# STRENGTH OF MATERIALS BY RAMAMRUTHAM EBOOK DOWNLOAD

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Strength of Materials by Ramamrutham - Ebook download as PDF File .pdf), Text File .txt) or read book online. Ramamurtham. Download or Read Online eBook strength of materials rk rajput in PDF Format of materials by ramamrutham pdf OF MATERIALS L free download conduct. Download Strength of Material Fourth Edition By Andrew Pytel, Ferdinand L. Singer Chapter Solution to Problems in Strength of Materials 4th Edition.

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 conventions—shear 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. The middle thirdrule— Core or kernel of a section -Circular section—Hollow — section Structural section— Walls and pillars— Solved Problems to — Problems for exercise. Masonary Dams Forces acting on a dam— Stress distribution on the base of a dam.

Now consider the section F. For instance. Alternatively S. Now consider the independent effect of the kg. Shear Force and Beading Momeot Fig. Treat sagging moments as positive and hogging moments as negative. Note that the moment due to every downward force is negative and the moment due to every upward force is positive. For instance if the S. Diagrams A shear force diagram for a structural member is a diagram which shows the values of shear forces at various sections of the member.

But the B. We shall now cons- truct these diagrams for mem- bers carrying different load systems. Suppose a cantilever AE is2 metres long and is subjec- ted to the forces shown in Fig.

Hence we find that the S. Consider a section X distance of x from the free end S. M diagrams. At any section between B and C distant x from E. In fact the compulation could be made by considering the left hand side of a section also. At any section between A and B distant x from E. In the above example the S. At m. Consider any section X distant x from the end B. Suppose we consider a section between B and C distant x from E. At x—0. Consider any section between D ana B dist nt xr from the free and B-.

The S F. Hence for this range the S. Let the intensity of loading X. Let Mu be the reacting moment or fixing moment at A. Jt is convenient to find the S.

Hence we come to a very important point. Calculations " S. Draw Shear force and Bending moment diagrams for the cantilever shown in Fig J Let Va be the Solution. Let the span of the beam be 1 and let the beam carry a concentrated load W at mid span. HM6iiAf S. Maximum bending moment occurs at mid span. Similarly the B M. It may be observed from the S F. JLet us construct the S. For the ei uilibrium of the beam. Taking moments of the forces on the beam about A. For the equilibrium of the beam.

M dia- grams that the maximum B. Let Va and Vn the vertical reactions at the supports A and B respectively. Since the loading is symmetrical on the span. The S. The B. Equating the S. For the equili- brium of the beam.

Let the S. At any section bet- ween C and B. At any section in 4 C distant x from 4 the B. Consider any sec- tion between A and C distant x from A.

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Consider any section between C and D distant x metres from A S. At any section in CD distant x from A.

W x kg. Let Va and Vu be the vertical reactions at A and B respectively.. Also find the position and magnitude cf the maximum bending moment. From D to B the B. Let the salient 4ookglm. Draw the shear force and bending moment diagrams for the beam shown in Fig. B Taking moments about the left support.

Between D and B. Between A and C. Reactions Since the loading is symmetrical reaction at each support equals half the total load 5 f5f x4 I It. Draw SF and BAf diagram for the beam. Analysis S. AnaHysis B. DfA6RAM fig. At u At Y-O. Taking moments about the end we have. The intensity of loading on a simply supported beam of 5 metres span increases uniformly from kg.

The trapezoidal loading on the beam consists of a uniformly distributed loading and a triangular loading as shown in Fig. Consider any section XX at a distance x from A. Also draw shear force and Bending Moment diagrams. Ka--Total load— Kb Find the position and magnitude of the maximum bending moment. Analysis B. Let the Solution.

Wa WM. Let us construct the Ya-4t ein- ivat J S. Let Va and Vb be the vartical reactions at A and B. Equaling the S.

The point 0 where the B M. S v Some interesting observations For the beam di. M is zero. It may be noted that the S. Taking moments about the end A. Let the beam carry a uniformly distributed load of w per unit run over the whole length. Maximum bending moment. Equating the general expression for shear force to zero.

At any section in CB distant. At A -O. At any section in AB distant x from E the S. For the portion BB. Let Vu and be the reactions at A and B. At any section in AB. Similarly the B. We find from the above discussion that the B.

I Hence the points of contrailexure Oi and O2 are at the distance -y from the middle point C of the beam. We know at any section in A distant x from E the B. For this case there will be two points of contrailexure Oi and Oi between A and B. Hence Me is positive and the B.

When I. The positions of the. M is given by. The beam is subjected to only hogging bending moments. Draw bending moment and shearing force diagrams. Hence for this case. The points of contraflexure Oi and O2 will coincide with C Fig. Let reactions at A and Bhe Va and Vh respectively. Calculate the reactions at the supports A and B of the beam shown in Fig. Taking moments about Ay we have. Determine also the points of contraflexure within the span AB and show their positions on the bending moment diagram- Solution.

Odcabrtioiis B. Catealatfcns S. D and S. Let the overhanging length of each side be a. The beam carries a uniformly distributed had of w per unit length over the entire length. Let this point bo X metres from A.

Point of contraflexure Oi between A and C. Point of contraflexture 2 between O C and B. One of them lies between 'A and C and the other lies between C and B. Let this point be X metres from B. Equating the bending moment to zero. A beam of length L is si nply supported on two intermediate supports.

Each vertical reaction 0-Wwl 0. M and S F diegrams for this position. The supports are so adjusted that the maximum B Si.

F I'roblem Make the values oj hanging beam carrying n-' S principal ordinates and locate the point of contraflexure. M and S. Equating the sagging B. F Calculations. Taking moments about. V -1 Let the B. Let the reactions at A and BbtVa and Vd respectively. A beam AB. Calculations S. Point of controflexurC. Problem 1 Calculations B. A we have. Mark the values of. Bending moment calculations. Taking moments about A.

Shear force calculations S. Also show the location and magnitude of the maximum bending moment. Taking moments about the left support. Let Va and Vh be the vertical reactions A and B. Determine the support reactions for the beam shown in Fig. Deter- nitne the location of the maximum bending moment and mark it clearly on each of the diagrams. Construct the bending moment and shear force diagrams. Calculate the reactions for the beam shown in Fig. Vr -4 tonnes 8!

Calcufations V S. Let this point of contraflexore be at x metres from C. Equating the B. It is obvious that the moment of the individual constituting the forces couple about any point in the plane of the couple is Fp, Hence at any section X in AC B. Draw the B. M, - n diagram for the cantilever shown '. Find the reaction at the fixed end of the cantilever loaded as shown in Fig. Draw also the shear force and the bending moment diagrams Solution.

Hence at A the fixed sup- port will provide a balancing or reacting moment of 29 tm. Hence the reaction at A will consist of an upward re- acting force ofi5 tonnes and an anticlockwise reacting moment of 29 tonne metres. BM Calculations B. Just on the left hand side of —7 tm.

Find the reaction at the fixed end and draw the shear force and bending moment diagrams for shown in Fig. Calculate the reactions at A and B for the hear: Let the reactions at A and B be Ko and K. Taking moments about A we have, WL F.

Beam with a couple at an intermediate point Fig. Let the vertical reactions at A and B be Va and Vb. Taking moments about A we have the following moments. From A to C, the S. Shear force just on the hand side of left — 3 tonnes. Shear force just on the right band side of tonnes. I Nov Solution. Consider the equilibrium of the portion of the beam between sections 1 — 1 and , dx apart, at a distance X from the left support. Ler the bending moments at the sections 1 and — 1 2—2 be and M respectively.

The forces and moments keep- ing the portion of the beam between the sections and 2—2 in equili- Fig. If instead of a uniformly distributed load of w per unit run there bad been a load varying according to some law the last. Ignoring higher powers of small quantities, and simplifying die above relation, we get,. Let the end A be hinged while the end B is placed on rollers. The various forces can be resolved into their vertical and hori- zontal components [Fig.

Taking moments about A, we get. A diagram which shows the variation of the axial load for all sections of the span is called ihe thrust diagram. A simply supprrtvd beam carries inclined loads kg.

If the. Bending Moment and Thrust diagrams. A I E Summer M. Thrust Calculations Thrust at any sjc. Draw S. For the equilibrium of the whol: This part is in equilibrium under the action of the following forces:. From 5 to C the S. The structure shown in Fig. Four loads are applied as shown. Let the reactions at D, E and F be Rd. Re and Rf. A vertical pile is hinged at the base and AB B subjected to the variable load due to earth pressure as shown. The pile is anchored by a tie pin connected at Draw the C shear force and bending moment diagrams for the pile AB.

These forces are shown in Fig. At any section m AC distant x from A, the S. At A, i. At any section i i CB distant x from A the B. To obtain the B. Let at a section distant jc from A the B. The diagram shown is the shear force diagram for a beam which rests on two supports one being at the left hand end. Area of the S. But the area of the S. Y- A-a. Area of the S,F. Hence at B there must be a concentrated load. This member is supported at F as a support and at D on the vertical member DC.

Solution Consider the equilibrium of the member DEF. Let V. Taking moments about F. Now DF can be considered as a simply supported beam. Now consider the member ABC. A beam 5 metres long. Point nj contraflvxure. Draw the S. Considering the equilibrium of this member. Examples in Chapter 4 1. Find the maximum S. Find the maximum shear force and bending moment and draw rhe S F. Sagging B. It carries a point load of it at the middle point of AB and a point load of it at C.

A beam 5 long has one support at the end A and the other support at B 3 metres from A. A beam simply supported and with equal overhangs. Member EFGH: Find the value of a. The beam caniet a uniformaly distributed load of w per unit run over the whole length. Find the normal force. Pure Beading Fig. Theory of simple bending Fig. The part of length being subjected to pure bending has deformed to the shape e M. This condition of the beam between A and B is called pure bending or simple bending.

It is possible offer resistance or stresses certain assumptions. Stresses introduced by bending moment are called Bending Stresses. In this chapter to estimate these stresses with we will discuss about these stresses..

This fibre has been shortened in its length. The material of the member will against these deformations. Let a point load If be applied at each end of the beam. The fibre BD oa the contrary has been elongated and has taken the shape BiDi.

A moment bends a bending member. It is easily seen that between A and li the B. I units apart. Consider the fibre GH distant y from the neutral layer. If now all the fibres between the two transverse sections AB and CD be considered. Let the radius of the neutral surface be R.

Change in length of the fibre GH. At a level between the top and bottom of the beam there will be a layer of fibres which arc neither shortened nor extended. The line of intersection of the neutral surface on a cross-section is called the neutral axis.

Fibres in this ayer arc not stressed at all. This layer is called the neutral layer or neutral surface. For the case explained above all fibres below the neutral layer aretsubjected to tensile stresses while those above the neutral layer arc subjected to a rnpressive stresses. Neutral axis As mentioned earlier. The stress distribution diagram is shown in Fig. Let R be the radius of curvature of the neutral layer at this section.

Hence the stress at any point di? Lc Va and be the end reactions. Moment of resistance Fig. Now consider an elemental area distant y from the neutrid axis. This part is in equilibrium under the action of the following forces: Hence the moment of resis- tance offered by the section is equal to the bending moment.

For the equilibrium of the part AX. This couple is called the moment of resistance. The two equal and opposite resistances C and T will form a couple Taior C at.

Now consider an elemental area Ba at a distance y from the neutral axis. In practice. Further the bendin. Practical application of bending equation The bending equation is based on the theory of y " R pure bending. If the plate section be 12 wide and 2 cms. The stress at any point on the section distant y from the neutral axis is given by The maximum stress occurs at the greatest distance from the neutral axis. A steel plate is bent of radius into a circular arc 10 metres.

Let the width and depth of the centrally situated rectangular hole be b and d Moment of inertia about the neutral axis Fig. Let the overall width and depth be B and D. Section modolos for various shapes of beam sections p Rectangular section.

Hence the maxinnim stress offered by a section is known we if can easily conipute the moment of resistance that can be offered by the sect on. Equating the moment of resistance to the max. A cast iron test beam 2 'em.

Let the distributed load on the cantilever be w kg. Let us first tem. Maximum bending non. What uniformly distributed load will break a cantilever of the same material 5 cm. The maximum flange stress in tlu material is not to exceed 1 kg. Moment of inertia about the neutral axis.

Equating the maximum bending moment to moment of resis- tance we have. The moment of inertia of a beam section 50 cm.

A rolled steel joist of 1 section has the foUovmg dimensions: X— metres. A timber beam is freely supported on supports 6 metres apart.

Taking moments about the left support A. If the stress in timber is not to exceed 80 kg. It carries a uniformly distributed load of kg.

The beam section is 23 cmx46 cm. But the M. A timber beam is 16 cm. Consider 1 metre run of the main. Weight of water for one metre run of the pipe -1 x1x kg. Area of the pipe section Av - i. Steel and water weigh kg. A water main of cms. L tn important observation: The depth h of the beam 2' 5 cm.

Di is very large compared litii d and hence the ratio will approach the value 2 -fy-. Th section may be conveniently split into three rectangular components] 1! The total depth of the section ij 28 cm. Total 2' 28 ' jf: Find the position of tk neutral axis and the moment of inertia of the section about the ncutrai axis and determine the maximum bending moment that should be im- posed on this section if the tensile stress in the top flange is not ta exceed kg.

The top flange is 20 cm. A cast iron beamsection is of I section with a flange 8 cms. Let the max. X kg. Area Centroidal Component a distance y ay I. Let us first determine the neutral axis and the moment of inertia about the neutral axis. X 20 cm. The section may be split up into three components— top flange..

PraUea Find the width and depth of the strongest beam is 50 dm can be eat out of a eytndrical log of wood whose diameter ems. Vi Problem Three beams have the same lengthy same allow- able bending stress and are objected to the same maximum bending moment. November Solation. For the conditions mentioned in the problem the three sections must have the same section modulus.

Let ABCD be the rectangular section cut out of tbe cylindrical log of wood. Let the circular section be of diameter d Let the square section be of side x Let the rectangular section be of width b and depth 2b..

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