MP3002/MP4102 - Mechanics of Deformable Solids, Part 2: Energy Methods and Theory of Elasticity

Tutorial 12 Theory of Elasticity (Application of equilibrium equations in Cartesian coordinates, applications of Lames

equations)

Q12.1 A beam of depth d, thickness b (measured perpendicular the plane of the paper), and length L is simply-supported at t he ends and carries a uniformly distributed load w units per unit length over the whole span as shown in Fig. 12.1.

Fig. 12.1

Assume that x due to bending moment M at any cr oss-section is distributed as x = My/I where I is the second moment of cross-sectional area about the centroidal axis.

a) Starting with the equilibrium equations in Cartesian coordinates, derive expressions for the stresses (y, xy).

Ans: IQd

IQy

xy 82

22

where dxdMQ ,

2486

323 dydyIw

y

b) Show that the maximum vertical normal stress y is 4/3( d/l )2 times the maximum bending stress x at the mid-span, and therefore in most cases may be considered insignificant in relation to the latter.

y

x

w/units length

FujitsuText Box

FujitsuText BoxPg 1/5

FujitsuText Box

FujitsuText Box Q12.1-Theory of Elasticity

FujitsuRectangle

FujitsuPencil

FujitsuPencil

FujitsuPencil

FujitsuPencil

FujitsuPencil

FujitsuRectangle

FujitsuText BoxPg 2/5

FujitsuText Box

FujitsuPencil

FujitsuPencil

FujitsuPencil

FujitsuText BoxPg 3/5

FujitsuText Box

FujitsuPencil

FujitsuPencil

FujitsuText BoxPg 4/5

FujitsuText Box

FujitsuPencil

FujitsuPencil

FujitsuPencil

FujitsuPencil

FujitsuPencil

FujitsuPencil

FujitsuPencil

FujitsuPencil

FujitsuPencil

FujitsuText BoxPg 5/5

FujitsuText Box

Q12.2 Theory of Elasticity Pg 1/2

Q12.2 In a pressure test on a hydraulic cylinder of 120 mm external diameter and 60mm internal diameter, the hoop and longitudinal strains are measured by means of strain gauges on the outer surface and found to be 266 x 10-6 and 69.6 x 10-6 respectively for an internal pressure of 100 MN/m2. Determine the experimental hoop stress at the outer surface and compare this result with the calculated value. Determine also the safety factor for the cylinder according to the maximum shear stress theory. Take y =280 MN/m2; E=208GN/m2; v=0.29

Given: op = atmospheric pressure; =ip 100 MN/m2; mmri 30= ; mmro 60= Actual Stresses (from Experiment at Outer Surface)

( )rzEEE +== 06266 (1) ( )rzz EEE

+== 066.69 (2) Note: Outer surface = atmospheric pressure 0=r Solved (1) and (2) to get:

Longitudinal stress =z 33.33 MN/m2

Hoop stress = 65 MN/m2

FujitsuPencil

FujitsuStamp

FujitsuPencil

FujitsuPencil

FujitsuText BoxQ12.2

FujitsuText Box12.2

FujitsuPencil

Q12.2 Theory of Elasticity Pg 2/2

Lame's Eqs

Radial stress: 2rBAr =

Hoop stress: 2rBA +=

Determination of parameters of Lame's Eqs by boundary conditions: At the inner and outer surfaces of the cylinder At r = 0.03m, r = - 100 MN/m2; At r = 0.06m, r = 0 MN/m2 Lame's Eqs 2r

BAr = A = 33.33 MN/m2, B = 0.12 MN Calculation of the hoop stress on outer surface

At r = 0.06m, 2rBA += 66.67= MN/m2

The calculated value is 2.6% higher than experiment value The worst stress situation occurs at the bore

At r = 0.03m, 2rBA += 166.67= MN/m2 r = -100 MN/m2

z = 33.33 MN/m2 (from experiment) Determination of safety factor Maximum shear stress in cylinder max = ( )r 5.0 = 133.3MN/m2 In simple tension, allowable shear stress max = y5.0 = 140 MN/m2 The safety factor 3.133140=SF = 1.05

MP3002/MP4102 - Mechanics of Deformable Solids, Part 2: Energy Methods and Theory of ElasticityQ12.3 A steel tube has an int ernal diameter of 25mm and an external diameter of 50mm. Another tube of the same material (steel), is to be shrink-fitted over the outside of the first tube so that the shrinkage stresses just produce a condition of yield at the inner surface of each tube. Assume that yielding occurs according to the maximum shear stress theory and that no axial stresses are set up due to shrinking. Take y = 414MN/m2; E = 207GN/m2.

Fig. 12.3

Determine the necessary difference in diameters of the mating surfaces before shrinking and the required external diameter of the outer tube.

Ans: 0.125 mm; 100 mm;

mm 25

0op

0ip

mm 05

Q12.3 Theory of elasticity

Shrink fit analysis Steel tubes

Given: ir = 12.5 mm; mr = 25 mm. Determine: 1) Outer radius or 2) Diametral interference Lames Equations

Inner Tube: 2rBArA = 2r

BAA += Outer Tube: 2r

DCrB = 2rDCB +=

Boundary Conditions: Bore:

ir =0.0125 m 0=rA ( ) yArA 5.05.0 =

A = -207 MN/m2 B = -0.032344 MN

Interface:

mr =0.025 m rBrA = ( ) yBrB 5.05.0 =

C = 51.75 MN/m2 D = 0.129375 MN

i.e All stresses are known, since the constants A, B, C & D are known.

FujitsuPencil

FujitsuPencil

FujitsuPencil

FujitsuPencil

FujitsuStamp

FujitsuPencil

FujitsuPencil

FujitsuPencil

FujitsuStamp

FujitsuStamp

FujitsuRectangle

FujitsuPencil

FujitsuStamp

Q12.3 Theory of elasticity

Find Outer Radius ( )or of the tube

Condition: At r= or , 20o

r rDC ==

Solve to get: or = 0.05m Find radial interference

BA += AmBm rr = ( )ABmEr

= From Lames eqn, at the interface: B = 258.75 MN; A = -258.75 MN Substitute to get = 0.0625 mm Diametral interference = 2 = 0.125 mm.

FujitsuStamp

FujitsuStamp

FujitsuStamp

FujitsuStamp

FujitsuStamp

FujitsuRectangle

FujitsuPencil

FujitsuPencil

FujitsuPencil

FujitsuPencil

FujitsuStamp

FujitsuStamp

FujitsuStamp

FujitsuStamp

FujitsuStamp

MP3002/MP4102 - Mechanics of Deformable Solids, Part 2: Energy Methods and Theory of ElasticityQ12.4 A steel tube of 30cm external diameter is shrunk on to a steel shaft of 8cm nominal diameter with a shrink fit allowance of 0.024mm. Under working conditions, the tube is subjected to a radial tensile stress of 12MPa on its outer surface, while the shaft carries an axial compressive stress of 9MPa. The axial stress in the tube is zero. The Young's Modulus is 208GPa and the Poisson's ratio is 0.3. You may proceed from Lame's equations.

Fig. 12.4

(a) Calculate the interface pressure, P, between the shaft and the tube. Ans: P = 47.22MPa

(b) Determine the maximum hoop and radial stresses in the assembly under such working conditions. Ans: rmax = -47.22 MPa, max = 80.3MPa

12MPa

Q12.4 : Theory of elasticity Page 1/4

Given: (Notations: s=shaft, t=tube )

Shink Fit Allowance st = = 0.024mm Tube: Axial stress ( ) 0=tz Radial stress ( )tor = 12 MN/m2 Outer radius ot = 0.15 m Inner radius it = 0.04 m Shaft: Axial stress ( )sz = -9 MN/m2 Radius sr = 0.04 m

Find: Interface pressure P, maximum hoop & radial stresses

At the interface, ( ) ( ) Psortir ==

Hoop strains ( )rzEEru +==

shaft: ( )04.0s

s = tube: ( ) 04.0 tti

= Lames equations

Tube 2rBAr = 2r

BA+=

Shaft 2rDCr = 2r

DC += Note: Unknowns: ( A, B, C , D, ts , & P ) Initially solve for (A, B, C & D) in terms of interface pressure P. Then use interference = 0.024m to solve for P.

FujitsuRectangle

FujitsuStamp

FujitsuStamp

FujitsuStamp

FujitsuStamp

Q12.4 : Theory of elasticity Page 2/4

Boundary conditions for tube ( Solve for A & B )

At outer radius: ( ) 215.012BAtor ==

At inner radius: ( ) 204.0BAPtir ==

Solve to get: PA 07656.092.12 +=

56.580

12 PB += Boundary conditions for the solid shaft (Solve for C & D)

At outer radius: ( ) 204.0DCPsor ==

At the centre: ( )0DCsir =

0=D for the radial stress ( )sir to be finite. Therefore: PC = Note: The constants A, B, C & D are determined in terms of the interface pressure P.

Hence, the hoop stresses and radial stresses for the tube and solid shaft are now known in terms of the interface pressure P.

Q12.4 : Theory of elasticity Page 3/4 Page 3/4

Solving for the interface P from the hoop strains equations & shrink fit allowance of 0.024 m. Hoop strain at the interface (Tube)

( ) ( ) ( ) ( )[ ]tiztirtitti EE +== 04.0

Note: ( ) 0=tiz ; ( ) ( )204.0BAti += ; ( ) ( )204.0

BAtir = Substitute and solve to get the displacement of the tube outwards as:

( )Pt 453.1837.2510*923.1 7 += Hoop strain at the interface (Solid Shaft)

( ) ( ) ( ) ( )[ ]sozsorsosso EE +== 04.0 Note: ( ) 9=soz ; ( ) PCso == ; ( ) PCsor == Substitute and solve to get the displacement of the shaft inwards as:

( )Ps 7.07.210*923.1 7 =

mmst 024.0== 2/22.47 mMNP =

( )PP 7.07.2453.1837.2510*923.110*024.0 73 ++=

FujitsuStamp

FujitsuStamp

FujitsuStamp

FujitsuStamp

FujitsuStamp

FujitsuStamp

Q12.4 : Theory of elasticity Page 4/4 Page 4/4

Determine maximum hoop and radial stresses for the assembly. Lames equations

Tube 2rBAr = 2r

BA+= Shaft PCr == PC ==

For the Tube:

Radial stress: ( ) 2 12 mMNtor = (Given)

( ) 2 22.47 mMNPtir == Maximum

Hoop stress: ( ) 2 07.21 mMNto =

( ) 2 3.80 mMNti = Maximum For the solid shaft

( ) ( ) ( ) ( ) 2 22.47 mMNPsosisorsir =====

Binder1.pdfQ11.1.pdfQ11.1-Theory of Elasticity

Binder2Binder3Q11.3.pdfQ11.3-Theory of Elasticity

Binder4Q11.4.pdfQ11.4-Theory of Elasticity