mechanical design engineer interview questions and answers
If you have done mechanical engineering degree or diploma and looking for any mechanical engineering job. Then this article definitely will help you to prepare and face interview. This article is about mechanical design interview questions and answers for freshers and experienced.This questions are frequently asked in mechanical design engineer interview.
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This is the first question asked as a mechanical design engineer .Machine design is defined as the use of scientific principles, technical information’s and imagination in the description of a machine or mechanical system to perform specific function with maximum economy and efficiency.
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Machine design is the creation of new and better machines and improving the existing ones. A new or better machine is one which is more economical in the overall cost of production and operations.
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1. Type of load and stresses caused by the load.
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Mass:It is the amount of matter contained in a given body and does not vary with the change in its position on the earth’s surface. The mass of a body is measured by direct comparison with a standard mass by using a lever balance.
Weight:It is the amount of pull, which the earth exerts upon a given body. Since the pull varies with the distance of the body from the center of the earth, therefore, the weight of the body will vary with its position on the earth’s surface (say latitude and elevation). It is thus obvious, that the weight is a force.
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Steel – 7850 kg/m^3
Aluminium – 2700 kg/m^3
Wrought iron- 7780 kg/m^3
Copper – 8900 kg/m^3
Cobalt – 8850 kg/m^3
Tungsten – 19 300 kg/m^3
Ans: The engineering materials are mainly classified as:
The metals may be further classified as:
(a)Ferrous metals, and (b) Non-ferrous metals.
The ferrous metals are those which have the iron as their main constituent, such as cast iron, wrought iron and steel.
The non-ferrous metals are those which have a metal other than iron as their main constituent, such as copper, aluminium, brass, tin, zinc, etc.
Ans: Heat treatment is operation or a combination of operations, involving the heating and cooling of a metal or an alloy in the solid state for the purpose of obtaining certain desirable conditions or properties without change in chemical composition. The aim of heat treatment is to achieve one or more of the following objects:
Following heat treatment processes commonly employed in engineering practice:
Ans: 1. Duralumin, 2.Y-alloy, 3.Magnalium, 4.Hindalium.
Ans: It is an alloy of copper, tin and zinc. It usually contains 88% copper, 10% tin and 2% zinc. This metal is also known as Admiralty gun metal. The zinc is added to clean the metal and to increase its fluidity.
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Metals- 1. Copper-base alloys, 2. Lead-base alloys, 3. Tin-base alloys, and 4. Cadmium-base alloys.
A bearing material should have the following properties:
Ans: According to mechanical design,The degree of tightness or looseness between the two mating parts is known as a fit of the parts.
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Ans: Hole tolerance
We know that hole tolerance = Upper limit of hole – Lower limit of hole
= 25.02 – 25 = 0.02 mm
Shaft tolerance We know that shaft tolerance = Upper limit of shaft – Lower limit of shaft
= 24.97 – 24.95 = 0.02 mm
Allowance We know that allowance = Lower limit of hole – Upper limit of shaft
= 25.00 – 24.97 = 0.03 mm
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When some external system of forces or loads act on a body, the internal forces (equal and opposite) are set up at various sections of the body, which resist the external forces. This internal force per unit area at any section of the body is known as unit stress or simply a stress.
It is denoted by a Greek letter sigma (σ). Mathematically,
Stress, σ = P/A
where P = Force or load acting on a body, and A = Cross-sectional area of the body.
In S.I. units, the stress is usually expressed in Pascal (Pa) such that 1 Pa = 1 N/m2.
In actual practice, we use bigger units of stress i.e. megapascal (MPa) and gigapascal (GPa), such that
1 MPa = 1 × 106 N/m2
= 1 N/mm2 and 1 GPa
= 1 × 109 N/m2 = 1 kN/mm2
When a system of forces or loads act on a body, it undergoes some deformation. This deformation per unit length is known as unit strain or simply a strain. It is denoted by a Greek letter epsilon (ε). Mathematically,
Strain, ε = δl / l or δl = ε.l where
δl = Change in length of the body, and l = Original length of the body.
Ans: It is in general, as the ratio of the maximum stress to the working stress. Mathematically, Factor of safety = Maximum stress ÷Working or design stress
In case of ductile materials e.g. mild steel, where the yield point is clearly defined, the factor of safety is based upon the yield point stress
And it is ,
In case of brittle materials e.g. cast iron, the yield point is not well define as for ductile materials. Therefore, the factor of safety for brittle materials is based on ultimate stress. ∴
Ans: 1.Rivet joint
2.Flange Joint
3.Spigot and socket joint
4.Welded Joints – a. Lap joint, b. Butt Joint
5.Bolted Joint
6.Screw Joint
7.Cotter and Knuckle Joint.
Ans: 1. Fusion Welding
Ans: 1. Sunk keys, 2. Saddle keys, 3. Tangent keys, 4. Round keys, and 5. Splines.
It is used to connect two shafts which are perfectly aligned. Following types of rigid coupling are important from the subject point of view:
(a) Sleeve or muff coupling. (b) Clamp or split-muff or compression coupling, and (c) Flange coupling.
It is used to connect two shafts having both lateral and angular misalignment. Following types of flexible coupling are important from the subject point of view:
(a) Bushed pin type coupling, (b) Universal coupling, and (c) Oldham coupling.
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Ans: 1. Flat belt, V belt and Circular or Rope.
Ans: Tolerance is allowable variation for any given size or the total amount by which a given dimension may vary in order to achieve proper function of product.
Tolerance for hole = Hole (MMC)- Hole (LMC)
Ans: According to mechanical design, The production of closely mating parts, with very small tolerances is theoretically possible, but economically its unfeasible as this will increase rejection rate and high precise tools and workforce will be required.so that deviation is required in parts.
Ans: Engineering drawing a type of technical document used to transfer technical information, to define requirements. Or its graphical language that communicates ideas and information from one mind to another.
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Ans: 1. First angle projections
2.Second angle projections
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It is the smallest value that can be measured by the measuring instrument.
Least count = Value of main scale division – Total number of vernier scale division
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Normal scale – Minimum distance measured by normal scale is 1 mm so that least count is 1 mm.
Least count of vernier caliper:
=Smallest reading on main scale ÷ No of divisions on vernier scale
=1/10
=0.1 mm so that vernier have least count 0.1 mm
Micrometre may have 0.01 mm least count.
Ans: As a mechanical design engineer you must know the basic behavior of brittle and ductile materials.
Thank you for reading interview questions and answers for mechanical design engineers.
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