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UNIT-I STRESS, STRAIN

1. A Member A B C D is subjected to loading as shown in fig determine the totalelongation. Take E= 2 x105 N/mm2

Given dataYoung’s modulus E= 2 x105 N/mm2

Area1=900mm2

Area2=400mm2

Area3=625mm2

Length1=1500mmLength2=1000mmLength3=750mmTo find

Total elongationSolutionConsider tensile force +ve [increasing length]Compressive force –ve [decreasing length]Step1Consider part AB:To calculate the total load acts on suction AB (P1)Left side load = right side load

Step2Consider part BC:To calculate the total load acts on suction BC (P2)

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Step3Consider part CD:To calculate the total load acts on suction CD (P3)

Step 4To calculate the change in length = + +

2. A reinforced concrete column 300 x 300 mm has 4 reinforcing steel bars of 25mmdiameter in each corner. Find the safe axial load on the column when concrete subjectedto a stress of 5N/mm2what is the corresponding stress in steel take Es/Es=18

Given dataArea of the column =300x300

Area of the steel bar=

Stress on concrete =5N/mm2

Es/Es=18To findLoad (P)Stress in steelSolutionStep: 1To calculate total load on steel=

3

Step: 2To calculate total load on concrete=

Step: 3To calculate total load on the beam= +

Step: 4Total Stress in steel=

3. A rod of made brass, copper and aluminum, as shown in fig is held between two rigidsupports at A and D calculate the stresses developed in each material when thetemperature of the system is raised by 40oC. Take Es=2.1x105 N/mm2. Ec=1.1x105

N/mm2. Ea=0.7x105 N/mm2 = 12x10-6/0C; = 17x10-6/0C.; = 21x10-6/0C;

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Given dataArea of the steel As=100 mm2

Length of the steel Ls=75mmArea of the copper Ac=200 mm2

Length of the copper Lc=150mmArea of the Aluminium Aa=400 mm2

Length of the Aluminium La=200mmRise in temperature T=400CYoung’s modulus of steel Es=2.1x105 N/mm2

Young’s modulus of copper Ec=1.1x105 N/mm2

Young’s modulus of aluminium Ea=0.7x105 N/mm2

Co- Efficient of linear expansions of steel = 12x10-6/0CCo- Efficient of linear expansions of copper = 17x10-6/0C.Co- Efficient of linear expansions of Aluminium = 21x10-6/0C

To Find1. Stress in steel2. Stress in copper3. Stress in aluminiumSolutionStep: 1To calculate elongation of steel =

Step: 2

To calculate elongation of copper =

Step: 3

To calculate elongation of aluminum =

5

Step: 4

To calculate total elongation = + +

Step: 5

Total force exerted in the bar P= = =

Step: 6

To calculate total elongation = + +

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4. Find the Young’s modulus and poisons ratio of a metallic bar of length 300mm breadth40mm depth 40mm when the bar is subjected to an axial load of 40KN decrease in lengthis 0.75 and increase in breadth in 0.03mm also find the modulus of rigidity of bar.

Given dataLength L=300mmBreadth b =40mmDepth t = 40mmAxial load P= 40kNDecrease in length δL =0.75mmIncrease in breadth δb=0.03mmTo findYoung’s modulus EPoisson’s ratio 1/mModulus of rigidity G

SolutionStep: 1To calculate the passion ratio

Passion ratio 1/m= =

Lateral strain (et) =

Lateral strain (et) =

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Step: 2To calculate the young’s modulus

Tensile stress σ= load /area

E= tensile stress /tensile (or) longitudinal strain

Step: 3To calculate the modulus of rigidity

Young’s modulus (E) = 2 1 +

UNIT –II SHEAR AND BENDING OF BEAMS

5. A cantilever 1.5m long is loaded with uniformly distributed load of 2KN/m run over thelength of 1.25 m from the free end. It also carries a point load of 3kNat a distance of0.25m from the free end. Draw the shear force and bending moment diagram for thecantilever beam.

Given data

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To draw: SFD and BMD.

SolutionStep: 1SF calculation draw: SFD and

BMD.

Step: 2BM calculations

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6. A simply supported beam is loaded as shown in fig Q. 12 (b). Draw the shear force andbending moment diagrams. (AU-APR-2005)

Given data

Refer figure Q. 12

To draw: SFD and BMD. Draw: SFD and BMD.

SolutionStep: 1SF calculation

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Step: 2BM calculations

7. A beam 8 meters long rest on two supports one at the right end and the other two metersfrom the left end. The beam carries uniformly distributed load of 15kN/m over its entirelength and point load of 80kN at a point 1.5m from the right end. Draw the SFD andBMD and find magnitude and location of maximum bending moment. Locate also thepoint of contra flexure

Given data

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To draw: SFD and BMD.

SolutionStep: 1SF calculation draw: SFD and BMD.

Step: 2BM calculations

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8. A timber beam of rectangular section is to support a load of 20 kN uniformly distributedover a span of 3.6 m, when the beam is simply supported. If the depth is twice the widthof the section and the stress in timber is not to exceed 3.5 N/mm2, find the dimensions ofthe cross section?

Given dataUDL load=20kN/mLength (L) =3.6mDepth=twice the widthStress= 3.5 N/mm2

To findDimensions of the cross section

Solution

Step: 1To calculate the bending moment

M= for SSB

Step: 2To calculate section modulus of the beam (Z)

Bending equation= == = y=d/2 for rectangular crass section

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Step: 3To calculate the dimensions of the beam

Section modulus of the beam (Z) = for rectangular crass section

9. A I section beam 350mm x 200mm has a web thickness of 12.5mm and a flangethickness of 25mm. it carries a shearing force of 20 tonnes at a section. To calculate shearstress a crosses the section

Given dataBreath b=350mmDepth d=200mmFlange thickness =25mmWeb thickness =12.5mmShear force (F) =20tonnes

To findShear stress of the beam

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Solution

Step: 1To calculate the center of gravity section from top face

. = + ++ +

Step: 2To cal calculate the total moment of inertia of the beam (I)= + +

Step: 3To calculate shear stress in upper flange with web (q)=

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Step: 4To calculate shear stress in web with upper flange (q)=

Step:5

To calculate maximum shear stress of the beam (qmax)

= 8 − +

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UNIT –III TORSION

10. A shaft is subjected to a torque of 1.6kNm. Find the necessary diameter of the shaft, ifthe allowable shear stress is 60Mpa, the allowable twist is 1ofor every 20 diameterslength of the shaft, take C=80Gpa.

Given dataTorque (T)=1.6 kNmAllowable shear stress ( )=60MpaAngle of twist =1o

Length l= 20dTo findDiameter of the shaft (d)SolutionStep:1To calculate the diameter considering shear stress

Torque (T) =

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Step:2To calculate the diameter considering angle of twist

Torsion equation = = for solid shaft

11. Calculate the power that can be transmitted at a300rpm by a hollow steel shaft of 75mmexternal diameter and 50mm internal diameter when the permissible shear stress for thesteel is 70N/mm2 and the maximum torque is 1.3 times of the mean torque. Comparethe strength of this hollow shaft with that of an solid shaft. The same material, weightand length of both the shafts are same.

Given dataSpeed N=300rpmExternal dia D=75mmInternal dia d=50mmShear stress ( ) = 70N/mm2

Torque Tmax =1.3Tmean

To findPowerStrength of hollow shaftStrength of solid shaft

SolutionStep: 1To calculate maximum torque (Tmax)

(Th) =(Tmax)= × ×

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Step :2To calculate mean torque (Tmean)

Tmax =1.3Tmean

Step: 3To calculate power

Power (P) =

Step :4Strength compressionArea of the hollow shaft = area of the solid shaft4 − = 4To calculate solid shaft diameter

Step : 5Torque transmitted to the solid shaft (Ts)

Torque (T) =

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Step :6

=

12. Hollow steel shaft of 75mm outside diameter is transmitting a power of 300kW at2000rpm. Find the thickness of the shaft if the maximum shear stress should not exceed40N/mm2

Given dataSpeed N=2000rpmExternal dia D=75mmPower (P)= 300KWShear stress ( ) = 40N/mm2To findThickness of the shaft

SolutionStep :1To calculate maximum torque (T)

Power (P) =

Step :2

(T)= × ×

20

Step :3To calculate thickness

t=

13. A closed coil helical spring of circular crass section wire 18mm diameter is loaded by aforce of 500N. the mean coil diameter of the spring is 125mm. the modulus of rigidity is80 kN/mm2 determine the maximum shear in the material of the spring. What number ofcoils must the spring have for its deflection to be 6mm and calculate the stiffness of thespring

Given dataWire diameter d=18mmMean diameter D= 125mmLoad P= 500NModulus of rigidity C= 80 kN/mm2

Deflection δ=6mmTo findShear stress (τ)Number of coils (n)Stiffness of the spring (K)Solution

Step: 1To calculate the Shear stress (τ)

Shear stress (τ) =

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Step: 2

To calculate number of coils

= 8

Step: 3To calculate the stiffness of the spring

= = 64

Step: 4

To calculate strain energy (U) = × × l=πDn

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14. An open coiled helical spring made of 5mm diameter wire 16 coils 100mm innerdiameter with the helix angle of 16o calculate the deflection, maximum direct and shearstresses induced due to an axial load of 300N. take G= 90Gpa and E=200Gpa

Given dataWire diameter d=5mmInner diameter Di= 125mmLoad P= 300NModulus of rigidity C= 90 GpaYoung’s modulus E=200GpaNumber of coils (n) =16Helix angle = 16To findMaximum direct& Shear stress (τ)Deflection δStiffness of the spring (K)

SolutionStep: 1

Deflection = +

Step: 2Calculate bending stress

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= 32

Step: 3Calculate shear stress

= 16

Step: 4Calculate maximum shear stress= 16

Step: 4Calculate maximum principal stress (direct stress)= 16 + 1

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UNIT –IV DEFLECTION

15. A cantilever AB, 2m long is carrying a load of 20kN at free end and 30kN at a distanceof 1 m from the free end. Find the slop and deflection at the free end. Take E= 200Gpaand I= 150 x 106 mm4

Given data

E= 200GpaI= 150 x 106 mm4

To findSlop and deflection of the beam at the free end

SolutionDouble integration method

Step: 1To calculate Slope at the free end due W1

= 2

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Step: 2To calculate Slope at the free end due W2

= −2

Step: 3To calculate total slope at the free end

= 2 + −2 = +

Step: 4To calculate the deflection at the free end

To calculate deflection at the free end due W1

= 3

Step :5

To calculate deflection at the free end due W2

= −3

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Step: 6To calculate total deflection at the free end

= 3 + −3 = +Macaulay’s method

Moment area method

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Conjugate beam method

16. A beam is simply supported as its ends over a span of 10m and carries two concentratedloads of 100kN and 60 kN at a distance of 2m and 5m respectively from the left support.Calculate (i) slope at left support (ii) slope and deflection at under the100 kN load.Assume EI=36x104 kN-m2

Given data

EI=36x104 kN-m2

To find(i) slope at left support(ii) slope and deflection at under the100 kN load

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SolutionStep-1Macaulay’s method

Step-2Slope at the left support

Step-3Deflection at 100kN load

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Step-4Slope at 100 kN load

17. An I section joist 400mmx 200mmx 20mm and 6m long is used as strut with both endsfixed. What is Euler’s crippling load for the column? Take E=200Gpa.

Given dataDepth (d) = 400mm

Breath (b) =200mmThickness (t) = 20mmLength l= 6mYoung’s modulus (E) = 2x 105 N/mm2

To findCritical loadSolutionStep-1To calculate the moment of inertia about x-x axis

= −

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Step-2To calculate the moment of inertia about y-y axis= −

Step-3To calculate the crippling load (P)

( ) = 4

18. Find Euler’s crippling load for a hollow cylindrical cast iron column of 20mm externaldiameter, 25mm thick and 6m long hinged at both ends. Compare the load with

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crushing load calculated from Rankin’s formula fc= 550N/mm2 Rankin’s constant=1/1600, E=1.2x 105 N/mm2

Given dataMajor dia D = 200mm

Minor dia d=(200-25-25)Thickness (t) = 25mmLength l= 6mσc =550N/mm2

Rankin’s constant =Young’s modulus (E) = 1.2x 105 N/mm2

To findCritical loadSolutionStep-1To calculate the moment of inertia for circular section

= 164 −

Step-2

Radius of gyration K= area (A) = −

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Step-3To calculate the crippling load (P)

( ) =

Critical load by using Rankin’s formula

Step-4

P=

UNIT-5 COMPLEX STRESSES AND PLANE TRUSSES

19. A steel cylinder shell 3m long which is closed at its ends, had an internal diameter of1.5m and a wall thickness of 20mm calculate the circumferential and longitudinal stressinduced and also change in dimensions of the shell if it is subjected to an internalpressure of 1.0 N/mm2 assume the modulus of elasticity and passion ratio for steel as 200kN/mm2 and 0.3 respectively

Given dataLength of the shell L= 3mDiameter of the shell = 1.5mPressure (P) = 1.0 N/mm2

Thickness of shell t= 20mmModulus of elasticity (E)=200 kN/mm2

Passion ratio (1/m) = 0.3To find

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Circumferential stressLongitudinal stressChange in dimensionsSolutionStep-1To calculate circumferential stress= 2

Step-2To calculate longitudinal stress= 4

Step-3To calculate maximum shear stress= −2

Step-4

To calculate circumferential strain = −

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Change in dia =

Step-5

To calculate longitudinal strain = −

Change in length =Step-6To calculate volumetric strain = +

Change in length =

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20. A spherical shell of 2m diameter is made up of 10mm thick plates. Calculate the changein diameter and volume of the cylinder, when it’s subjected to an internal pressure of1.6 Mpa. Take E= 200Gpa and 1/m= 0.3

Given dataDiameter of the shell = 2mPressure (P) = 1.6 N/mm2

Thickness of shell t= 10mmModulus of elasticity (E) =200 GpaPassion ratio (1/m) = 0.3To findChange in dimensionsSolutionStep-1Change in dia = 4 1 − 1

Step-2Change in dia = 8 1 − 1

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21. The state of stress at a certain point in a strained material is shown in fig calculate theprincipal stress inclination of the principal plane normal, shear and resultant stress onthe plane MN

Given dataDirect stress σ1=200N/mm2

Direct stress σ2= − 150N/mm2 (compressive)Shear stress q= 100 N/mm2

To find1. Principal planes and principal stresses2. Maximum shear stress and planeSolutionStep-1To calculate principal plane2 = 2−

Step-2

Major Principal Stress = + − + 4

Step-3

MinorPrincipal Stress = − − + 4

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Step-4

Maximum shear Stress = − + 4A steel cylinder shell 3m long which is closed at its ends, had an internal diameter of and wallthickness 20mm calculate the circumferential and longitudinal stress induced and also chanein dimensions of the shell if it is subjected to an internal pressure 1.0 N/mm2 assume themodulus of elasticity and passion ratio for steel as 200kN/mm2 and 0.3 respectively.