SOM PTU QUESTION PAPER SOLUTION CLASS NOTES

SOM PTU QUESTION PAPER

SOLUTION CLASS NOTES

PTU SOM Question answers improves

clarity of the subject. It also increases

the confidence in handling practical

problems. It helps in the selection of

material for a certain application. It is the

ultimate aim of the subject.

2-5 marks question answers

Q1

(a). What is slenderness ratio? State the limitations of Euler’s formula.

ANS: slenderness ratio = k = L_e/k _min

Where L­_e is the effective length

k _min is the minimum radius of gyration

Firstly L_e = L for both ends hinged column

Secondly L_e =2 L for a cantilever column (One end fixed and the other end is free)

Thirdly   L_e = L/ for one end hinged and the other end fixed column

Fourthly   L_e = L/2 for both ends fixed column

k _min =    I _min/A

LIMITATIONS OF THE EULER FORMULA

1. The column is initially straight.

2. The load applied is truly axial.

In reality these assumptions cannot be true. Thus these are the limitations of Euler’s formula.

(b). Differentiate between the Centroid and the center of gravity.

ANS; Centroid is the center for a 2-dimensional figure like a rectangle, square, circle, triangle, ellipse etc.

Center of gravity is the center of a 3-dimensional object. At the center of gravity the whole weight of the object acts. For example we talk of center of gravity for a chair, table, television, any object.

(c) What is the practical application of the quantity ‘Moment of inertia’.

ANS:  It’s symbol is “I”. It is the second moment of the area about a centroidal axis. Its units are m⁴.

Mathematically

I =   ∫dA y²

It is a resistance to bending. If moment of inertia is high, bending will be less.

(d) What is the radius of gyration of an area and what is the practical use of it?

ANS: Radius of gyration is an imaginary radius about which the whole body is supposed to rotate.

It is property of an area. It’s symbol is “k”.  k = (I/A)^0.5. Its least value is used to calculate the maximum buckling load for a long and medium column.

(e) Define the moment of inertia of an area about an axis lying in the plane of the area.

ANS: It is the second moment of the area. Mathematically

I =   dA y²

I _xx  = (1/12) bd³ for a rectangle with width b and depth d.(xx is the axis parallel to width b.)

I _yy  = (1/12) db³ for a rectangle with width b and depth d. (y axis is the axis parallel to depth ‘d’.)

I _xx  =  I _yy  =  d⁴ for a circular area.

(f) Write the assumptions made in the theory of bending.

1. Beam is initially straight.

2. Beam is homogeneous i.e. composition is same throughout.

3. Material is continuous i.e. no voids or no foreign material is present.

4. Beam is isotropic material i.e. properties in x, y, z directions are same.

5. A transverse plane (perpendicular to the axis) remains a transverse plane before and after the bending moment is applied.

6. A constant bending moment is applied on the beam.

7. No shear force is acting on the beam

(h) What is Torsional rigidity and Polar modulus?

ANS: Torsion equation is

T/J = τ/r = Gϴ/L

Torsional rigidity is the quantity which produces minimum angle of twist for a given torque.

Hence GJ is torsional rigidity,

greater GJ,  lesser will be ϴ.

J is the polar modulus, J _{solid shaft} = d⁴      m⁴

J _{hollow shaft} =d_o⁴ –d_i⁴)    m⁴

(i) How failure of the short and long column takes place?

Failure of short column is by contraction.

Failure of long columns is by buckling.

Buckling is bending due to axial load.

(j) What is crippling load?

ANS:  It is a load for a long column. If the value of the external load is greater than the crippling load, the column will buckle.

P_cr = ²E I _min/L_e²⁺

L_e is the effective length

Q1.

(a).What is Factor of safety?

For Ductile materials            FOS = σ_yp /σ_all

For Brittle Materials               FOS = σ_ult /σ_all

σ_yp = yield point stress

It is possible only for ductile materials. Its value is very near to the value of proportional limit stress.

σ_all =  allowable stress = 1/3 to 2/3 of elastic limit stress

σ_ult  = ultimate stress

(b) How stress concentrations affect the life of the material?

ANS:

Stress concentration reduces the life of a brittle material because the failure will be sudden and unexpected due to crack propagation.

(c)How fatigue strength of a shaft can be improved?

By relieving the residual stresses by annealing.

(d) A load of 400 N has to be raised at the end of a steel wire. If the stress in the wire must not exceed 80 MPa, what will be the diameter of the wire?

ANS:

Area =( /4)d² = load/stress=400/80 = 5

d=2.47 mm

(e)What is modulus of rupture?

ANS:

It is energy required to fracture a material under an impact load.

(h) Give the relationship between modulus of rigidity and Young’s modulus.

ANS:

E = 2G(1+μ)

Where μ is the Poisson’s ratio, E=Young’s modulus and G= Modulus of rigidity

(i) What is the principal stress and a principal plane?

ANS:

Principal stress is normal stress with a zero shear stress.

Plane on which there is only normal stress with zero shear stress is called the principal plane.

(j) Write maximum strain energy theory.

ANS:

(1/2E)(σ₁²+σ₂²–2μσ₁σ₂) σ_all²/ 2E

Q. What is the energy of distortion or shear strain energy?

ANS: Every object in universe is under a number of forces. Thus, it is under complex loading. From the complex loading, one can determine the three principal stresses σ₁, σ₂ and σ₃ respectively. ASSUME

(i) these principal stresses as tensile and

(ii) σ₁> σ₂ > σ₃

Then energy of distortion OR shear strain energy is

u_sse=(1/12G)[ (σ₁–σ₂)² + (σ₂—σ₃)² + (σ₃—σ₁)²]

NOTE: if any of σ₁, σ₂ and σ₃ is not tensile, then respective sign must be used in the above relation.

Q. Explain the necessity of the theories of failure.

ANS: When identical jobs (Same shape and same size) of different materials are tested in the lab under identical loading, it is found that failure of different materials take place differently. Therefore we need a way to explain these different types of failures under identical loading. Then only these theories of failure help us. Thus these are needed to explain different failures under identical testing.

Q. Define stiffness of a spring or define a spring constant.

ANS: Spring is a highly elastic body. Force required to cause unit deflection is called spring stiffness.

K = W/δ    N/mm

Where k is the stiffness

W is the load acting on the spring

δ  is the deflection under the load W

Q. For what purpose, cylindrical and spherical vessels are used?

ANS: Cylindrical vessel are used as process vessels as boilers for producing steam.

These vessels are used as storage vessels as LPG cylinders and Oxygen cylinders.

Spherical vessels are used only for storage vessels as these can store more volume in a given space.

Q. State Lame’s equations.

ANS: Lame’s equations are for stresses in a thick vessel. The three stresses are hoop stress

σ_h, radial stress, σ_r and longitudinal stress σ_l

Lame’s equations are

σ_h =α + β/r²

σ_r =α — β/r²

Where α and β are Lame’s constant. These are found from the boundary conditions like

use σ_r =p_i at r = r_i

and σ_r =p_o at r = r₀

σ_l =α  and σ_l is negligible.

Q. List the types of stresses produced in a rotating disc of uniform thickness.

ANS. Variable hoop and radial stresses are produced in a rotating disc of uniform thickness.

Q What is a trapezoidal section?  Why is this used in initially curved beams as in a hook of a crane?

ANS: A trapezoidal section is a quadrilateral with two parallel and two non- parallel sides. This is used in a crane hook

because its area is variable which makes stress nearly uniform in the innermost and outermost fibers

Q. Where does the maximum shear stress occurs in an I section?

ANS: It occurs in the center of the web section of the I section.

Q. Explain the importance of the full length leaf in a leaf spring.

ANS: One full length leaf is necessary for mounting of the spring on the vehicle. Any extra full length leaf takes the transverse shear force which is maximum at the ends.

Q.  What is the purpose compound cylinders?

ANS: These are used to resist high pressures.  It is difficult to make a cylinder from a single thick plate. It is made with number of thin cylinders by hoop shrinking. Such a thick vessel made from thin vessels is called a compound cylinder.

Q. Distinguish between longitudinal and hop stresses in a thick cylinder.

ANS:

Sr. No.	Longitudinal stress	Hoop stress
1.	This stress is in the length direction.	Hoop stress is in the tangential direction.
2.	The magnitude of this stress is constant.	Magnitude of hoop stress is variable.
3.	Symbol of this stress is σl.	Its symbol is σh.
4.	Magnitude of this stress is quite small.	Its  magnitude is quite large. It is greater than the fluid pressure
5.	This stress is tensile.	It is also tensile.

Q. For the crane hook, locate the plane which is severely stressed.

ANS: The innermost curve of the crane hook is severely stressed.

Q. Define strain energy.

ANS: When an external force acts on a body, strain is produced in the body. Thus force moves through some distance. Hence the work is done. This work is stored in the body as long as it is strained. This work is strain energy. Its symbol is U. Its units are Joules. It is given by the formula

U =(1/2) F δl = σ²x volume/2E=( σ²/2E)x Volume

SECTION- Numerical question answers

Q.2 A hollow steel cylinder of 300 mm length, 150 mm inside diameter and 3mm uniform wall thickness. It is filled with concrete. It is compressed between two rigid parallel plates at the ends by a load of 600 k N. Find the compressive stress in the each material. Also find total shortening of the cylinder.

Given

Young’s modulus of steel =2 x 10⁵ N/mm²

Young’s modulus of concrete is = 2 x 10⁴ N/mm².

ANS:

Area of steel = A_s= ( /4)(156² –150²) = 1836 mm²

Area of concrete =A_c=( /4)(150²) = 17671.4586764                mm²

σ_s = P E_s /(A_s E_s+ A_c E_c)

= 600 x2x10⁵/( 1836x2x10⁵  + 17671.4586764x  2 x 10⁴ )

= 8.71  N/mm²                                                                                                          

σ_c = P E_c /(A_s E_s+ A_c E_c)

=600 x2x10⁴/( 1836x2x10⁵  +17671.4586764x  2 x 10⁴)                = 0.871 N/mm²

    Q.3 Two principal stresses at a point in a bar is 200 N/mm² (tensile) and 100 N/mm² (compressive). Determine the resultant stress the magnitude and direction on a plane. Plane is inclined at 60⁰ to the axis of the major principal stress. Also determine the magnitude of the maximum shear stress.

ANS: σ_R =180.27 N/mm², θ=46⁰6’, (σ_t)_max =150 N/mm².

σ_n=σ₁cos²θ + σ₂sin²θ provided σ₁and σ₂ are tensile.

σ_n=200cos²30⁰ — σ₂sin²30=200 x 3/4—100×1/4=125

τ_θ= (σ₁—σ₂)sin60⁰/2=(200-(–100))x 1/4=75

σ_R=( σ_n² + τ_θ²)^0.5=177 N/mm²

tan α = τ_θ/ σ_n =75/125=0.6

α=31⁰

(σ_t)_max= (σ₁—σ₂)/2=(200—(-100))/2=150 N/mm²

4. Two principal stresses at a point in a bar are 200 N/m m ² (tensile) and 100 N/m m ² (compression). Determine the resultant stress in magnitude and direction on a plane inclined at 60⁰ to the axis of the major principal stress. Also determine the maximum intensity of the shear stress.

ANS:

Firstly   σ_R =180.27 N/m m ², θ=46⁰6’, (σ_t)_max =150 N/m m ².

Secondly   σ_n=σ₁cos²θ + σ₂sin²θ provided σ₁and σ₂ are tensile.

Thirdly    σ_n=200 cos²30⁰ – σ₂ sin²30 =200 x 3/4—100 x 1/4=125

τ_θ= (σ₁—σ₂) sin60⁰/2=(200-(-100))x 1/4=75

σ_R=( σ_n² + τ_θ²)^0.5=177 N/mm²

tan α = τ_θ/ σ_n =75/125=0.6

α=31⁰

(σ_t)_max= (σ₁—σ₂)/2=(200—(-100))/2=150 N/mm²

5. A timber beam 0.3 m depth is symmetrical. It is simply supported at the ends with a span of 8 m. What uniformly distributed load (including its own weight) it can carry if the maximum permissible stress is 8 N/ m m ². The moment of inertia of the section of the beam is 450 x 10⁶ m m ⁴. Find the maximum bending stress and the radius of curvature at section 1 m from a support. Modulus of elasticity for timber = 12.6 x 10³ N/ m m ².

ANS:

• M/I = σ/y=σ_max/ y _max

We get    w L²/8I =8/150

w x 8000 x 8000/8 x 450 x 10⁶ = 8/150

ANS: w = 3 k N/m

• M _{1m from the support} = [(wL/2)x1 – wx1x0.5] x 10⁶ N mm

= [(3×8/2)x1—3x1x0.5] x 10⁶ N mm

=10.5×10⁶ N mm

Using         M/I = σ/y

10.5×10⁶/450 x 10⁶ = σ/150

σ= 3.17 N/ mm²

© Using M/I = E/R

10.5×10⁶/450 x 10⁶ = 12.6 x 10³/R

R =540 m

6. A solid circular shaft transmits 75 kW power at 200 RPM. Calculate the shaft diameter. The twist in the shaft is not to exceed 1 degree in a length of 2 meters.   Shear stress is limited to 50 MPa. Take modulus of rigidity = 100 x 10³ MPa.

ANS;

P =75 kW= 75000 W

N = 200 RPM

ϴ=1⁰ =

L = 2 m = 2000 mm

τ= 50 MPa

G = 100 x 1000 MPa

Power = 2 NT/60 = 2 200 T/60

T = 3.5809 x 10⁶ N mm

Diameter from strength

We know T/J = τ/r

T=(π/16) d³ τ

3.5809 x 10⁶ =(π/16) d³ 50

d = 72 mm

Diameter from stiffness (angle of twist)

T/J = Gϴ/l

J =(π/32)d⁴ = Tl/Gϴ = 3.5609 x 10⁶ x 2000/(100 x 10³ x 0,01745)

d =80.4 mm

ANS larger diameter = 80.4 mm

Section-C

Q.7  A short concrete column of 250 mm x 250 mm is under an axial load of 300 k N. The column is reinforced with steel bars. The total area of steel is 5600 mm². If the modulus of steel is 15 times the modulus of concrete, find the stresses in concrete and steel. If the stress in the concrete is not to exceed 4 N/mm², find the area of steel required so that the column may support a load of 600 k N.

ANS:

A _column=250 x 250, A _steel=5600 mm², A _con = 56900 mm²,

E _steel=15 E_con

σ_con=P E_con/(E_s A_s + E_c A_c) =300 x 1000 x E_con/(15 E_con5600 +56900 E_con) =2.1 N/mm²

σ_steel=PE _steel/(E_sA_s + E_cA_c) =300 x 1000 x 15 E_con/(15 E_con5600 +56900 E_con) =31.5     N/mm²

(B) given σ_con=4 N/m m ²,

Using σ_con=PE _con/(E _s A_s + E _c A_c)=600 x 1000 x E_con/(15E _con A_s +(62500–A_s) E_con) =4

600 x 1000/(15A_s + 62500 –A_s)=4

A_s=6250 mm²

 Q.8 Compare the strength of solid circular column and hollow circular column. Solid is of diameter 200 mm.  Hollow circular column of the same cross sectional area and thickness 30 mm. The other parameters are same for both the sections.

ANS:

Assume these columns as long columns.

A _solid = (π /4) d² =(π /4) 200²

A _hollow = ( π /4)( d_o² –(d_o –60)²)

Given A _solid= A _solid

(π /4) 200² = (π /4)( d_o² –(d_o –60)²)

d_o = 363.33 mm

I _solid = ( π /64) d⁴

I _hollow = (π  /64)( d_o⁴ –d_i⁴) =( π /64)( d_o⁴ –(d_o –60)⁴)

P _{Crippling solid} = ² EI _solid/L_e² 

P _{Crippling hollow} = ² EI _hollow/L_e ²

Ratio of strengths = P _{Crippling solid}/ P _{Crippling hollow}

=I _solid/I _hollow

Ratio of strengths =( π /64) d⁴/( π /64)( d_o⁴ –(d_o –60)⁴)

= 200⁴/[(363.33⁴ –(363.33 -60)⁴]

=0.1786

Q.9 A 4 m long simply supported wooden beam at its ends. It is carrying a point load of 7.25 k N at its center. The cross sectional area of the beam is 140 mm wide and 240 mm deep. If E = 6 x 10 ³ M Pa, find the deflection at the center using the double integration method.

ANS:

Prove the article for y_c = WL³/48EI

Deflection at the center = WL³/48EI

= 7250 x 4000³/48 x6x10³x1/12×140 x 240³

=9 mm

Q. Compare the strains produced in a body subjected to same amount of load when applied gradually and suddenly.

ANS:

 gradually applied load =P

 area of cross section of the body=A

Then stress = P/A

 Gradually applied load is twice of suddenly applied load.

Let P is suddenly applied load.

Stress = 2P/A

Now stress is two times.  Strain is also two times.

PROOF

 Gradually applied load is two times the suddenly applied load

W.D. in suddenly applied load = P _sudden  δl

W.D. under gradually applied load= 0.5 P_GAL δl

where GAL is gradually applied load

In order to find the equivalent gradually applied load, the W.D. and extension are same. Therefore equate the W.D. in the two cases.

P _sudden  δl =  0.5 P_GAL δl

P_GAL = 2P_sudden

Hence proved.

2. For a thin cylindrical shell, the length diameter ratio is 3 and its volume is 20 m³. The safe tensile stress for the shell material is 100 M Pa. Determine the cylinder diameter and wall thickness. It contains water at an absolute pressure of 2 M Pa.

ANS: Given L/D =3 and volume V =D²L= D² 3D= 20

D³ = 20

D = 2.04 m

Now hoop stress is the highest stress which taken equal to the safe stress.

σ_h=p_i D/2t = 100

t = 2x 2.04 x 1000/2×100 = 20 mm

3. Distinguish between energy of dilation and energy of distortion.

ANS

Every object in universe is under a number of forces. Thus it is under complex loading. From the complex loading, one can determine the three principal stresses σ₁, σ₂ and σ₃ respectively. ASSUME

• these all principal stresses as tensile and

• σ₁> σ₂ > σ₃

Energy of dilation deals with stresses that cause a change in size (length, width, thickness, depth, area, volume). Definitely these are tensile and compressive stresses. Principal stresses are also tensile and compressive stresses. Then the energy of dilation is given below

3 Dimensional

u _3d=(1/2E)[ σ₁² + σ₂² + σ₃² -2μ σ₁σ₂-2μ σ₂σ₃-2μ σ₃σ₁]

2-Dimensional

u _2d=(1/2E)[ σ₁² + σ₂² + -2μ σ₁σ₂]

where μ is the Poisson’s ratio

It will be used in Maximum Strain Energy Theory of Elastic Failure.

Then energy of distortion OR shear strain energy theory (sse)

applies when there is no change of size but there is twist or distortion in the body. It is given by the relation given below.

3-Dimensional

u _sse=(1/12G)[ (σ₁–σ₂)² + (σ₂—σ₃)² + (σ₃—σ₁)²]

2 Dimensional

u _sse=(1/12G)[ (σ₁–σ₂)²]

It is used in Maximum Strain Energy of Distortion Theory of Elastic Failure.

NOTE: if any of the three principal stresses σ₁, σ₂ and σ₃ is not tensile, then respective sign must be used in the above relation.

4. State maximum principal stress theory.

ANS: Every object in universe is under a number of forces. Thus it is under complex loading. From the complex loading, one can determine the three principal stresses σ₁, σ₂ and σ₃ respectively. ASSUME

• these all principal stresses as tensile and

• σ₁> σ₂ > σ₃

 Maximum principal stress theory

states that a material will fail if largest principal stress in the material exceeds the allowable stress.

Therefore when σ₁ = σ_allowable =σ_elastic/Factor of safety

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