tvnovellas.info Theory STRENGTH OF MATERIALS BOOK BY RAMAMRUTHAM

STRENGTH OF MATERIALS BOOK BY RAMAMRUTHAM

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This book on the Strength Of Materials deals with the basic principles of the tvnovellas.info topics have been introduced in a simple manner. The book has been. Results 1 - 16 of 22 Strength of materials by S. Ramamrutham and r. Narayan. 1 January by S Ramamrutham & R Narayanan. Currently unavailable. Download Strength of Material By S. Ramamrutham – This book on the Strength of Materials deals with the basic principles of the tvnovellas.info topics have been.

Simple Stresses and Strains Introduction Definitions, stress, strain, tensile and compres- sive stresses Sheat stress Plastic limit Hookes law poissons ratio Modulus of Elasticity -Modulus of Rigidity, Bulk Modulus Bars of varying section 'Extension of a tapering rod Composite section -modular ratio- Bar of uniform strength- Equivalent area of composite sections Temiterature stresses Hoop stress Stresses on oblique sections State of simple shear - Relation between the Elastic constants Volumetric Strain Rectangular blodt subject to normal stresses Diagonal tensile and diagonal compressive stresses Solved problems 1 to 71 Problems for exercise. Strain Energy Impact Leading Strain Energy Elastic, plastic and rigid members Stresses due to different types of axial loading Gradually applied loads Suddenly applied loadj Impact loads Solved pro- blems 72 to 84 Problems for exercise. Shear Farces and Bending Moments Definitions -Cantilevers, simply supported beam, fixed beam, continuous beams -Conception of Shear Force and Beading Moment Sign conventionsshear force and Bending Mo- ment diagrams for cantilevers, beams supported at ends. Beams with overhangs Point of contrafiexure -Member sub- jected to couples Members subjected to Oblique loading Miscellaneous types of members and corresponding S. Chapter Pages simple binding Practical application of bending equation Sectioi modulus Section moduli for different shapes- Rectangular, triangular, circular, I-section, T-section Normal force on a partial area of a beam section Moment of resis- tance of a partial area of a bmm section -Fliiched beams Equivalent section Beams of uniform strength Shear stress distribution on a beam section Shear stress distribution on rectangular, circular, triangular, I and T sections Shear stresses in bolts connecting components in laminated beams. Direct and Bending Stresses Stress distribution of the section of an eccentrically loaded rectangular column.

Since the rigid bar remains horizontal, FIs. Strain in steel Strain in brass. But i4. Cdnader the equilibrium of the rimd rodt see Fig. Hence the load must be on the bar at.

An axial load of 60, kg. I0 0I5CW I 4ca I "T f56cn Y. EquWaleot Area of a Compomid Sectioa Suppose a compound column consists of a concrete oolumn reinforced with steel bars.

Area of steel. Stress in concrete.

Materials strength by of ramamrutham book

A steel strip of cross-section X 10 mm. There are two bolts on the line of the pull. Assume E for steel is twice that of copper. Figure 30 shows the details of connection of the members mentioned in the problem. Let the load applied on the connection be kg. Let the load trans- ferred to the bolt AbtPi kg. Hence between the two bolts the load in the steel plate will be P-Pi kg. This load will be transferred to the bolt B. The load Pi transferred to the bolt A will be tratisferred to the two copper plates between the two bolts.

Now consider the plates between the wo bolts. Load on the steel plate P Pi kg. ProUem Two copper rods and a steel rod, together support SL The stKsm. Young s Modulus for steel is twice that of cooner. FroUen Two copper rods and one steel rod together support a load of kg. Find the stresses Us the rods. Each rod will be compressed by the same amount.

Let the strain in copper and steel be and respectively. Let the stresses in and copper be p, and pc respectively. Young's Modulus for steel is twice that load that cat be supported. Each rod will. Let the strain in steel and copper be e and e respectivdy. Three vertical rods equal in length and each 12 mm. If now an additional load of kg. The initial stress in each bar ,. A steel rod 18 mm. The tube is 75 cm. The nuts are tightened until the compressive load on the tube is 2 tonnes. Calculate the stresses in the tube and the rod.

Strength Of Materials By Ramamrutham

There are 4 threads per cm. When the nuts arc tightened the tube will be com- pressed and the rod will be elongated. Since HP external forces have been applied, the compressive load on the tnbe must be equal to the tensile load on the rod. Let the stresses in the rod and tube be p. Obviously ft is a compressive stress and ft is tensile, and.

Ax 75 cm. Consider any sectton XX of the rod distant x from the lower end. Weight of the rod below the sectioa XX where 41 sectional area of the rod. Let the weight per unit volume of the member be w Ginsider the equilibrium of the strip EFGH Total force acting upwards Total force acting downwards Let p be the uniform stress intensity. Vthe w tfthe bw metres hng.

Of ramamrutham strength by materials book

Fig, shows a P rigid square platform of negligible weight and of side I supported by four identical elastic pillars each of height h If a load P be applied at a point distant a and h from the adjacent sides AB and AD find the pressure on each pillar and the dep- ression of the centre of the plat- form, Soiotion.

SaoiF, 8a AlPa. ProHea If the three ropes are of the same sectioned ana. If a load W be placed at a distance ,Klfrom the Uft endt find the tensions in the three wires neglecting the weight of the beam. Let the tension. A uniform rope of length I units hangs vertically- Find the extension of the first a units of length of the rope from the top due to the weight of the rope itself Find also the total exten- sion of the rope.

Consider an elemental length dx of the rope at a distance x from the botto n of tne rope. Let the weight per unit volume of the rope be p.

Strength of Materials by Ramamrutham

Stress on the section of the elemental part. II aUll d Temperature Stresses. When the temperature of a material changes there will be corres- ponding change in the dimension. When a member is free ta expand or contract due to rise or fall of temperature, no stresses will be induced in the member. But, if the natural change in length due to rise or fall of temperature be prevented, stresses will be offered.

Let the temperature rise by T. If the member is allowed to freely expand no stresses will be induced. But if the member is prevented from expanding, compressive stresses will be induced.

This can be realized as follows. Let the rod B be of length t Let its ends A and B be fixed. IZf perature. The rod tends to expand by Let a be the sectional area of the rod. Stress , E Youngs modulus Strain. Suppose a rod of length L when subjected ro a rise of tem- perature is permitted to expand only by 5, the temperature strain. Take kg. O' cm. A 15 mm. The tube is closed at each end by rigid plates of negligible thickness.

The nuts are tightened lightly home on the projecting parts of the rod. If the temperature of the assembly is raised by 6 fC. Calculate the stresses developed in copper and steel. Actual expansions of steel - Actual expansion of copper. Area of steel tube. But, since.

Let S cm, be the final contraction of each rod. The free contraction of the gun metal rod is greater than 5, while the free contraction of the steel tube is less than S. Hence the steel tute will be subjected to compressive stress while the gun metal rod will be subjected to tensile stress.

Let and pt, be the stresses in steel and gun metal. For the equilibrium of the whole system. Total compressive force in steel Total tension in gun metal. Problem 48 A steel bar placed between two copper bars each having the same area and length as the steel bar at A5C. At this stage they are rigidly connected together at both the ends. Determine the original length and the final stresses in the bars.

Let the sectional area of the steel component be A cm. Steel is in tension and copper is in compression. For the equilibrium of the system Tension in steel compression in copper. Actual expansion of steel Actual expansion of copper. Now consider say the steel bar. Actual expansion of the steel bar. A 12 mm- diameter steel rod passes centrally through a copper tube 48 mm.

The tube is closed at each end by 24 mm. The nuts are tightened until the copper tube is reduced in length by O' mm.

Area of the copper tube cm. Me tightening to the nuts. When the nuts are tightened the steel K rod will be subjected to tcnsiic and me tensile stress ana the Fig. Stresses due to rise of temperature. If the two members had been free to expand. But since the ends of the rod are provided with washers and nuts the members are not free to expand fully Final expansion of each of the members will be the same.

Let this final expansion be 5. The free expansion of copper IS greater than 8 while the free expansion of steel is less thanl. Let f, and f. A steel rod 20 mm. Take x l fi kg. E - l2xl0-x80x2xl0A: Case ay When the ends yield by O il cm. A steel tube cm. The bar and the tube are rigidly connei ed together at the ends at a tem- perature of 30"C. Area of copper bar -Jr- cm-. Let be the actual expansion of each component. Hence steel is in tension and the copper bar ms in compression.

For the equiBbrium of the system. A weight of 20 tomes is supported by three short pillars,each -5 cm. The pillars are so adjusted that at a tem- perature of J5C each carries equal load. The temperature is then raised to 'C.

May, Solution. Area of each pillar. Stresses due to rise of temperature alone. Let the stresses due to rise of temperature aloue be pc kg Icm. X5 -prXlO p. A flat bar of aluminium alloy 24 mm, wide and 6 mm. The three bars are fastened together at their ends when the temperature is 1 PC.

Strength of Materials by S Ramamrutham

Find the stress in each of the materials when the temperature of the whole assembly is raised to SOTC. If at the new temperature a tensiie load of kg. Let this expansion be The free expansion of aluminium is greater than 8 while the free expansion of steel is less than Hence steel is subjected to tensile stress while aluminium is subjected to compressive stress. Let ps and Pa be the stresses in steel and aluminium. Hi Stresses due to enternal load of kg. Two steel rods one of 8 cm. The other end of each rod is rigidly fixed and there is inltialty a small tension in the rods.

If the effective length of each rod is 4 metres, find the increase in this tension when the turn buckle is turned by one-quarter of a turn. On the rod of bigger diameter there are IS meads per centimetre while there are 2 threads per centimetre on the other rod. Neglect the extension of the turn buckle, also what rise in temperature would nullify the increase in tension. Cross-sectional area of the stnalif- r bar.

Extension of the bigger bar cm. Total extension of the two rods cm. Total extension in the two bars. In order this tension must be nullified by rise of tempomture 7 total expansion of the two rods mustbeequal to.

A steel. Both the rod and the tube are 40 cms. Solution, a i When the distance between the stops remains constant. Strain in steel , expansion prevented ,, -r original length a.

Let the expansion of the composite member be 8 cm. The glial -tyie is heated so that its diameter exceeds D. Hence it will grip the wheel. Hence a pensile stress u induced circumferentially along the tyre. Sodi a sireas it called a hoop stress. Prahlem A rigid wheel is 3 metres in diameter. It is derired to shrink on to the wheel a thin steei tyre. Find the internal dia- meter of the tyre if qfier fitting the hoop stress in the tyre is kg.

Find also the minimum temperature to which the tyre is to be raised so that it can befitted over the wheel. Ratio St and W be the decrease in width and depth. When the deformation of the member is within the elastic limit it is found that the ratio of the lateral strain to the longitu- dinal strain is a constant for a given material.

This ratio is called. Volumetric Strain When a member is subjected to forces deforming it, it undergoes changes dimensions and hence its volume will be subjected to in its changes. The ratio of the change in volume to the original volume is called the volumetric strain. This is usually denoted by ev Change in vo lume Original volume Let its diameter be d. Let the diameter of a solid sphere hfd. Original volume of the sphere. A steel bar SO mm. Umg is subjected to an axial pull of kg.

Find the change in die lengtht wUth, thickness and the volume of the bar. Take J? Longitudinal strain. A bar of wAform rectangular seetbm A is stdtjectei to an axial tensile load P.

Show that the volumetric stndn is given by. Strain of the length et. PMMem A steel rod cms. Img and 20 mm. Let reference axes O. E mE mE Since for the metals we come across m lies between 3 and 4 the. This is no doubt possible if the stresses py and pt may not all be like stresses Regarding tensile stresses as positive and com- pressive stresses as negative the volumetric strain can be determined.

According to the above convention if c: Fig 54 a shows a rectangular bar of cross-sectional area A subjected to axial tensile load P Suppose we consider a normal cross-section 1 1 secticn normal to the axis of the member. P- intensity of stress on this normal cross-section 1 The. The atea corresponding to the section 2 2 -A sec 6 Intensity of stress on this section. Now the poll P applied can be resolved into a normal compo- nent PnP cos 6 and a tangential component P sin 6.

The above expressioss for the normal and tangential stress intensities may also be obtained as follows ; Consider unit area of the section Suppose the magnitude of the tensile force P be increased till failure occurs on the section The failure may be due to excessive normal stress pn or due to excessive tangential stress pi.

It is easily seen that there are two planes perpendicular to each other carrying the greatest shear stress and these planes are at 45 with the plane carrying the maximum normal stress. Element in a State of Simple Shear Fig. Let shear stresses of intensity q be set up on the faces A D and BC. Hence the forces acting on these faces, each of which equals q. AD, will form a couple. If the block should be in equilibrium, shear stresses of intensity say q' must be set up on the faces BA and DC.

The forces acting on these faces, each of which equals q' AB will form a restor- ing couple. This principle is called the principle of oomplementary shear stresses. Element in a State of Simple Shear. Consider an elemental rectangular block ABCD fig. Consider a plane BE at angle 0 with the face BC. Consider the equilibrium of the wedge BEC. This is subjected to the following forces, Fi?. Hence, we come to a very important conclusion.

When an element is in a state of simple shear, maximum direct stresses are induced on mutually perpendicular planes which are at 45 to the planes of pure shear.

One of the maximum direct stresses is tensile while the other maximum direct stress is compressive. These direct maximum tensile and compressive stress intensities are of the same magnitude as the intensity of shear stress on the planes of pure shear. Consider the forces on the part ABD. Hence when there is equilibrium Ihe force V2a9 is resisted. The pure direct tensile and compressive stresses acting on the diagonal planes BD and AC are called diagonal tensile and diagonal compressive sti esses.

Suppose the elemental block ABCD is of a material very poor in offering tensile stresses, then, as the magnitude of the shear stress will fail due to excessive diagonal q goes on increasing the block tensile stress. On the contrary if the material is very poor in offering compressive stresses, then a failure may occur by crushing due to excessive diagonal compressive stress.

Consider a square block ABCD of side a and of thickness unity perpendicular to the plane of the drawing Fig. Let the block be subjected to shear stresses of intensity q as shown in the figure.

Due to these stresses the block will be subjected to a deformation such that the diagonal idC is elongated and the dta gnnal BD is shortened. Consider the diagonal AC. The increase in length of the dia- gonal can be computed by considering the effect of the diagonal tensile and diagonal compressive stresses.

We know Fig, The strain of the diagonal AC may also be determined from the geometry of the distorted shape of the block Fig. Bulk Modulus. We find that the ratio of thr. When the deformation is within a certain limit, this ratio is called the bulk modulus and is usually denoted by K.

Let the faces of the cube be subjected to a direct stress of intensity p. Let E be the Youngs Modulus, and the Poissons ratio. Let us now consider the strain of one of the edges, say, AB. A rectangular block 25 cm.

X8 cm. Also calculate the change in the volume of the block due to the application of the loading specified above. Z The stresses in the directions of these axes are,. The modulus of rigidity of a material is 0'8 X l fi kg. For a given material the Young's Modulus is tonneslcm. A bar of metal 10 cms. X 5 cms. X25 cms. What change must be made in the tonnes load in order that there shall beno change in volume of the bar.

Let reference axes X, Y and Z be taken as sbown in Fig. Hence the compressive load on the 25 cm. A bar of steel is of square section 60 mm. It is subjected to axial load of 30 m tonnes. The lateral strain is prevented by the application of uniform.

If and E2xl lfi kg. When only half the lateral strain is prevented. Lateral strain J? A bar of steel is 4 cm. Strain along the F-axis. Let p the axial stress acting alone longitudinally i. A 60 mm. Find the norntal and tangential stress intensities across planes at 30, 45 and 60 with the normal section of the bar. Area of the normal section of the bar. Stress on the normal cross-section. The normal stress on any plane at an angle 0 with the normal section is given by.

The measured extension on gauge length of 20 cms. Calculate the Poisson's ratio aid the vcdues of the three moduli. A steel bar 40 mm. A cylindrical bar is 2 cm. Find the change in volume, when the bar is subjected to a hydrostatic pressure' of kg.

Find the elongation in cm. The value of the modulus of elasticity of the material is unknown. Exanqilcs on Oapter 1 1 rod of steel 6 cm. A It extends by 0 cm. Find also wbat would be the stress in the top and bottom washers when the nuts are tightened so as to produce a tension of kg.

Find the intensity of stress and the strain when it carries an axial compressive load of 95 tonnes. Take Fig. J 7 A rrd circular in section tapers from 2 cms. On applying an axial pull of kg. StniB Energy. When a load is applied on a member, the member is deformed.

The member offers a resistance against this deformation. We know that the internal resistance offered by the member is the total stress in the member. It is very important to note that only when the member has the capacity to offer a resistance to the deformation, a stress will be induced in the member, when the member is subjected to a deformation.

Suppose the top end A of the rod is fixed and the bottom end B is pulled down, so that the rod is extended by. In this position if R is the resistance offered by the member against the exten- sion. Stress intensity p" I. Since the bar is in equilibrium after Pip. The resistance at any instant depends upon the extension at that instant. This w ork done on the member will be stored by the mem- ber as energy and is called strain energy.

Strain energy is the energy absorbed or stored by a member when work is done on it to deform it. Suppose a member is of such a material that it docs not offer any resistance at all against the deformation. Such a member can- not regain its original shape after the load producing the deforma- tion IS removed, in oihcr words when the member allows itself to be deformed without oHering any resistance at all.

A member iraich allows itself to be deformed without any resistance is said to teptoyc. When a member is in a plastic stage, no stress will be uiduoed by its section due to any deformation of the member. Amcmber which allows itself to be deformed, bat will offci a mistanoe to the iteformation is said to be elastic.

When the external load producing the dc- focmatiofi is mraber will regain its original dimension. Strucniral members, we come across, are neither plastic nor rigid.

In general, structural members are elastic allowing themselves to be deformed and offering a resistance against the deformation when the deformation is within the elasic limit. If the member is deformed beyond the elastic limit, the member will certainly allow itself to be deformed without offering any more resistance.

Let P be Che load causing the deformation. Let p bd the stress in the member when the full extension S has taken place. Stress X Strain x Volume of the member. Strain Enefgy stored by the member. Equating the strain energy stored by the member to the work done by the loading. Suddenly applied load. Let the load P be suddenly applied. Let p be the maximum stress induced. Equating the strain energy stored by the member to the work done.

In this case the load P is dropped from a height h before it commences to stretch the bar. Let p be the maximum intensity of stress produced in the bar. Extension of the bar. Fig VI. Pkrobleni A steel rod 5 irn. Find the maximum instantaneous stress induc ed when a pull of 10 tonnes is suddenly applied to it.

Find aho the instantaneous elongation. An unknown weight fails hv J cms. If the maximum instantaneous extension is found to be 0 cm. Maximum stress Ex Max. Let P be the load. Equating the loss of petential energy to the strain energy stored by the rod, we have,. Calculate the strain energy stored in the specimen at this stage. If the load at the elastic limit for the.

Load at elastic limit kg. Elongation due to kg. A bar cm. Stress in the smaller part. An object of weight 10 kg. The top of the bar is rigidly fixed to a support.

I - 10 metres cm. The top of the bar is rigidly fixed to a ceiling. Calculate the maximum stress and the strain induced in the bar. Maximum elongation kg. ProMem An' anvils the weight of which may he neglected, is secured to the wire I' 8 m. The wire is to be tested allowing a weight bored to slide over the wire to drop freely from I metre above the anvil Cidadate the weight required to stress the wire to kg.

Take 21 X 70 kg. X CJ. Let P be falling load. Equating the loss of potential energy to the strain energy stored by the wire, we have,. Units on bending stress is explained excellently and can find unfamiliar problems useful for GATE. If exams are the main concern, please download this book. This book discusses absolutely " nothing" about concepts and is a book written by a theoretical fellow who knows nothing about what really goes on in the industry.

If however, you want to understand SOM please don't waste your money in this book. It si absolute thrash. Gunny mama Feb, This book really provides a good explanation. It contains a very good and effective theory material for university exams. Anurag Dutta Jul, Too many question to solve. Not recommended for GATE. Have doubts regarding this product? Post your question.

Safe and Secure Payments. Easy returns. Back to top. It then discusses the different types of stresses and strains acting on an object. The book also focuses on Spherical and Cylindrical Shells and provides an overview of the subject of Reinforced Concret.

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