Strength of Materials

Strength of Materials
Fluid Mechanics & Fluid Machines
Design of Machine Elements
Thermodynamics
Materials Engineering
Metrology & Instrumentation
Manufacturing Technology
Heat Transfer
Internal Combustion Engines and Gas Turbines
Advance Manufacturing Technology
Kinematics & Theory of Machines
Manufacturing Processes
Engineering Mechanics
Solid Mechanics
Applied Thermodynamics

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1.

The Young’s modulus of elasticity of a material is 2.5 times its modulus of rigidity. Find the Poisson’s ratio for the material?

Answer:

$E=2G(1+\mu )\phantom{\rule{0ex}{0ex}}or,1+\mu =\frac{E}{2G}\phantom{\rule{0ex}{0ex}}or,\mu =\frac{E}{2G}\u20131\phantom{\rule{0ex}{0ex}}or,\mu =\frac{2.5}{2}-1\phantom{\rule{0ex}{0ex}}\therefore \mu =0.25$

2.

The moduli of elasticity and rigidity of material are 200 GPa and 80 GPa, respectively. What is the value of Poisson’s ratio of the material?

Answer:

$E=2G(1+\mu )\phantom{\rule{0ex}{0ex}}or,\mu =\frac{E}{2G}\u20131\phantom{\rule{0ex}{0ex}}or,\mu =\frac{200}{2\times 80}-1\phantom{\rule{0ex}{0ex}}\therefore \mu =0.25$

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1.

A hydraulic jump forms at the downstream end of spillway carrying 18${m}^{3}/s$ discharge.** **If the depth before jump is 0.80m, determine the depth after the jump and energy loss

Answer:

Given:

Discharge Q= 18m/s

Depth before jump, ${d}_{1}=0.8$

Taking width b = 1m

Discharge per unit width, q = $\frac{18}{1}=18$

Let, ${d}_{2}$= Depth after jump

${h}_{L}=$Loss of energy

Hence using the expression

${d}_{2}=$$-\frac{{d}_{1}}{2}+\sqrt{\frac{{d}_{2}^{1}}{4}+\frac{2{d}^{2}}{g{d}_{1}}}$

$=-\frac{0.8}{2}+\sqrt{\frac{0.{8}^{2}}{4}+\frac{2\times {18}^{2}}{9.81\times 0.8}}$

$=-0.4+9.09$

$=8.69m$

$\therefore $ Depth after jump = 8.69m

Using the equation for loss of energy

${h}_{L}=\frac{({d}_{2}-{d}_{1}{)}^{3}}{4{d}_{1}{d}_{2}}=\frac{(8.69-0.8{)}^{3}}{27.808}=17.66m$

$\therefore $Energy loss = 17.66m

2.

Explain head-discharge curve for centrifugal pump

Answer:

The head-discharge curve (H-Q curve) relates the head produced by a pump to the volume of water pumped per unit time (i.e., discharge of the pump). Generally, the head produced by a pump steadily decreases as the discharge of the pump increases as shown in the figure. The values of the head and the discharge corresponding to the maximum efficiency are known as the design head or normal head and the design discharge or normal discharge of a pump. The shape of the H-Q curve varies with the specific speed and impeller designs of the pump.

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1.

Two rods are connected by means of a cotter joint. The outside diameter of the socket collar and the inside diameter of the socket are 100 and 50 mm respectively. Rods are subjected to a tensile force of 50kN and the cotter is made of steel 30C8$\left({S}_{yt}=400N/m{m}^{2}\right)$ and the factor of safety is 4. If width of the cotter is 5 times of thickness. Determine width and thickness of the cotter on the basis of shear failure.

Answer:

${S}_{yt}=400N/m{m}^{2},\left({f}_{s}\right)=4,$

$P=(50\times {10}^{3})N$

${d}_{4}=100mm$

${d}_{2}=50mm$

Permissible stress for cotter,

${\sigma}_{t}=\frac{{S}_{yt}}{\left({f}_{s}\right)}=\frac{400}{4}=100N/m{m}^{2}$

$\tau =\frac{{S}_{yt}}{\left({f}_{s}\right)}=\frac{0.5{S}_{yt}}{\left({f}_{s}\right)}=\frac{0.5\left(400\right)}{4}=50N/m{m}^{2}$

The width and thickness on the basis of bending failure,

$b=5t$

$P=2bt\tau Or,50\times {10}^{3}=2\left(5t\right)t\left(50\right)$

$Thicknessofcottert=10mmand$

$Widthofcotterb=5t=50mm$

2.

A gib and cotter joint is used for strap-type big end of the connecting rod. The rod is subjected to a maximum pull of 50kN and the diameter of the circular path of the rod adjacent to strap end is 60mm. The strap gib and cotter are made by plain carbon steel of grade 30C8 $({s}_{yt}=400N/m{m}^{2})andthe\left({f}_{s}\right)is10$. Calculate the strap width and the cotter thickness.

Answer:

$P=50\times {10}^{3}N\phantom{\rule{0ex}{0ex}}d=60mm\phantom{\rule{0ex}{0ex}}{S}_{yt}=400N/m{m}^{2}\phantom{\rule{0ex}{0ex}}\left({f}_{s}\right)=10\phantom{\rule{0ex}{0ex}}Permissiblestresses\phantom{\rule{0ex}{0ex}}{\sigma}_{t}=\frac{{S}_{yt}}{\left({f}_{s}\right)}=\frac{400}{10}=40N/m{m}^{2}\phantom{\rule{0ex}{0ex}}\tau =\frac{{S}_{yt}}{\left({f}_{s}\right)}=\frac{(0.5{S}_{yt})}{\left({f}_{s}\right)}=\frac{0.5\left(400\right)}{10}=20N/m{m}^{2}\phantom{\rule{0ex}{0ex}}Widthofstrap{B}_{1}=d=60mm\phantom{\rule{0ex}{0ex}}Thicknessofcottert=\left(\frac{{B}_{1}}{4}\right)=\left(\frac{60}{4}\right)=15mm$

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1.

What do you understand by quality of energy?

Answer:

The quality of entropy is used to explain the quality of energy. However, it can be explained in terms of high quality energy and low quality energy. High quality energy means electrical energy, mechanical energy and chemical store energy. Low quality energy means low temperature heat. High quality energy is used to produce low quality energy but it is almost impossible to to produce high quality energy from low quality energy. Therefore, study of high quality energy is more important than high quality energy. Thus conservation of high quality energy is very curtailed.

2.

What is pure substance? Draw the phase equilibrium diagram for a pure substance on P-V, T-S and P-T plot with relevant constant property lines.

Answer:

It is a single substance or a mixture of substances, which has the same chemical composition throughout the system and does not change chemically, is called a pure substance. In other words it is a homogeneous substance or substances, whose molecular weight does not vary. Examples of pure substances are:

a) Gases such as oxygen (O_{2}), nitrogen (N_{2}), carbon dioxide (CO_{2}), superheated steam etc.

b) Mixture of gases such as air, oxygen and gaseous fuel etc., because these has a uniform chemical composition throughout.

c) combination of liquid and vapour of same substance like combination of water and steam, combination of liquid and vapour oxygen etc., as long as the combination is homogeneous. However, a mixture of oil and water is not considered as a pure substance, as the oil is not soluble in water rather, it floated on the top of water surface and also generates two chemically dissimilar regions.

A phase is a chemically or physically homogeneous portion of a mixture such as melting ice is a mixture of two phases i.e., solid phase and liquid phase. The individual pure substance making of a mixture is called component or constituent. A mixture of ice water and steam can be treated as a single component system. A mixture of different gasses may be considered as one component system provided the chemical composition of the mixture is uniform. If one of the gas condenses the mixture is no longer be considered as a one component system, since the condense phase has a different composition than the original mixture. The phase equilibrium diagram for a pure substance on P-V, T-S and P-T has been shown with relevant constant property lines.

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1.

Determine the volume and theoretical density of the FCC unit cell of an atom having an atomic radius of 0.1 nm and an atomic weight of 60.0 g/mol. Take ${N}_{A}=6.023\times {10}^{23}atoms/mol.$

Answer:

$Giventhat\phantom{\rule{0ex}{0ex}}R=0.1\times {10}^{-9}m.\phantom{\rule{0ex}{0ex}}ThevolumeofFCCunitcellisgivenby{V}_{c}=16\sqrt{2}{R}^{3}=0.0226n{m}^{3}\phantom{\rule{0ex}{0ex}}NumberofatomsperunitcellforFCCstructurearen=4;MolarweightM=60g/mol.\phantom{\rule{0ex}{0ex}}Therefore,densityofunitcellisgivenby\rho =\frac{n\times (M/{N}_{A}}{{V}_{c}}=17.63g/c{m}^{3}$

2.

Rhenium has an HCP crystal structure with an atomic radius of 0.137 nm and a c/a ratio of 1.615. Compute the volume of the unit cell for Rhenium.

Answer:

$ThevolumeofunitcellforHCP{V}_{C}=c\times basearea\phantom{\rule{0ex}{0ex}}ThebaseareaforHCPstructureis6{R}^{2}\sqrt{3}\phantom{\rule{0ex}{0ex}}Giventhat,c=1.615a\phantom{\rule{0ex}{0ex}}Anda=2R\phantom{\rule{0ex}{0ex}}\therefore c=323R\phantom{\rule{0ex}{0ex}}Again,giventhatR=0.137nm\phantom{\rule{0ex}{0ex}}\therefore {V}_{C}=3.23\times 0.137nm\times 6\times (0.137nm{)}^{2}\times \sqrt{3}=0.0863n{m}^{3}$

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1.

A Vernier scale consists of 25 divisions on 12 mm spacing and the main scale has 24 divisions on 12 mm. What is the least count?

Answer:

25 divisions of vernier scale = 24 divisions of main scale

Smallest division on main scale $=\frac{12}{24}mm=0.5mm$

Therefore, the value of smallest division on vernier scale $=\frac{24}{25}\times 0.5mm$

Thus, least count = value of smallest division on main scale - value of smallest division on vernier scale $=\frac{12}{24}-\frac{24}{25}\times 0.5=0.02mm$

2.

Define relative error, random error and systematic errors.

Answer:

Relative error: It is the quotient of the absolute error and the value of comparison used for the calculation of that absolute error.

Random Error: Random errors are non-consistent. They occur randomly and are accidental in nature. Such errors are inherent in the measuring system. It is difficult to eliminate such errors.

Systematic error: They are regularly repetitive in nature. They are of constant and similar form these are also called controllable errors.

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1.

Two cubical castings of the same metal and sizes of 2 cm side and 4 cm side are molded in green sand. If the smaller casting solidifies in 2 minutes then find the expected time of solidification of large casting.

Answer:

$Module=\frac{V}{A}\phantom{\rule{0ex}{0ex}}Forcube1,{M}_{1}=\frac{{2}^{3}}{6\times {2}^{2}}=\frac{1}{3}\phantom{\rule{0ex}{0ex}}Forcube2,{M}_{2}=\frac{{4}^{3}}{6\times {4}^{2}}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}Solidificationtime\propto {M}^{2}\phantom{\rule{0ex}{0ex}}Hencewehave\frac{{T}_{1}}{{T}_{2}}=\frac{{M}_{1}^{2}}{{M}_{2}^{2}}\phantom{\rule{0ex}{0ex}}Given{T}_{1}=2min\phantom{\rule{0ex}{0ex}}{T}_{2}=2\times {\left(\frac{2}{3}\right)}^{2}\times {\left(\frac{3}{1}\right)}^{2}=8min$

2.

With a solidification factor of $0.97\times 10s/{m}^{2}$, find out the solidification time (in seconds) for a spherical casting of 200 mm diameter.

Answer:

$Solidificationtime,t=Q{\left(\frac{V}{A}\right)}^{2}\phantom{\rule{0ex}{0ex}}t=0.97\times {10}^{6}\left(\frac{{\displaystyle \frac{4}{3}}\pi (0.1{)}^{3}}{4\pi (0.1{)}^{2}}\right)\phantom{\rule{0ex}{0ex}}t=1077.78sec\phantom{\rule{0ex}{0ex}}t=1078sec$

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1.

Determine the heat transfer coefficient from a 120mm diameter steel pipe placed horizontally in ambient at 25℃ . Nusselt number is 30 and thermal conductivity of air is 0.05W/mK.

Answer:

Given

$D=0.12m\phantom{\rule{0ex}{0ex}}Nu=30\phantom{\rule{0ex}{0ex}}k=0.05W/mk\phantom{\rule{0ex}{0ex}}Heattransfercoefficient(W/{m}^{2}K)isgivenby\phantom{\rule{0ex}{0ex}}Nu=\frac{hD}{k}\phantom{\rule{0ex}{0ex}}h=\frac{(Nu\times k)}{D}=12.5W/{m}^{2}K$

2.

A fin has a 5mm diameter and 100mm length. The thermal conductivity of fin material is 400W/mk. One end of the fin is maintained at 130℃ and its remaining surface is exposed to ambient air at 30℃. If the convective heat transfer coefficient is $40W/{m}^{2}K$ then find the heat loss (in W) from the fin.

Answer:

$Givend=0.005m\phantom{\rule{0ex}{0ex}}l=0.1m\phantom{\rule{0ex}{0ex}}k=400W/mk\phantom{\rule{0ex}{0ex}}{T}_{o}=130\mathbb{C}\phantom{\rule{0ex}{0ex}}{T}_{\infty}=30\mathbb{C}\phantom{\rule{0ex}{0ex}}h=40W/{m}^{2}K\phantom{\rule{0ex}{0ex}}Therefore,cross-sectionalareaandperimeterare\phantom{\rule{0ex}{0ex}}A=\frac{\pi {d}^{2}}{4p}=\pi d\phantom{\rule{0ex}{0ex}}Assuminglongfin,heattransferis\phantom{\rule{0ex}{0ex}}\dot{Q}=\sqrt{hpkA}({T}_{o}-{T}_{\infty})\phantom{\rule{0ex}{0ex}}=\sqrt{40\times \pi \times 0.005\times \pi \times \frac{0.{005}^{2}}{4}}\times (130-30)\phantom{\rule{0ex}{0ex}}=7.02W$

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1.

A gas turbine plant receives air at 1bar and 290K and compresses it to 5 bar. If the temperature of the air after compression is 1000k, find the thermal efficiency of the turbine. (Take γ for the air as 1.4)

Answer:

$Given:\phantom{\rule{0ex}{0ex}}{p}_{3}={p}_{4}=1bar;{T}_{4}=290K\phantom{\rule{0ex}{0ex}}{p}_{1}={p}_{2}=5bar\phantom{\rule{0ex}{0ex}}{T}_{2}=1000\phantom{\rule{0ex}{0ex}}{K}_{y}=1.4\phantom{\rule{0ex}{0ex}}Weknowthatpressureratio\phantom{\rule{0ex}{0ex}}r=\frac{{p}_{1}}{{p}_{4}}\phantom{\rule{0ex}{0ex}}=\frac{{p}_{2}}{{p}_{3}}\phantom{\rule{0ex}{0ex}}=\frac{5}{1}\phantom{\rule{0ex}{0ex}}=5\phantom{\rule{0ex}{0ex}}\therefore Thermalefficiencyoftheturbine,\phantom{\rule{0ex}{0ex}}{\eta}_{th}=1-{\left(\frac{1}{r}\right)}^{\frac{\gamma -1}{\gamma}}\phantom{\rule{0ex}{0ex}}=1-\frac{1}{5}\phantom{\rule{0ex}{0ex}}=1-(0.2{)}^{0.286}\phantom{\rule{0ex}{0ex}}=1-0.631\phantom{\rule{0ex}{0ex}}=0.369or36.9\%$

2.

A gas engine has a piston diameter of 150mm, length of stroke 400mm and mean effective pressure of 5.5 bar. The engine makes 120 explosions per minute. Determine the mechanical efficiency of the engine, if its B.P. is 5kW.

Answer:

$Given:\phantom{\rule{0ex}{0ex}}{D}_{p}=150mm=0.15m\phantom{\rule{0ex}{0ex}}L=400mm=0.4m\phantom{\rule{0ex}{0ex}}{p}_{m}=5.5bar\phantom{\rule{0ex}{0ex}}n=120B.P=5kW\phantom{\rule{0ex}{0ex}}Weknowthatareaofthepiston,A=\frac{\pi}{4(0.15{)}^{2}}=0.0177{m}^{2}\phantom{\rule{0ex}{0ex}}Indicatedpower,I.P.=\frac{100{p}_{m}LAn}{60}\phantom{\rule{0ex}{0ex}}=\frac{100\times 5.5\times 0.4\times 0.0177\times 120}{60}\phantom{\rule{0ex}{0ex}}=7.79kW\phantom{\rule{0ex}{0ex}}Weknowthatmechanicalefficiencyoftheengine,\phantom{\rule{0ex}{0ex}}{\eta}_{m}=\frac{B.P.}{I.P.}\phantom{\rule{0ex}{0ex}}=\frac{5}{7.79}\phantom{\rule{0ex}{0ex}}=0.642or64.2\%$

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1.

Define the following:

Flexible manufacturing system, Transfer lines, project shop, Job shop, Cellular Manufacturing.

Answer:

Flexible manufacturing system: It is a manufacturing system in which there is some amount of flexibility that allows the system to react in case of changes, whether predicted or unpredicted.

Transfer lines: It is a manufacturing system which consists of a predetermined sequence of machines connected by an automated material handling system and designed for working on a very small family of parts.

Job shop: Job shops are typically small manufacturing systems that handle job production that is custom / bespoke or semi-custom / bespoke manufacturing process such as small to medium-size customer orders or batch jobs. Project shop: It is used when a product is one of a kind which is too large to be moved. It is highly flexible and finite duration often with deadline.

Cellular manufacturing: This manufacturing is designed to take advantages of mass production layout and techniques in smaller batch - production system. It is also called group technology.

2.

What are the different levels of automation?

Answer:

Automation is generally defined as the preprocess of having machines follow a predetermined sequence of operations with little or no human involvement, using specialized equipment and devices that perform and control manufacturing process. Automation is achieved through the use of variety of devices, sensors, actuators and specialized equipment that are capable of (a) observing all aspects of the manufacturing operation, (b) making decisions concerning the changes that should be made and (c) controlling all aspects of the operation.

It has been implemented successfully in the following basic areas of production:

i) Manufacturing processes and operations

ii) Material handling

iii) Inspection

iv) Assembly

v) Packaging

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1.

What is the maximum transmission angle when the crank angle is fixed with a link?

Answer:

The angle μ between the output link and the coupler is called as transmission angle. In fig. the input link is AB, DC the force applied to the output link is transmitted through the coupler BC. For the particular value of force in the coupler rod, torque transmitted through the coupler BC. For the particular valve of force on the coupler rod, the torque transmitted to the output link is maximum when the transmission angle μ is 90°. If BC and DC links become coincident, the transmission angle is zero and the machine lock or jam. Hence μ is kept more than 45°. The best mechanisms, therefore, have a transmission angle that does not deviate much from 90°.

Appling cosine law to triangles ABD and triangle BCD

${a}^{2}+{d}^{2}-2adcos\theta ={k}^{2}\dots \left(i\right)\phantom{\rule{0ex}{0ex}}{b}^{2}+{c}^{2}-2bccos\mu ={k}^{2}\dots .\left(ii\right)\phantom{\rule{0ex}{0ex}}Andfrom\left(i\right)and\left(ii\right),weget\phantom{\rule{0ex}{0ex}}{a}^{2}+{d}^{2}-2adcos\theta ={b}^{2}+{c}^{2}-2bccos\mu \phantom{\rule{0ex}{0ex}}Or,{a}^{2}+{d}^{2}-{b}^{2}-{c}^{2}-2adcos\theta +2bccos\mu =0\phantom{\rule{0ex}{0ex}}Thevaluesofthetransmissionangleofmaximumorminimumcanbefoundbyputtingd\mu /d\theta =zero.\phantom{\rule{0ex}{0ex}}Differentiatingtheaboveequationonrespectto\theta ,\phantom{\rule{0ex}{0ex}}2ad\mathrm{sin}\theta -2bc\mathrm{sin}\mu .\frac{d\mu}{d\theta}=0\phantom{\rule{0ex}{0ex}}\frac{d\mu}{d\theta}=\frac{ad\mathrm{sin}\theta}{bc\mathrm{sin}\mu}\phantom{\rule{0ex}{0ex}}Or,thusifd\mu /d\theta tobezero,thetermad\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}hastobezeromeansiseither0\xb0or180\xb0.\phantom{\rule{0ex}{0ex}}Itisseenthat\mu ismaximumwhen\theta is180\xb0.$

2.

What is the inversion of the quick-return mechanism?

Answer:

We make a cylinder of an oscillating cylinder engine in the form of a guide and we make the piston in the form of a slider, arrangement is shown in the fig. is obtained. If the crank rotates about A, guide 4 oscillates about B. Point c on the guide, link 5 is pivoted, the other end of which is connected to the cutting tool through a pivoted joint.

Fig. b shows the extreme positions of the oscillating guide 4. The time of the forward stroke is proportional to the angle θ whereas, for the return stroke, it is proportional to angle β, provided the crank rotates clockwise.

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1.

A casting of size $100\times 100\times 100m{m}^{3}$ solidifies in 20 minutes. Find out the solidification time for casting of size $100\times 100\times 50u{m}^{3}$ under the same condition.

Answer:

$Castingsite=100\times 100\times 100m{m}^{3}\phantom{\rule{0ex}{0ex}}Solidifiestime=20minutesSolidificationtime={t}_{2}\phantom{\rule{0ex}{0ex}}Forsize=100\times 100\times 50m{m}^{3}\phantom{\rule{0ex}{0ex}}Solidificationtime{t}_{s}=K[V{]}^{2}\phantom{\rule{0ex}{0ex}}WhereK-SolidificationareaA=surfacearea\phantom{\rule{0ex}{0ex}}Forsamecondition=T{s}_{1}={\left[\frac{100\times 100\times 100}{6\times {100}^{2}}\right]}^{2}\phantom{\rule{0ex}{0ex}}K=0.072.\phantom{\rule{0ex}{0ex}}T{s}_{2}=\frac{100\times 100\times 50}{2\times 100\times 100+2\times 100\times 50+2\times 100\times 50}\phantom{\rule{0ex}{0ex}}=11.25\approx 11.3minutes$

2.

What is centrifugal casting?

Answer:

Centrifugal casting a casting process in which molten metal poured in cavity and allows to solidify while the mould is revolving is called the centrifugal casting process. The casting produced under this centrifugal force is called centrifugal casting. This process is especially employed for casting articles of V shape.

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1.

A block of mass m rests on an inclined plane and is attached by a string to the wall as shown in the figure. The coefficient of static friction between the plane and the block is 0.25. The string can withstand a maximum force of 20 N. The maximum value of the mass (m) for which the string will not break and the block will be in static equilibrium is.

(Take cosθ=0.8 and sinθ=0.6 and acceleration due to gravity )

Answer:

$R=mg\mathrm{cos}\theta \dots .\left(1\right)\phantom{\rule{0ex}{0ex}}mg\mathrm{sin}\theta =T+\mu R\phantom{\rule{0ex}{0ex}}Or,T=mg\mathrm{sin}\theta -\mu R\phantom{\rule{0ex}{0ex}}=mg\mathrm{sin}\theta -\mu mg\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}=mg(\mathrm{sin}\theta -\mu \mathrm{cos}\theta )\phantom{\rule{0ex}{0ex}}Or,m=\frac{T}{g(\mathrm{sin}\theta -\mu \mathrm{cos}\theta )}\phantom{\rule{0ex}{0ex}}=\frac{20}{10(0.6-0.25\times 0.8)}\phantom{\rule{0ex}{0ex}}=\frac{20}{10(0.6-0.2)}\phantom{\rule{0ex}{0ex}}=\frac{2}{0.4}\phantom{\rule{0ex}{0ex}}=\frac{20}{4}\phantom{\rule{0ex}{0ex}}=5kg$

2.

If equal Angle is made by the force A with X and Y in the given figure, then find the angle, which is made by force A with the other forces X and Y?

Answer:

$Inequilibrium\frac{80}{sin\alpha}=\frac{40}{sin(180\u20132\alpha )}\phantom{\rule{0ex}{0ex}}sin\alpha =2sin2\alpha \phantom{\rule{0ex}{0ex}}sin\alpha =2[2sin\alpha cos\alpha ]\phantom{\rule{0ex}{0ex}}cos\alpha =(1/2)\alpha =co{s}^{-1}\left)\right(1/4)\phantom{\rule{0ex}{0ex}}\alpha =75.52\xb0$

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1.

A concentrated load of P acts on a simply supported beam of span L at a distance $L/3$ from the left support. The bending moment at the point of application of the load will be how much?

Answer:

Reaction at the left end of the beam can be found out by taking moment about the other end of the simply supported beam

$R\times L-P\times (L-L/3)=0\phantom{\rule{0ex}{0ex}}R\times L=P\times (L-L/3)\phantom{\rule{0ex}{0ex}}R=\frac{2P}{3}$

Therefore, bending moment at the point of application due to the reaction at left end is given by

$M=R\times \frac{L}{3}\phantom{\rule{0ex}{0ex}}=\frac{2PL}{9}$

2.

Two beams, one with a square cross-section and the other with a circular cross-section, are subjected to the same bending moment. If the material and the cross-sectional area in, both beams are same then, proof that the bending stress in circular beam is more than that of the square beam.

Answer:

If a is the size of the square (s) section and d is the diameter of circular (c) section, then

${a}^{2}=\frac{\pi}{4}{d}^{2}\phantom{\rule{0ex}{0ex}}a=\frac{\pi}{2d}\phantom{\rule{0ex}{0ex}}Ratioofsectionmodulusis\phantom{\rule{0ex}{0ex}}\frac{{Z}_{s}}{{Z}_{c}}=\frac{\pi {d}^{3}/32}{{a}^{3}/6}\phantom{\rule{0ex}{0ex}}=1.18164\phantom{\rule{0ex}{0ex}}Bendingstressisgivenby\phantom{\rule{0ex}{0ex}}\sigma =MZ\phantom{\rule{0ex}{0ex}}Materialofbeamisthesame.Theratioofstressesis\phantom{\rule{0ex}{0ex}}\frac{{\sigma}_{c}}{{\sigma}_{s}}=\frac{{Z}_{c}}{{Z}_{s}}=1.18164\phantom{\rule{0ex}{0ex}}Therefore,\phantom{\rule{0ex}{0ex}}{\sigma}_{c}{\sigma}_{s}\phantom{\rule{0ex}{0ex}}$

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1.

If an engine of 40 % thermal efficiency drives a refrigerator having a coefficient of performance of 5. Determine the heat input to the engine per kJ of heat removed from the cold body of the refrigerator.

Answer:

$Giventhat\phantom{\rule{0ex}{0ex}}\eta =0.4\phantom{\rule{0ex}{0ex}}COP=5\phantom{\rule{0ex}{0ex}}{Q}_{2}=1kJ\phantom{\rule{0ex}{0ex}}Theheatengineandrefrigeratorareshowninthefigure.\phantom{\rule{0ex}{0ex}}\eta =\frac{Output}{Input}\phantom{\rule{0ex}{0ex}}\eta =\frac{W}{{Q}_{1}}\phantom{\rule{0ex}{0ex}}W=\eta \times {Q}_{1}\dots ..\left(1\right)\phantom{\rule{0ex}{0ex}}COP=\frac{HeatRemoved}{Workinput}\phantom{\rule{0ex}{0ex}}COP=\frac{{Q}_{2}}{W}\phantom{\rule{0ex}{0ex}}W=COP\times {Q}_{2}\dots \dots \left(2\right)\phantom{\rule{0ex}{0ex}}CombiningEquation\left(1\right)and\left(2\right),weget\phantom{\rule{0ex}{0ex}}\eta \times {Q}_{1}=COP\times {Q}_{2}\phantom{\rule{0ex}{0ex}}Heatinputtotheengineisgivenby\phantom{\rule{0ex}{0ex}}{Q}_{1}=\frac{{Q}_{2}}{COP\times \eta}\phantom{\rule{0ex}{0ex}}=\frac{1}{5\times 0.4}\phantom{\rule{0ex}{0ex}}=0.5kJ$

2.

For a heat engine operates on the Carnot cycle; the work output is 1/4th of heat rejected to the sink. Find the thermal efficiency of the engine.

Answer:

$w=\frac{1}{4}{Q}_{2};againw={Q}_{1}-{Q}_{2}\phantom{\rule{0ex}{0ex}}\frac{1}{4}{Q}_{2}={Q}_{1}-{Q}_{2}\phantom{\rule{0ex}{0ex}}{Q}_{1}=\frac{5}{4}{Q}_{2}\phantom{\rule{0ex}{0ex}}\eta =\frac{W}{{Q}_{1}}=20\%$

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