DME 316357 ASSIGNMENT WITH SOLUTIONS ( IMPOARTANT QUESTIONS)
Assignment No. 1
Year: 2025-26
Program Name : - Mechanical Engg. Course Name:- DME [316357]
Name of Faculty:-
Prof. P.P.BHIRUD Weightage: 12 Marks
UNIT I : FUNDAMENTALS OF DESIGN
CO1: Use fundamental
concepts of design of machine elements for given
application
|
Qu |
DESCRIPTION |
M |
L |
EXAM |
|
1 |
State the composition of the
following materials: a)Fe E 230 ,b) X20Cr18Ni2 ,
c) 35C8 ,d) 40Ni2Cr1Mo28 ,e) 40C8
f)35Mn2Mo28, g) 30Ni4Cr1, h)
25Cr3Mo55,g) FeE220,h)20C8 i) X10Cr18Ni9Mo4Si2
j) |
4 M |
R |
W- 2016,17 |
|
2 |
State the composition of
material a)FG200 b) FeE230
c) 35C8 d) X20Cr18Ni2 |
4 M |
R |
S19,22W18 |
|
3 |
Define
stress concentration. State causes and methods of reducing stress
concentration with neat sketches. |
6 M |
U |
W24,18 S18,19 |
|
4 |
Explain
the following types of stresses:
a)Transverse shear stress b) Compressive
stress c) Torsional shear stress |
6 M |
U |
W-16, W-18 |
|
5 |
Explain
maximum shear stress theory with equations. |
4 M |
U |
W-18 |
|
6 |
Draw stress–strain diagram for ductile material
stating salient points. |
4 M |
A |
S16 ,19 |
|
7 |
Draw
stress–strain diagram for brittle material. |
2 M |
A |
W-17 |
|
8 |
Define
endurance limit and draw typical S–N curve for steel. |
4 M |
U |
W-18 |
|
9 |
Define endurance
limit (fatigue limit)? Explain its importance in design. |
4 M |
U,A |
S-19 |
|
10 |
Define
factor of safety. State the factors affecting on the selection of f.o.s |
6 M |
U.R |
S17W22 |
|
11 |
Define
factor of safety w.r.t. ductile and brittle material. |
4 M |
U |
W-18 |
|
12 |
Define
creep? Explain creep curve with stages . |
3 M |
U,A |
W-17 |
|
13 |
State
the theories of elastic failure. Explain any two. OR Explain
maximum normal stress theory and maximum shear stress theory with equations. |
8 M |
U,A |
W-17 |
|
14 |
Define
machine design. |
2 M |
U |
W-17 |
|
15 |
State
and explain main considerations in machine design. |
6 M |
R |
W-16 |
|
16 |
2 M |
R |
W-17 |
|
|
17 |
State
&Explain general design procedure. |
4 M |
R,U |
S24,25 |
|
18 |
Explain importance of aesthetics consideration in
design with examples. |
6 M |
U |
W17,18 |
|
19 |
State
the meaning of colour codes in aesthetic design: a)Red
b) Orange c) Green d) Blue |
4 M |
R |
W16,S23 |
|
20 |
State
general equation for bending moment & state meaning of each term. |
2 M |
R |
S-18 |
|
21 |
State
torsion equation and state meaning of each
term |
2 M |
R |
S-18 |
|
Qu |
DESCRIPTION |
M |
L |
EXAM |
||||||||||||
|
1 |
State the composition of the
following materials: a)Fe E 230 : It is a plain
carbon steel with Min. Yield
strength is 230 b) X20Cr18Ni2 : It is a High alloy steel with
carbon of average 0.20%, c)
35C8 : It is a Plain carbon steel with
avg. 0.35% of carbon, manganese d) 40Ni2Cr1Mo28 : It is a Alloy steel with Avg 0.4 % average of Carbon , e) 40C8 :
It is a
Plain carbon steel with Avg. 0.4 % of Carbon, manganese 0.8% f)35Mn2Mo28: It is a Alloy steel with 0.35 % average of Carbon , Manganese g) 30Ni4Cr1: It is a Alloy steel with average
0.3 % of Carbon , Nickel h)
25Cr3Mo55 : It is a Alloy steel with average 0.25% of Carbon ,
chromium g) FeE220:
It is a plain carbon steel with Min. Yield strength is 220 N/mm2 h)20C8: It is a Plain carbon steel carbon 0.2% of average, manganese 0.8% , i) X10Cr18Ni9Mo4Si2 : It is a High alloy steel with carbon
0.10% of average , j) X77W18Cr4V1: It is a High alloy steel
carbon 0.77% of average, tungsten |
4 M |
R |
W- 2016,17 |
||||||||||||
|
2 |
State the composition of
material a)FG200 b) FeE230
c) 35C8 d) X20Cr18Ni2 a)
FG200 : It is a grey cast Iron with min. UTS 200 N/mm2 c)35C8 : It is a Plain carbon steel with average 0.35% carbon, manganese 0.8% b)Fe E230 : It is a Plain
carbon Steel with yield strength of 230N/mm2 d)X20Cr18Ni2 :
It is a High alloy steel with average 0.20% carbon, chromium 18%, Nickel
2% |
4 M |
R |
S19,22W18 |
||||||||||||
|
3 |
Define stress concentration.
State causes and methods of reducing stress concentration with neat sketches. During the design of any machine component,
discontinuities in any machine part are there due keyway, threaded grooves
and steps are present on the component which is functional requirement to
perform their functions. 2. Such type of
discontinuity alters the stress distribution in the vicinity of the
discontinuity and elementary stress equations no longer describe the state of
stress in the component. 3. The stresses induced in the neighbourhood of the
discontinuity are much higher than the stresses in the other part of the
component. 4. This
concentration of high stresses due to the discontinuities or abrupt change of
cross-section is called stress concentration.  Causes of stress
concentration are as under. i) Abrupt changes in cross-section like in
keyway, steps, grooves, threaded holes results in stress concentration. ii) Poor surface finish – The surface
irregularities is also one of the reason for stress concentration. iii) Localized
loading – Due to heavy load on small area the stress concentration occurs
in the vicinity of loaded area. iv)Variation in
material properties – Particularly defects like internal flaws, voids,
cracks, air holes, cavities also results in stress concentration. Remedies to reduce
stress concentration are as follows: i. Introducing additional notches and holes in tension
member like use of multiple notches and drilling additional holes. ii. Fillet radius, undercutting and notch for member in
bending. iii. Reduction of stress concentration in Threaded
members. iv. Change in cross-section should be gradual. v. By
improving surface finish  |
6 M |
U |
W24,18 S18,19 |
||||||||||||
|
4 |
Explain the following types of
stresses: a)Transverse shear
stress b) Compressive stress c) Torsional shear stress i)Transverse Shear
Stress: - When a mechanical component is subjected to two equal and
opposite forces acting tangentially across the resisting area resulting in
shearing off of the section, the stress induced in such a case is known as
transverse shear stress. ……………………01 mark From figure
Mathematically transverse shear stress is represented as, τ =F / A Where, F = Tangential
force applied A
= Area of cross section d = Diameter of rivet. ……………………01 mark
 ii )Compressive Stress: When a body is
subjected to two equal & opposite axial pushes ,then the internal
resistances set up in the material is called as compressive stress .……………………01 mark It is denoted by σc =P/A Where ,P: Axial compressive force , A : Cross Sectional Area.……………01 mark iii) Torsional
stress: When a machine component is under the action of two equal and
opposite couples i.e. twisting moment or torque, then component is said to be
torsional and the stresses set up due to torsion are called as torsional
shear stress. ……………………01 mark Consider a component of circular cross-section. ‘d’ in
diameter, subjected to torque T, Torsional shear stress is given by, basic torsion equation τ / r
= T / J
=Gθ/L Where, r = distance of outer fibre from neutral axis =
d/2 J = Polar moment
of inertia of cross- section ……………………01 mark |
6 M |
U |
W-16, W-18 |
||||||||||||
|
5 |
Explain maximum shear stress
theory with equations. Maximum Shear Stress Theory: This theory states that
failure occurs when the maximum shear stress from a combination of stresses
equals or exceeds the value obtained for the shear stress at yielding in the
simple tensile test. Mathematically, τ max = τyt
/FoS ...
(i) where , τmax = Maximum shear stress in a bi-axial stress system, τyt = Shear stress at yield point as determined from
simple tension test, and FoS = Factor of
safety. Since the shear stress at yield point in a simple tension
test is equal to one-half the yield stress in tension, therefore the
equation (i) may be written as
τmax = 2 x σ yt x foS This theory is mostly used for designing members of ductile materials |
4 M |
U |
W-18 |
||||||||||||
|
6 |
Draw stress–strain diagram for
ductile material stating salient points.
|
4 M |
A |
S16 ,19 |
||||||||||||
|
7 |
Draw stress–strain diagram for
brittle material.
|
2 M |
A |
W-17 |
||||||||||||
|
8 |
Define
endurance limit and draw typical S–N curve for steel. Endurance Limit: It is defined as
maximum value of the completely reversed bending stress which a polished
standard specimen can withstand without failure, for infinite number of
cycles (usually 107 cycles).It is known as endurance or fatigue
limit ( ϭe).  The plot of fatigue strength Sf Vs stress cycle N on log paper. This plot
is popularly is known as S-N Diagram .For the steel material ,the graph
becomes horizontal at 106 cycles, indicating that the fatigue
failure will not occurs below this stress, whenever may the number of
cycles.this stress is known as endurance limit Sf of the material. |
4 M |
U |
W-18 |
||||||||||||
|
9 |
Define endurance
limit (fatigue limit)? Explain its importance in design. Endurance Limit: It is defined as
maximum value of the completely reversed bending stress which a polished
standard specimen can withstand without failure, for infinite number of
cycles (usually 107 cycles).It is known as endurance or fatigue
limit ( ϭe). Need of Endurance
Limit in Machine Design: Endurance limit is used to describe a property
of materials: the amplitude (or range) of cyclic stress that can be applied
to the material without causing fatigue failure. |
4 M |
U,A |
S-19 |
||||||||||||
|
10 |
Define factor of safety. State
the factors affecting on the selection of f.o.s Factor of Safety:
Factor of safety: It is defined as ratio of Maximum stress to the working
stress ( permissible /design stress OR it is the ratio of failure load to
allowable or working load  Selection of
factor of safety (fs) depends upon following factors i)
Effect
of failure: Time, finance, danger to human life decides the value of fs. ii)
Type of
load : Static load– low fs, Dynamic load and Impact Load-high fs, iii)
Degree
of accuracy in force analysis: Accurate - low fs, iv)
Material of component: Homogeneous material or ductile low fs, iv)
Reliability
of component: increases for higher reliability of component. vi) Cost of
component: Cost of component is directly proportional to value of factor of safety.
vii) Testing of machine element: Actual testing conditions known - low
fs viii)Service conditions:
Operating conditions like temp, corrosion, humidity add to value of factor of
safety. ix) Quality
of Manufacture: High Mfg. quality leads to low fs Selection of value of factor of safety
is normally low for ductile materials and high for brittle materials. It is
essentially a compromise between the associated additional cost and weight
and the benefit of increased safety or/and reliability. |
6 M |
U.R |
S17W22 |
||||||||||||
|
11 |
Define
factor of safety w.r.t. ductile and brittle material. Factor of Safety:
Factor of safety: It is defined as ratio of Maximum stress to the working
stress ( permissible /design stress OR it is the ratio of failure load to
allowable or working load  |
4 M |
U |
W-18 |
||||||||||||
|
12 |
Define creep? Explain creep
curve with stages . Creep: When a machine component is subjected to
constant stress (load) at high temperature for a long period of time, it will
undergo a slow and progressive permanent deformation called creep.  Creep curve has three stages: 1. Primary
creep 2. Secondary
creep 3. Tertiary
creep First Stage: Primary Creep
1. It
is called primary creep. 2. It
is shown by curve AB. 3. In
this stage, the creep rate decreases with time. Second Stage: Secondary Creep
1. It
is called secondary creep. 2. It
is shown by curve BC. 3. In
this stage, the creep rate remains constant with time. 4. This
stage covers the maximum period of the total life of the
component. Third Stage: Tertiary Creep
1. It
is called tertiary creep. 2. It
is shown by curve CD. 3. In
this stage, the creep rate increases rapidly with time. 4. At
the end of this stage, necking occurs and the component fails. Examples :
·
Creep behavior is important for materials used
at high temperature and under constant stress for long time. ·
Examples: turbine blades, boilers,
pressure vessels. |
3 M |
U,A |
W-17 |
||||||||||||
|
13 |
State the theories of elastic
failure. Explain any two. OR Explain maximum normal stress
theory and maximum shear stress theory with equations. Maximum Principal
Stress Theory: This theory states that failure occurs when the maximum
principal stress from a combination of stresses equals or exceeds the value
obtained for the direct stress at yielding in a simple tension test.
Maximum Shear
Stress Theory: This theory states that failure occurs when the maximum
shear stress from a combination of stresses equals or exceeds the value
obtained for the shear stress at yielding in the simple tensile test. Mathematically, τ max = τyt
/FoS ... (i) where , τmax = Maximum shear stress in a bi-axial stress system, τyt = Shear stress at yield point as determined from
simple tension test, and FoS = Factor of
safety. Since the shear stress at yield point in a simple tension
test is equal to one-half the yield stress in tension, therefore the
equation (i) may be written as
τmax = 2 x σ yt x foS This theory is mostly used for designing members of ductile materials |
8 M |
U,A |
W-17 |
||||||||||||
|
14 |
Define machine design. Machine Design : It is creation of new or
modified plan for the machine to perform desired functions.
Or It is
the process of selection of the material, shape, size and arrangement of
mechanical element so that the result machine will perform the prescribed
task. |
2 M |
U |
W-17 |
||||||||||||
|
15 |
State and explain main
considerations in machine design. Following
main considerations are important during machine design · Type of load and stresses · Motion of parts (Kinematics) · Selection of materials · Form and size of parts · Friction and lubrication · Convenience and economy · Use of standard parts · Safety of operation · Workshop facilities · Number of machines to be produced · Cost of construction · Assembling |
6 M |
R |
W-16 |
||||||||||||
|
16 |
State four types of loads
acting on machine elements. 1. Dead or steady load. A load is said to be
a dead or steady load, when it does not change in magnitude or direction. 2. Live or variable load.A load is said to
be a live or variable load, when it changes continually. 3. Suddenly applied or shock loads. A load
is said to be a suddenly applied or shock load, when it is suddenly applied
or removed. 4. Impact load. A load is said to be an
impact load, when it is applied with some initial velocity |
2 M |
R |
W-17 |
||||||||||||
|
17 |
State &Explain general
design procedure.  1. Recognition of
need. First of all, make a complete statement of the problem, indicating
the need, aim or purpose for which the machine is to be designed. 2. Synthesis
(Mechanisms). Select the possible mechanism or group of mechanisms which
will give the desired motion. 3. Analysis of
forces. Find the forces acting on each member of the machine and the
energy transmitted by each member. 4. Material selection. Select the
material best suited for each member of the machine. 5. Design of
elements (Size and Stresses). Find the size of each member of the machine
by considering the force acting on the member and the permissible stresses
for the material used. It should be kept in mind that each member should not
deflect or deform than the permissible limit. 6. Modification.
Modify the size of the member to agree with the past experience and judgment
to facilitate manufacture. The modification may also be necessary by
consideration of manufacturing to reduce overall cost. 7. Detailed
drawing. Draw the detailed drawing of each component and the assembly of
the machine with complete specification for the manufacturing processes
suggested. Prepare assembly drawing giving part numbers, overall dimensions
and part list. The component drawing is supplied to the shop flow for
manufacturing purpose, while assembly drawing is supplied to the assembly
shop 8. Production.
The component, as per the drawing, is manufactured in the workshop. |
4 M |
R,U |
S24,25 |
||||||||||||
|
18 |
Explain importance of
aesthetics consideration in design with examples. Aesthetics means the appearance and look
of a product. It is important because people like products that look good. Importance with examples: 1. Attracts
customers 2. Increases
product value 3. Gives
good impression 4. Improves
user satisfaction 5. Helps
in market success Aesthetic design makes the product attractive, acceptable in the market,
and successful. |
6 M |
U |
W17,18 |
||||||||||||
|
19 |
State the meaning of colour
codes in aesthetic design: a)Red b) Orange
c) Green d) Blue
|
4 M |
R |
W16,S23 |
||||||||||||
|
20 |
State general equation for
bending moment & state meaning of each term.
|
2 M |
R |
S-18 |
||||||||||||
|
21 |
State torsion equation and
state meaning of each term
|
2 M |
R |
S-18 |
