1.1 DEFINITIONS

Stress: All bodies offer an equal internal resistance to the externally applied forces. The magnitude of the resisting force per unit area is called stress. Its SI units are N/mm2 or N/rn2.

Tensile (or ultimate) strength: It is the ratio of maximum load to original area of cross-section.

• Yield strength: It is the stress of which a material exhibits a specified deviation from proportionality of stress and strain.

• Compressive strength: It is the maximum compressive stress based upon the original area that a material is capable of withstanding.

Normal Stress: The stress developed on a plane normal to it is called normal stress. It is equal to the force acting on the body per unit normal area. Thus, a1, = -. The normal

Stress may be tensile or compressive depending upon the force to be either of the pull or push type. Tensile and compressive stresses together are called direct stresses.

Shear Stress: It is defined as the ratio of shear force to area parallel to the force.

                       

Conventional (or engineering) Stress: It is defined as the ratio of load P to the original

Strain: It is defined as the change in length per unit length.

Conventional (or engineering) Strain: It is defined as the change in length per unit

Natural (or logarithmic) Strain: It is defined as the change in length per unit

Instantaneous length. Thus,   

True Stress: True Stress it is defined as the ratio of load P to the instantaneous area of cross-

Normal Strain: It is the strain produced under the action of direct stresses.

Shear Strain: It is the strain produced under the action of shear stresses.

Percentage Elongation: It is the change in length per unit original length expressed as percentage, i.e. ji0 x 100, where 1 = final length and 10 = original length.

Percentage Reduction of Area: It is defined as the change in area per unit original area of cross-section A0 expressed as a percentage, i.e. — 0 x ioo where A = final area of cross-section.

• Gauge length: It is the specified length of the test piece on which elongation is measured during the test. The gauge length is generally 5.65 where S0 is the original area of cross-section of the specimen.

• Ductility: It is the ability of a material to deform plastically before fracture. This enables the material to be drawn into whirls.

• Malleability: It is the property of metal and alloys to deform plastically under Compression without rupture. This enables the material to be rolled into thin sheets.

• Yield point: The first stress in a material at which an increase in strain results without an increase in stress. Mild steel has two yield points: Upper and lower Yield points. This term refers to ductile materials only.

• Necking: A form of shape of the localized reduction in cross-section occurring in a ductile under tension before fracture is called necking.

• Permanent set: The plastic deformation which persists after the removal of the applied load is called the permanent set.

Poisson’s Ratio, v It is the ratio of lateral strain to longitudinal strain for a circular—(Mid) bar of diameter d and length 1 subjected to tensile load,

Poisson’s ratio, v ‘= (bl/1)

Hooke’s Law: This law states that within elastic limits, stress and strain are proportional.

Modulus of Elasticity, E: It is defined as the ratio of normal stress to normal strain y P10

within the elastic limits. E = —z.

Modules or Rigidity, G: It is defined as the ratio of shearing stress to shearing strain.

Bulk Modulus, K: It is the ratio of hydrostatic stress to volumetric strain.

                                

Free Body Diagram: It is the diagram of only that member as if made free from the rest, with all the internal and external forces acting on it.

 

1.2 CONCEPT OF STRESS

 

All engineering structures must be adequately strong to withstand the required loads. One of the main problems of structural analysis is to investigate the internal resistance of a body, i.e., the nature of the forces set up within a body to balance the effect of the externally applied forces. For this purpose, a complete diagrammatic sketch of the member to be investigated is prepared showing all the external forces acting on the body, including the reactive forces caused by the supports and the weight of the body itself due to its mass. Such a sketch is called the free body diagram. Since the body at rest is in equilibrium, the forces acting on it satisfy the equations of static equilibrium.

                   

Consider a body subjected to the action of external forces P1 to P5 as shown in Fig. 1.1 (a). If a section 1-1 is passed to divide the body into two parts I and II, then the free body diagrams of both the parts are shown in Fig. 1.1 (b) and (c) respectively. The internal forces F1 to F3 developed at the cut must balance the external forces applied on the body.

In general, the internal forces acting on infinitesimal areas of a cut are of varying magnitudes and directions. They vary from point to point and are inclined with respect to the plane of the section. It is advantageous to resolve these forces perpendicular and parallel to the section considered. Now again consider part I of the body with the force AP acting on an infinitesimal area A as shown in Fig. 1.2 (a). The components of this force are shown in Fig. 1.2 (b) along the x-axis perpendicular to the cutting plane and the other two perpendicular aces y and z, lying in the plane of the section. By definition, stress along the x-axis is:

and is called the normal stress in the x-direction at a point. It is customary to refer to normal stresses that cause tension on the surface of a section as tensile stresses. On the other hand, those that are pushing against it are called compressive stresses. The stresses lying in the plane of the section are called shear stresses. By definition,

                                            

Stress is a tensor quantity. The first subscript indicates the plane perpendicular to the axis and the second subscript designates the direction of the stress component. Thus we find that stress is the internal resistance offered by the body and is the force per unit area. The units of stress are N/rn2 or MPa.

 

Example 1.1 A circular bar of 15 mm diameter is subjected to an axial tensile load of 3 kN Calculate the tensile stress developed in the bar

 

                  

 

Example 1.2 A strut of rectangular cross-section 80 mm x 100 mm is subjected to a compressive load of 200 kN Calculate the compressive stress developed an the strut

 Solution, P = 200 kN

                               A =80x1008000rnm

                                                 

 

Example 1.3 A concrete pier of 1 m diameter and 2 m height is loaded at the top with a uniformly distributed ‘load of 20 kN/m2. Calculate the stress at the top and the bottom of the pier. Concrete weighs 25 kN/m3.

 

Example 1.4 An axially loaded connecting link having a T cross- section (Fig. 1.3) is subjected to a uniform tensile stress of 150 MPa. Calculate the magnitude of the applied force.

 

Example 1.5 The head of a cylinder is held by 8 studs of 6 mm diameter each. The diameter of the cylinder is 200 mm. If the in external pressure in the cylinder is 1.5 MP. Calculate the tensile stress in each stud.

 

Example 16 Calculate the force required to punch a hole of 10mm diameter through a mild steel plate 4 mm thick. The maximum shear strength of mild steel is 250 MP Also find the compressive stress in the punch.

 

Example 1.7 The connecting rod of an engine is of I-cross- section as shown in Fig. 1.4. During the return stroke, the compressive force in the rod was 50 kN. Calculate the Compressive stress developed in the rod.

 

 1.3 BAR OF VARYING CROSS-SECTION 

Consider a bar of varying cross-section shown in Fig. 1.5 (a). The free body diagrams for the various parts are shown in Fig. 1.5 (h). The stresses in the various parts are:

 

Example 1.8 A steel bar 20 mm diameter is loaded as shown in Fig. 1.6 (a). Determine the stresses in each part.

 Solution, The free body diagrams of the three positions are shown in Fig 1.6 (h).

 

Example 1.9 A steel bar of 25 mm diameter is loaded as shown in Fig. 1.7 (a). Calculate the stress in each portion and the total elongation. Take E = 200 GPa.Solution. The free body diagram of various portions is shown in Fig. 1.7 (h).

 

Example 1.10 A stepped bar is loaded as shown in Fig. 1.8 (a). Calculate the stress in each part and total elongation. E = 200 GPa.

 Solution. The free body diagram is shown in Fig. 1.8 (h).

 

Example 1.11 A pillar is shown in Fig. 1.9. The total contraction of the pillar is 0.25 mm. If E5 = 200 GPA and Ed = 120 GPA, find the value of load P.

 

 

1.4 CONCEPT OF STRAIN

 If L0 is the initial gauge length and L is the observed length under a given load, the gauge elongation, 1L = L - L0. The elongation (or contraction) per unit of the initial gauge length is given as:

                          

This expression defines the tension (or compression) strain. Since this is associated with the normal stress, it is usually called the normal strain. It is a dimensionless quantity (mm/mm or jIm/m or rn/rn). This is also called the nominal or engineering strain.

In some engineering applications where strains may be large, the total strain is defined as the sum of the incremental strains.

                                     

Where L is the current gauge length of the specimen when the increment of elongation (or contraction) iSL occurs. If L0 is the initial length and Lf the final length, then

                                              

This strain, obtained by adding up the increments of strains, which are based on the current dimension of a specimen is called a natural or true or Logarithmic strain.

 

1.5 TRUE STRESS

The true stress is related to the instantaneous cross-sectional area A. For the applied force P,

 

1.6 LONGITUDINAL AND LATERAL STRAINS

Consider a circular test specimen of diameter d and length L If a tensile load P is applied to the specimen its length increases. The increase in length per unit length is called the longitudinal strain. Since the volume of the specimen remains constant, therefore, the increase in length is accompanied by a decrease in diameter. This decrease in diameter per unit diameter is called the lateral strain. The longitudinal and lateral strains are of opposite nature.

 

1.7 POISSON’S RATIO

The ratio of the lateral strain to the longitudinal strain is called the Poisson’s ratio. It is denoted by v (Greek letter Mue). Thus

Poisson’s ratio for most of the materials varies from 0.25 to 0.40.

 1.8 VOLUMETRIC STRAIN

It is the ratio of the change in volume V of the body to its original volume V0, when subjected to hydrostatic stress.

 1.9 HOOKE’S LAW AND ELASTIC MODULIAccording to this Law, stress is directly proportional to strain, within the elastic (strictly speaking proportional) limits.

 

 

 

 

 

 

 

 

 

  1.10 EXTENSION OF A TAPERED BAR

Consider a bar of length L tapering from d2 to d1 and subjected to axial tensile load P. as shown in Fig. 1.10, At a distance x from diameter d2, the diameter of the bar is

 1.11 EXTENSION OF A CONICAL BAR

Consider a conical bar of base radius r, height h and of mass density p. as shown in Fig. 1.11, hanging under its own weight. Consider an elementary strip at a distance x of width dx from the base.

 

  

 

1.12 EXTENSION OF UNIFORM BAR UNDER ITS OWN WEIGHT

Consider a bar of uniform area of cross-section A and length L as shown in Fig. 1.12. Consider a strip of the bar of thickness dx at a distance x from the bottom. The downward force acting on this strip is due to the weight of the bar that lies below this strip and is equal to Ax pg. where p is the density of its material.

                                   

Therefore, the total elongation of the bar of uniform cross section produced by the self weight of the bar is equal to that produced by a load of half of its weight applied at the lower end.

 1.13 STRESS-STRAIN DIAGRAM FOR MILD STEEL

The stress-strain diagram for a ductile material like mild

 

Steel is shown in Fig. 1.13. The curve starts from the origin, showing thereby that there is no initial stress of strain in the specimen. Up to point A, Hooke s law is obeyed and stress is proportional to strain.

Therefore, OA is a straight line. Point A is called the limit of proportionality. Upto point B, the material remains elastic, i.e. on removal of the load, no permanent set is formed. AB is not a straight line. Point B is called the elastic limit point. Beyond point B, the material goes to the plastic stage until the upper yield point C is reached. At this point the cross-sectional area of the material starts decreasing and the stress decreases to a lower value to point D, called the lower yield point. Between DE, the specimen elongates by a considerable amount without any increase in stress. From point E onwards, the strain hardening phenomena becomes predominant and the strength of the material increases thereby requiring more stress for deformation, until point F is reached. Point F is called the ultimate point and the corresponding stress is called the ultimate strength. At point F, necking of the material begins and the cross-sectional area decreases at a rapid rate. The apparent stress deceases but the actual or true stress goes on increasing until the specimen breaks at point C, called the point of fracture. The fracture of ductile material is of the cup and cone type. The phenomenon of yielding and necking is not exhibited by brittle materials. The ultimate strength is calculated at 0.2 per cent of maximum strain.

 

1.14 FACTOR OF SAFETY

It is the ratio of the maximum permissible stress to which a member can be subjected to the allowable or working stress. For ductile materials,

 

1.15 COMPOSITE SYSTEM OF EQUAL LENGTHS SUBJECTED TO LOAD

Consider a composite system shown in Fig. 1.14 consisting of different materials and

area of cross-section, subjected to load P.

 

1.16 COMPOSITE SYSTEM OF UNEQUAL LENGTHS SUBJECTED TO LOAD

Consider a composite system of unequal lengths shown in Fig. 1.15 subjected to lord

P. Before both the members start sharing the load,

 

Example 1.12 A steel rod 20 mm diameter is passed through a brass tube 25 mm internal diameter and 30 mm external diameter. The tube is 1 m long and is closed by thin rigid washers and fastened by nuts, screwed to the rod, as shown in Fig. 1.16. The nuts are tightened unitil the compressive force in the tube is 5 kN. Calculate the stresses In the rod and the tube. E5 = 200 GPa, Eb =80 GPa.

 

Example 1.13 A rigid beam BD is suspended, from two rods AB and CD as shown In Fig. 1.17. Rod AB is of steel of 20 mm diameter and rod B is of copper of 25 mm diameter. At what distance from rod AB the load P be applied if the beam is to remain burizontal ? Calculate the stresses in the rods if P = 25 kN.

Take E5 =200 GPa and E =100 GPa

 

Solution. Let P5, P = forces in rods AB and CD respectively. Taking moments about rod AB, we have

 

Example 1.14 Determine the stresses in the rods loded as shown in Fig. 1.18 and having the same area of cross-section of 5 cm2.

 

Example 1.15 A rigid bar is suspended from three rods and loaded -as shownin Fig. 1.19. Determine the stresses in the rods.

The given data is:

 

Example 1.16 A rigid beam is placed on there columns of identical cross-sectional areas of 200 cm2 each as shown in Fig. 1.20. Calculate the stresses in the columns if there was a gap of 2 mm between the beam and the middle column before the load was applied. Take E = 18 GPa.

 

Solution. Initial compression of columns 1 and 3 are:

 

Example 1.17 A solid steel bar 0.5 m long and 50 mm diameter is placed inside an aluminium tube having 60 mm inside and 80 mm outside diameter. The aluminium tube is 0.2 mm longer than the steel bar. An axial load of 500 kN is applied to the bar and tube through rigid cover plates as shown in Fig. 1.21. Find the stresses developed in steel bar and aluminium tube. E5 = 220 GPa, Ea = 70 GPa.

 

 

Example 1.18 A horizontal beam AB is supported on two cables CE and DF as shown in Fig. 1.22 (a). Calculate the values of P when the strain in DF is 3 x Assume that beam AB does not bend. Aa = A5 = 600 mm2, Ea = 70 GPa, E5 = 210 GPa.

 

Solution. From Fig. 1.22 (h), we have

 

Example 1.19 A compound bar loaded as shown in Fig. 1.23 has a gap of 1.0 mm. Calculate the stresses in the two bars E = 200 GPa, Ec1 = 105 GPa.

 

Solution. The steel bar is subjected to a tensile load of 100 kN. Its free extension is:

 

Example 1.20 A rectangular steel bar 25 mm x 12 mm and 0.5 m long is subjected b an axial tensile load of 10 kN. Calculate the change in dimensions if F = 200 CPa ai.d Poisson’s ratio 0.30.

-ye sign indicates decrease in dimension.

 

Example 1.21. In a tensile test on mild steel, the gauge length was 50 mm and diameter 10 mm. The load at the proportional limit was 30 kN and at the upper yield point 32 kN. The extension on 50 mm gange length was 0.028 mm under a load of 8.8 kN. Calculate the (a) proportional limit, (b) upper yield point stress and (c) modulus of elasticity.

 

Example 1.22 A steel column 4 m high, 250 mm external diameter and 200 mm inside

diameter is subjected to an axial compressive load of 40 kN at the top. The density of

steel is 7470 kg/rn3. Calculate the stress at the top and bottom of the column.

 

Example 1.23 A short reinforced concrete column has 600 cm2 area of cross-section.

The columia reinforced with 4 steel rods arranged symmetrically, each having 10

2 cross-sectional area. The column is subjected to an axial load of 800 kN. = 200

CPa and E 20 GPa. Calculate the load shared by concrete and steel.

 

 

Example 1.24.A steel bolt 15 mm diameter and 0.5 m long passes through a copper tube of external diameter 20 mm and internal diameter 16 mm. The pitch of the bolt

threads is 4 mm. The bolt is tightened by th of turn of the nut. A tensile load of 50 kN

is applied at the ends. Calculate the resulting stress in the bolt and the tube. Eç = 200

CPa and E = 105 CPa.

 

Example 1.25 A load of 2 kN is suspended from two rods as shown in Fig. 1.24. Rod AB is of steel of 15 STEELTh mm diameter and rod CD of copper having 10 mm diameter. Calculate the distance x so that the bar BD

remains horizontal. Also calculate the stresses produced

in the rods. E = 200 GPa, = 110 GPa.

 

Solution. For the bar BD to remain horizontal,

 

 

Example 1.26 A rigid bar ABC is hinged at A and suspended at two points B and C by two bars BD and CE made of aluminium and steel respectively as shown in Fig. 1.25. The

aluminium bar is of 6 mm diameter and steel bar of 4 mm diameter. Calculate the stress developed in each bar. Eai =70 GPa, E9 =200 GPa.

 

Solution. Let al and P5 be the loads shared by BD and CE respectively.

 

 

Example 1.27 A bar of 20 mm diameter is subjected to a pull of 30 kN. The measured extensions over a gauge length of 200 mm is 0.1 mm and the change in diameter is

0.0035 mm. Calculate the Poisson’s ratio, modulus of elasticity and bulk modulus.

 

1.17 TEMPERATURE STRESSES AND STRAINS

When the temperature of a material is changed, its dimensions change. A stress is setup in the material if this change in dimension due to temperature change is prevented. This is called the temperature stress.

When the temperature increases, length increases. Since this increase in length is prevented, compressive stress is developed in the material. The reverse phenomena occurs when the temperature is decreased and tensile stress is developed.

 

1.18 COMPOSITE SYSTEM OF EQUAL LENGTHS SUBJECTED TO VARIATION OF TEMPERATURE

For the composite system shown in Fig 1 26 let o. > cx2 then on temperature rise by

At, member 2 will be under tension and member 1 under compression If P is the common

force developed then,

                                 

 

1. 18. 1 More than Two Members

Consider three members shown in Fig. 1.27 subjected to temperature rise.

 

Example 1.28 A steel tube 0.75 m long, 25 mm external diametr and 20 mm internal

diameter encloses a copper rod of the same length and 15 mm indiamter. The tube is

firmly joined to the rod at both the ends and its temperature is raised by 120° C. Calculate

the stresses in tube and rod. Also calculate the increase in length of the composite system and the external force required to prevent this increas in length. Es = 210 GPa, Ec = 100 GPa, a = 10 x 10-6 per °C, cz = 15 x 10_6 per °C.

 

 

Example 1.29 A steel rod 1 in long and 15 mm diameter is held between rigid supports as shown in Fig. 1.28. The temperature of the rod is increased by 40°C. The coçfficient of linear expansion for steel is 12 x 1O/°C. Calculate’ the stress developed in the rod, E9200GPa.

 

 

Example 1.30 Three identical vertical wires of 1 m length each and 4 mm diamter are suspended from a horizontal support as shown in Fig. 1.29. A load of 5 kN is applied t their lower ends by means of a rigid cross bar. E = 210

GPa, E 105 Gla., If the cross bar remains horizontal, find

 

 

Example 1.31 A compound bar is shown in Fig. 1.30. Its temperature is raised by

1200 F. Calculate the stresses in each metal and the change in’length. 210 GPa,E,,

1O5GPa,a5z6.5x1O4per0F,9.5x104per0F.

 

Example 1.32 A composite rod of steel and copper of lenghts 1. m and 1 m respectively is clamped at the ends, as shown in Fig. 1.31. Its temperature Is then raised by 500 C . Calculate the stresses in the two metals of rod. E5 210 GPa, E = 105 GPa, cx 10 10 per °C and a = 15 x 1O per °C.

 

Example 1.33 Determine the stresses in the bar shown in Fig. 1.32 if its temperature is increased by 30°C. F5 = 2E = 200 GPa, cz = 12.5 x 10per °C, a = 16.5 x 10-6 per °C.

 

Example 1.34 A load of 45 kN is transmitted through a slab toa composite solid steel cylinder (% —15cm2) and hollow copper cyllndè (A 20 cm2). Afithe application of the load the temperature was increased by 30°C. Determine the distribution of load before and after the change in temperature. E5 2E 200 GPa, a5 12.5 x 1o per °C, ra1i16.5*1O4per°C.