J = π/32 ((D)4 – (⅞D)4) = π/32 x D4 (1– ( ⅞)4) = 0.04 D4 r = D/2

D

d

T = 10 x 1000 x 1000 = 107 N. mm e.l = 60 N/mm2

DJ = π/32 D4 r = D/2

r J Te.l=

e.l

D/2

60 π/32 (D)4 107=

Outer diameter (D) = 94.7 mm

r J Te.l=

e.l

D/2

60 0.04 (D) 4 107=

Outer diameter (D) = 127.7 mm inner diameter = 111.8 mm

τmax

τmax

(a) (b)

6 ton6 ton 6 ton3 ton

3 ton

( ) Determine the number of rivets required to transmit a load of 6 ton in two cases of riveted connections shown in figures (a) and (b), If the diameter of rivet is 16 mm and the allowable shear stress in rivet is 0.84 ton/cm2

Case (a) Single shear

16

= P

n A= 6

n x π/4 (1.6)2

0.84

n = 4

= P

2n A

= 6

2n x π/4 (1.6)2

0.84

n = 2

Case (b) Double shear

( ) Determine the number of rivets to transmit a load of 12 ton from two angles to a plate as shown in figure. If the diameter of rivet is 20 mm and the allowable shear stress in rivet is 0.8 ton/cm2. Assume the failure is in the rivets and the rivets carry the same load.

Double shear

( ) A shaft is transmitting 130 Hp at 180 rpm. If the allowable stress in the material is 600 kg/cm2. Determine the suitable diameter for the shaft. Take G= 8 x 105 kg/cm2

17

12 ton12 ton

= P

2n A

= 12

2n x π/4 (2)2

0.8

n = 3

( ) A shaft 120 mm in diameter and 0.75 m long has a concentric hole drilled for a portion of its length. Find the diameter and maximum length of the hole so that when the shaft is subjected to a torque of 1.67 KN.m, the maximum

18

G = 8 x 105 kg/cm2e.l = 600 kg/cm2

DJ = π/32 D4 r = D/2

2πnT 60 x 75

P =

= 517.3 N .m = 5173 kg.cm130 x 60 x 75

2π x 180 T =

r J Te.l=

e.l

D/2

600 π/32 (D) 4 5173=

Diameter (D) = 3.53 cm

shearing stress will not exceed 7.5 MN/m2 and the total angle of twist will not be greater than 1.5 (Take G= 80 GN/m2)

For part (II)

For the total shaft

( ) A hollow steel shaft of 400 mm external diameter transmits 9 MW at 120 rpm. If the angle of twist measured over a length of 2 m is 0.45° and G = 80 GN/m2, estimate the internal diameter of the shaft and the maximum

19

L = 750 cm

T = 1.67 x 106 N.mm

G = 8 x 104 N/mm2

l.e =mm/N 5.7 2

dar 5.1 = θ

mm

021

= D

mm 057 = L

L h d hIII

= r mm 06 = 2/021 = 2/D = J π/23(021)[ 4d) – h(4 ]

rl.e

J T l.e=

Lθ G l.e=

J T l.e

Lθ G l.e=

rl.e

J T l.e=

06

5.7π/23(021)[ 4d) – h(4]

01 x 76.1 6=

d hmm 29 =

θtotθ = 5.1 = Iθ + II

T L G J

θ=

L – 057) h( 01 x 76.1) x 6(

π/23(021) 4 x 01 x 8 45.1 = +

L) h( 01 x 76.1) x 6(π/23(021)[ 4(29) – 4] x 01 x 8 4

L hmm =

shearing stress. Find the diameter of the solid shaft, which will transmit the same power at the same maximum stress.

For the solid shaft

20

L = 2000 mm J = π/32 ((400)4 – (d) 4) 400 mm

d

G = 8 x 104 N/mm2θ = 0.45°

r = D/2 = 200 mm

= 7.2 x 108 N .mm 9 x 106 x 60

2π x 120 T =

e.l = ???

2πnT60

P =

J

T L

G θ=

7.2 x 108

π/32 ((400)4 – (d) 4)=

8 01 x 4) x 54.0 x π/081 (

0002

mm 812 = (d) retemaid lanretnI

rl.e

J T l.e=

=r x T

J=

2.7 01 x 8 002 x

π/23(004)) 4(812) – 4 (48.26mm/N 2=

=r x T

J=

2.7 01 x 8 2/D x

π/23D) 4(48.26mm/N 2=

mm 4.78 = (D) retemaiD

4 static shear

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