Simple Stresses and Strains

Simple Stresses and Strains CHPTER OJECTIVES In this chapter, we will learn about: Various types of (a) stresses as tensile and compressive stresses, positive and negative shear stresses, complementary shear stresses, volumetric stress; (b) strains as longitudinal strain, lateral strain, shear strain; and (c) strain energy absorbed by a body during deformation. Difference between gradual, sudden and impact loads. Elastic constants of a material as Young s modulus, ulk modulus, Shear modulus, Poisson s ratio as deformation in a body depend on these elastic constants. Stresses developed in a bar due to its own weight. ar of uniform strength in which stress due to self-weight remains constant. Variation of stress in a stepped bar, uniformly tapered bar and uniformly tapered flat. Statically indeterminate problems in which an additional equation of deformation is made for solution of unknowns. Difference between resilience and toughness, i.e., strain energy absorbed by a body within the elastic limit and strain energy absorbed by a body until fracture. Mild steel being the most commonly used engineering material, tensile test on mild steel to study mechanical properties as ultimate strength, yield strength, ductility, resilience and toughness. Introduction In strength of materials, we study the effect of forces and moments on the deformation of a body. Figure.(a) shows a bar subjected to an axial compressive force P which is perpendicular to the section of the bar. This is known as direct force producing change in length of the bar. shear force acting parallel to the top plane of the block is shown in Fig..(b); bottom plane of block is fixed. This type of force produces a change in the shape of the body, that is, a rectangular plane (abcd) of the body is converted into a parallelogram (a b cd). Shear force produces shear strain f in block. ending moment

Chapter Length P Section a P a b b φ M C g M (a) xial compressive load d P c (b) Shear force (c) ending moment T f T q (d) Twisting moment Figure. M causes angular rotation g in the bar as shown in Fig..(c). Then, in Fig..(d), a twisting moment acts on the bar, causing angular twist in the bar as defined by angle q. Stresses and strains in bars due to direct and shear forces, angular rotation and bending stresses in beams due to bending moment and angular twist and shear stresses in shaft due to twisting moment are studied in detail in various chapters of the book. Tensile and Compressive Stresses Figure.(a) shows a bar of length L and diameter d subjected to an axial tensile force P. Note that a force is applied along the axis of the bar and in a direction perpendicular to all cross-section. The effect of this tensile force (acting in a direction away from the plane) on bar is to extend its length and reduce its diameter as shown in Fig..(b). Note that the bar is of uniform section and on each cross-section the same force P acts producing the same stress on all the sections. P xial stress in the bar, rea of cross-section d d P d (a) d P L L L > L d < d (b) Figure. xial tensile force

Simple Stresses and Strains xial strain in the bar, P 4P π d πd 4 L δ ε L L L L Change in length Original length Strain s d d δd Change in diameter Lateral strain in the bar, εl d d Original diameter The axial load which produces extension in the length of the bar is termed as tensile load and the stress due to tensile load is termed as e, Strain tensile stress. Hooke s law s the load on the bar is gradually increased, the stress s and the strain e also gradually increase in the bar. The relationship between s Figure. and e is linear, that is, s a e (stress is proportional to strain). This relationship between the stress and the strain, that is, s a e, is known as Hooke s law (given by Professor Hooke) as shown in Fig... or s E e where E is the constant of proportionality. E ε xial tensile stress xial strain E is also called the Young s modulus of elasticity (given by Professor Young) of the material. Similarly, let us consider a bar subjected to an axial compressive force P, acting along the axis of the bar but perpendicular to the section of the bar. Due to this compressive force, the axial length of the bar is reduced from L to L while the diameter is increased from d to d (Fig..4). xial stress in the bar, P π 4P d πd 4 L δ xial strain in the bar, ε L L L L δl ε L, where dl is change in length Lateral strain in the bar, δ ε d d d l d + d Change in diameter Original diameter. Note that if the tensile stress is taken as a positive direct stress, then the compressive stress will be a negative direct stress. Moreover, we have learnt that if the axial strain is positive in the bar, then the lateral strain will be negative or vice versa. L (a) L L L < L d > d d Compressive force (b) Figure.4 d P

4 Chapter Example. steel bar of a diameter of 0 mm and a length d 0 mm of 400 mm is subjected to a tensile force of 40 kn. Determine (a) the tensile stress and (b) the axial strain developed in the bar if the Young s modulus of steel E 00 kn/mm (Fig..5). 400 mm Diameter of the bar, d 0 mm Cross-sectional area ( p/ 4) d π 0 00π mm 4 Figure.5 Example. xial load, P +40 kn 40,000 N Tensile stress in the bar, 40, 000 00π 7. N/mm 7. MPa ( MPa 0 6 N/0 6 mm N/mm ) Young s modulus of steel E,00,000 N/mm xial strain in the bar, ε /E (as per Hooke s law) 7. 0. 66 0, 00, 000 There are no units of strain, as strain is change in length per unit length. Exercise. 00-mm-long copper bar is subjected to a compressive force such that the stress developed in the bar is 50 MPa. If the diameter of the bar is 5 mm, what is the axial compressive force? If E for copper is 05 kn/mm, what is the axial strain in the bar? Shear Stress and Shear Strain rectangular block is fixed at the bottom plane and a force F is applied on the top plane as shown in Fig..6(a). Equal and opposite reaction F develops on the bottom plane. pplied force F and reaction F constitute a couple, tending to rotate the body in a clockwise direction. This type of shear force is a positive shear force and the shear force per unit area of the plane on which it acts is a positive shear stress. Figure.6(b) shows another block on which the shear force F applied on the top surface and the reaction F from the bottom surface produce anticlockwise or counter clockwise moment (the arm of the couple is equal to the height of the block). This type of shear force that tries to rotate the body in the anticlockwise (ccw) direction is termed as a negative shear force producing a negative shear stress in the block. The shear angle f shown in the figure is very small for any metal within the elastic limit. Shear strain tan φ D However, f is very small, much less than ; therefore, tan φ sin φ φ Shear angle f can be termed as shear strain because the angle f is very small within the elastic limit of the material.

Simple Stresses and Strains 5 y f y D D F C + f D F C C + F D C (a) x F (b) x Shear stress is proportional to shear strain, that is, t a f t Gf, where G is the constant of proportionality. τ Shear stress The shear modulus or the modulus of rigidity, G φ Shear strain Example. rectangular block of aluminium of size 60 mm 50 mm 40 mm is subjected to a shear force of 45 kn on the top surface as shown in Fig..7. What is the shear stress developed in the block? If G for material is 5 kn/mm, what is the shear strain? The bottom surface of the 40 block is fixed to the ground. F 45 kn 45,000 N rea of the top surface, 60 50,000 mm 45, 000 Shear stress, τ 5 N/mm, 000 ( negative shear stress as it tends to rotate the body in anticlockwise direction.) Shear modulus, G 5,000 N/mm Shear strain, Figure.6 (a) Positive shear force and (b) negative shear force τ φ 5 G 5, 000 0. 6 0 50 mm 60 mm 45 kn F Exercise. 50-mm cubical brass block is subjected to a shear force of 50 kn on the top surface as shown in Fig.8. What is the shear stress developed in the block? If G for brass 8 kn/mm, what is the shear strain developed in the block? F Figure.7 Example. 50 50 50 Figure.8 Example. F

6 Chapter Complementary Shear Stresses Whenever any external shear force applied on the body produces a shear stress +t, a shear stress of nature -t is developed in the body on perpendicular planes to maintain equilibrium. Consider a rectangular block of size l b h as shown in Fig..9. The bottom face of the block is fixed to the ground. On the top face, a shear stress t is applied. Stress force on the top face τ b l. Reaction at the bottom face τ b l. oth the forces constitute a couple of arm h as shown in Fig..9. Moment of the couple, C t l b h (cw) t b h t b h H G t b E t t F b t h C t b h h t D C t b t, E h, C b Figure.9 Rectangular block subjected to shear stress To balance this couple, an anticlockwise couple of the same magnitude must act on the body. Say the shear stress developed on vertical faces is τ as shown in the figure. Shear forces on the vertical faces τ h b rm of couple l Moment of the couple, C τ h b l However, couple C C to maintain equilibrium or τ lbh τlbh Negative shear stress τ positive shear stress t τ is the complementary shear stress to applied shear stress (numerically) t, acting at an angle of 90. Example. 40-mm cube with its lower face fixed to the ground. positive shear force of 6.4 kn acts on the top surface of cube. Draw the shear stress distribution on the vertical face of the cube (fig.0). Shear force, F 6.4 kn 6,400 N rea of the top surface 40 40,600 mm D 40 mm 40 mm 40 G 6.4 kn 6.4 kn Figure.0 Example. C F E

Simple Stresses and Strains 7 Shear stress, τ +6, 400/, 600 + 4 N/mm (as shown) Shear stress on the vertical face CEF t - 4 N/mm Shear stress distribution on vertical surface is t - 4 MPa (uniform) Shear stress tends to rotate the body in anticlockwise direction. Figure. shows the shear stress distribution on face CD. Students should learn how to draw complementary shear stress. Many a times in interviews, they fail to explain the concept of complementary shear stresses. 4 4 N/mm Figure. 4 4 MPa Shear stress distribution Exercise. rectangular block of dimensions 60 mm 40 mm 0 mm is fixed to the ground on face of size 40 mm 0 mm. On the top face, a negative shear stress of -0 N/mm is applied. Draw the complementary shear stress distribution on a vertical surface of the block (Figure.). Stresses on an Inclined Plane Consider a bar of rectangular section bh as shown in Fig... plane abcd is inclined at an angle a with the axis of the bar. xial load P applied on bar is acting at an angle a on the inclined plane. There are two components of this force P: Normal component, P n P sina (a tensile force; the arrow of the force is pointing away from the plane abcd) Tangential component, P t P cosa (a positive shear force, tending to rotate the part I of the bar in the clockwise direction. Take a point and consider the effect of P t ) h Side ab sin α t +0 MPa 0 40 mm 60 mm t +0 MPa 0 Figure. Exercise. Side ad b P n d P I b h a a P t b a c II P Figure. Forces on inclined plane

8 Chapter Cross-sectional area of the inclined plane, ad ad Normal stress on the plane abcd, hb sin α Pn P sin α (a tensile stress) (.) bh Shear stress on the plane abcd, Pt P cosα sin α P sin α τ + bh bh (.) (a positive shear stress) Example.4 Consider Fig... Take b 50 mm, h 40 mm, P 40 kn and a 0. Determine the normal and shear stresses on the inclined plane. ngle, a 0, cosa 0.866, sina 0.5 Force, P 40 kn 40,000 N Normal component, P n Psina 40,000 0.5 0,000 N Tangential component, P t Pcosa 40,000 0.866 4,640 N Side h 40 ab 80 mm sin α 0. 5 Side ad 50 mm rea of inclined section 80 50 4,000 mm 0, 000 Normal stress on the inclined plane, n + 5 N/mm 4, 000 4, 640 Shear stress on the inclined plane, τ s + 8. 66 N/mm 4, 000 Exercise.4 circular bar of a diameter of 40 mm is subjected to an axial force of 50 kn. plane is inclined at an angle of 45 to the bar. What are the normal and shear stresses on the plane, if the axial force is compressive? [Hint: Inclined plane is an ellipse with major axis d/sin a and minor axis d. rea of ellipse p /4 major axis major axis.] ars of Varying Cross-Sections Till now we have considered bars of uniform section throughout. Let us consider a stepped bar of diameters d, d and d with axial lengths L, L and L, respectively, as shown in Fig..4. The bar is subjected to an axial tensile force P as shown. Each section of the bar is subjected to the same force P, but the cross-sectional area of portions I, II and III are different. Minimum axial stress occurs in portion I and maximum axial stress occurs in portion III. xial stresses in portions I, II and III, respectively, are

Simple Stresses and Strains 9 P I d II III d d P L L L Figure.4 ar of varying cross-sections 4 P I πd 4 P II πd 4 P III πd If E is the Young s modulus of the material of the bar, then the strain in each portion can be obtained by using Hook s law. xial strains in portions respectively are Total change in length of the bar I 4P εi E πed ε ε II III II 4P E πed III 4P E πed 4PL 4PL 4PL δl δli + δlii + δliii + + πed πed πed 4P L L L + + π E d d d Example.5 Consider a bar in three steps, with diameters d 0 mm, d 5 mm and d 0 mm and axial lengths L 50 mm, L 75 mm and L 00 mm as shown in Fig.4. The bar is subjected to an axial tensile force of 8 kn. Determine the stress in three portions of the bar and the total change in length of the bar of E 67,000 N/mm.

0 Chapter Stresses are 4P 4 I 8, 000 5. 46 N/mm πd π 0 II III 4P 4 8, 000 45. 7 N/mm πd π 5 4P 4 8, 000 0. 86 N/mm πd π 0 E 67,000 N/mm 4P L L L Total change in length, δl + + π E d d d 4 8, 000 50 75 00 + + π 67, 000 0 5 0 0.5 (0.5 + 0. + ) 0.5.458 0.6 mm Exercise.5 circular stepped bar of diameters of 6, and 8 mm and axial lengths of 40, 60 and 70 mm, respectively, is subjected to an axial compressive force of 4.8 kn. The bar is made of steel. If E for steel 08 kn/mm, determine the axial strains in three portions of the bar and the total change in the length of the bar. Longitudinal Strain, Lateral Strain and Poisson s Ratio There is an axial (or longitudinal) strain in the bar due to an external axial load, there is a lateral strain of opposite nature in the bar. This is due to Poisson s effect. Many students may feel that if the length is increased, the diameter is decreased as the volume may remain constant. However, in reality, there is a change in volume within the elastic limit of the material. Figure.5 shows a bar of length L and diameter d subjected to an axial tensile force P. Length is increased to L and diameter is decreased to d. Note that the diameter of the bar is lateral (perpendicular) to its length. Therefore, the strain or change in diameter per unit diameter is termed as lateral strain. L P d d P L Figure.5

Simple Stresses and Strains L L δl xial (longitudinal) strain in the bar, εa L L Since L > L, it is a positive strain. d d δd Lateral strain bar, εl d d Since d < d, e l is a negative strain. Ratio of lateral strain to longitudinal strain is εl δd L a negative ratio εa d δl This is known as Poisson s ratio, given by Professor Poisson, a French scientist. e l -ve a, where v is Poisson s ratio. Though the ratio is negative, the negative sign is not generally attached with this ratio. P xial stress in the bar, a 4 πd xial stress in the bar, ε as per Hooke s law, where E is the Young s modulus of the material. a Lateral strain εl vεa v, E where v is Poisson s ratio. For common engineering materials, the values of elastic constants (E and v) are given in Table.. Figure.6 shows variation of axial stress s a and axial strain e a for a bar subjected to axial load. O to is a straight line and only up to, material obeys Hooke s law. In strength of materials, all loadings are confined up to point, such that the stress developed in the body is less than the elastic limit stress, s e. If the bar is unloaded within the elastic limit, then the strain developed in the body is fully recovered. However, if the load is removed in the plastic region, then unloading curve is along C, line C is parallel to O and residual strain equal to OC remains in the bar after unloading. Table. Elastic constants a a E xial stress s a Elastic region unloading Plastic region xial strain, e a Material Steel luminium rass Cast iron Glass Copper Young s modulus 00 0 67 70 00 05 00 05 00 0 0 MPa, E Poisson s ratio, v 0. 0. 0.5 0.5 0.0 0.4 O C s e Figure.6 xial stress versus axial strain graph

Chapter 0 P P 0 500 mm Figure.7 Example.6 copper bar of a rectangular section of 0 mm 0 mm and a length of 500 mm is subjected to an axial compressive load of 60 kn. If E 0 kn/mm and v for copper 0.4, determine the changes in the length and the sides of the bar (Fig..7). ar dimensions l 500 mm b 0 mm h 0 mm Cross-sectional area 0 0 600 mm 60, 000 Compressive stress, a 00 N/mm 600 E 0 kn/mm 0,000 MPa a 00 xial strain, εa 0. 98 0 ε 0, 000 Lateral strain, e l -ve a 0.4 0.98 0 - +0. 0 - Changes in dimensions, dl e a l -0.98 0-500 -0.49 mm Change in length, dh e l h +0. 0-0 +0 0 - mm Changes in sides, db e a b 0. 0 0 6.66 0 - mm. Exercise.6 steel bar of square section 0 0 mm and a length of 600 mm is subjected to an axial tensile force of 5 kn. Determine the changes in dimensions of the bar. If E 00 kn/mm, v 0.. Tapered ar Consider a circular bar uniformly tapered from diameter d at one end and gradually increasing to diameter d at the other end over an axial length L as shown in Fig..8. Take an elementary disc of diameter d x and length dx at a distance of x from end. Diameter d x of elementary disc d d d d + x d + kx L

Simple Stresses and Strains d P d x d P x dx L Figure.8 Circular tapered bar where constant d d k L Cross-sectional area, π π x ( dx) ( d + kx) 4 4 xial stress, xial strain, Change in length over dx, Total change in length, x P 4P π( d + kx) x x 4P ε x E πe( d + kx) δd 4P dx x πe( d + kx) δ L L O P πe 4P dx πe( d + kx) 4 d as d + kl d + ( d + kx) k L 4P πek d + kl d O 4P πek d d d L L d 4P P d d Therefore, δ L πek d d 4 ( ) πekdd d d Putting the value of k, we get L

4 Chapter δl PL Total change in length δl 4 πedd 4 P( d d ) L πe ( d d ) ( d d ) Example.7 brass bar uniformly tapered from diameter 0 mm at one end to diameter 0 mm at the other end over an axial length 00 mm is subjected to an axial compressive load of 7.5 kn. If E 00 kn/mm for brass, determine (Fig..9): (a) the maximum and minimum axial stresses in bar and (b) the total change in length of the bar. xial load, P 7.5 kn. Maximum stress occurs at end with minimum cross-sectional area: (a) (b) max min 4P 4., 7 5 000 95. 94N/mm at end π( 0) π 00 4P 4., 7 5 000. 87N/mm at end π( 0) π 0 PL δl 4 as the load is compressive πedd, 4 7. 5, 000 00 0. 4 mm π 00, 000 0 0 0 mm f P P 0 mm f 00 mm Figure.9 Exercise.7 400-mm-long aluminium bar uniformly tapers from a diameter of 5 mm to a diameter of 5 mm. It is subjected to an axial tensile load such that stress at middle section is 60 MPa. (a) What is the load applied? (b) What is the total change in the length of the bar if E 67,000 N/mm? [Hint: t middle, the diameter is ( 5 + 5) / 0mm]

Simple Stresses and Strains 5 Tapered Flat c t Consider a flat of same thickness t throughout its length, tapering uniformly from a breadth at one end to a breadth b at the other end over an axial length L. The flat is subjected to an axial force P as shown in Fig..0. Thickness of elementary strip abc is t at a distance of x from end. b readth, bx b+ x ( b + kx), L b where k L Cross-sectional area of elementary strip, x b x t P b b x x dx b a L Tapered flat Figure.0 P (b + kx)t P P Stress at elementary strip, x x ( b + kx) t Change in length over d x, Total change in the length of flat, Putting the value of k, Pdx δd x Et( b + kx) L Pdx δ L 0 Et( b + kx) P δl ln( ) ln( b) Etk [ ] P ln Etk b PL δl Et b b ( ) ln Example.8 steel flat of an axial length of 600 mm and a uniform thickness of 0 mm has a uniformly tapered width of 40 mm at one end to 0 mm at the other end. If the axial force is P 8 kn and E 08 kn/mm, then what is the change in axial length of flat? Load, readth, P 8,000 N 40 mm b 0 mm Length, L 600 mm Thickness, t 0 mm Young s modulus, E 08,000 N/mm

6 Chapter Change in length, PL δl Et b b ( ) ln 8, 000 600 40 08, 000 0 ( 40 0) ln 0 0.596 ln.0 0.596 0.69 +0.8 mm Exercise.8 n aluminium flat of a thickness of 8 mm and an axial length of 500 mm has a tapered width of 5 mm tapering to 5 mm over the total length. It is subjected to an axial compressive force P, so that the total change in the length of flat does not exceed 0.5 mm. What is the magnitude of P, if E 67,000 N/mm for aluminium? ars Subjected to Various Forces stepped bar CD of diameters d, d and d and axial lengths L, L and L is subjected to axial forces P, P, P and P 4 as shown in Fig... If E is the Young s modulus of the material, let us determine the change in the length of three portions as shown. The bar is in equilibrium. C D P P d d P d P 4 L L L P I P II P + P C P 4 + P C D III P P Figure. Stepped bar subjected to various forces

Simple Stresses and Strains 7 Therefore, for equilibrium forces P + P P + P 4. Let us consider three portions independently (for equilibrium). Portions I: P forces on one side P forces on the other side (for equilibrium) Portion II: Forces at end, P - P Forces at end C, P 4 - P Portion III: P forces at end C P forces at end D (for equilibrium) For equilibrium of portions II Net force at end is P + P - P P Net force at end C is P 4 + P - P P 4 Note that force in each portion is compressive. Stresses in three portions 4 P I πd ( P + P ) II πd 4 P III πd P4 + P πd 4 4( ) Knowing the values of Young s modulus E, strains and change in the lengths of three portions can be worked out. Example.9 Considering Fig.., let us take d 0 mm, d 4 mm, d 0 mm, L 00 mm, L 40 mm, L 50 mm. Forces are P 0 kn, P 6 kn, P 0 kn, P 4 4 kn. If E 00 kn/ mm, determine the change in length in each portion and the overall change in length. P 0 kn Portion I Portion II P + P 50 kn P 6 kn P + P 4 50 kn δl 4 0, 000 00 I due to P π 0 E 9, 549 9, 549 E 00, 000-0.0955 mm 4 δl II 50, 000 40 due to P + P π 4 00, 000-0.455 mm

8 Chapter 4 6, 000 50 Portion III δl III π 0 00, 000-0.056 mm Overall change in length -0.0955-0.455-0.056-0.856 mm due to P Exercise.9 stepped circular steel bar of a length of 50 mm with diameters 0, 5 and 0 mm along lengths 40, 50 and 65 mm, respectively, subjected to various forces is shown in Fig... If E 00 kn/mm, determine the change in its length. C D 0 kn 0 f 5 5 f 5 0 f 0 kn f f f 40 mm 45 mm 65 mm Extension in ar Due to Self-weight Problem of stresses or strains exists in long hanging cables where the length of the cable becomes large; self-weight of the cable causes tensile stress in cable with increasing magnitude from top to bottom. bar of a cross-sectional area of and a length of L is suspended vertically with its upper end rigidly fixed in ceiling as shown in Fig.. Say the weight density of the bar is w. Consider a section yy at a distance y from the lower end. Weight of the lower portion abyy yw Cross-sectional area at yy yw Normal stress of yy, y wy ( tensile) (weight is acting downwards on section yy) Figure. Exercise.9 wl Strain in section yy, e y wy/e, where E is the Young s modulus of material. L y y dy Change in length over small length dy of element wydy δdy E L wydy Total change in length, δ L O E y a b O Figure. ar under self-weight

Simple Stresses and Strains 9 wl wl δl E E ( wl) L E W E where W is the total weight of the bar. Stress in the bar or cable due to self-weight gradually increases from bottom to top as shown in Fig... Example.0 stepped steel bar is suspended vertically. The diameter in the upper half portion is 0 mm, while the diameter in the 0 mmf lower half portion is 6 mm. What are the stresses due to self-weight m in sections and as shown in Fig..4. E 00 kn/mm ( ). Weight density, w 0.7644 N/mm. What is the change in its length? w 0.7644 0 - N/mm 6 mmf m Cross-sectional area, π 0 78 54 4. mm ( ) C Cross-sectional area, π 6 8. 7 mm 4 Figure.4 Example.0 Weight of upper portion, W l w 78.54,000 0.7644 0-60.0 N Weight of lower portion, W 8.7,000 0.7644 0 -.6 N Change in length at W W + 60. 0 +. 6. 095 N/mm 78. 54 at W. 6 0. 7644N/mm 8. 7 at, at, W L. 0 000 δl E 8. 7 00 0 W δl 0. 9 0 + E 60. 0 000 0. 9 0 + 78. 54 00 0 0.9 0 - + 0.9 0 - mm 0.8 0 - mm due to self-weight 0. 9 0 mm

0 Chapter dl change in length of upper part due to weight of lower part. W, 000 E. 6, 000 78. 54 00 0 7.57 0 - mm 0.757 0 - mm Over all change in length (0.8 + 0.757) 0-0.596 0 - mm Exercise.0 steel wire of a diameter of 6 mm is hanging freely over a length of 0 m. The weight density of the wire is 0.7644 0 - N/mm. What is the maximum stress developed in the wire and what is the change in its length due to self-weight? ar of Uniform Strength bar of varying cross-sections such that under the action of axial load F applied at the lower end produces the same stress throughout its length L as shown in Fig..5. The addition of stress due to selfweight is compensated by the increased area so that the stress developed remains the same. The area of bottom section is and the area of top section is if the self-weight of the bar is w F + W c d L dy y h e g f h e s s + d g f a b F Figure.5 ar of uniform strength thus F F + W Reaction at the upper end of bar is F + W. Consider an elementary thin disc efgh with area at ef as and area at gh as + d. If w is the weight density of the bar, the weight of elementary thin disc is wdy (.)

Simple Stresses and Strains For the equilibrium of forces Integrating both sides of Eq. (.4), we get or at any distance y from lower end + wd y ( + d) wd y d d w d y (.4) d L w d y 0 w ln L, e wl e wy /. Example..5-m-long vertical circular bar is subjected to a uniform stress of 0.4 N/mm throughout its length. The diameter at the bottom edge is 60 mm. What is the diameter at the top section if the weight density w 0.07644 N/mm? Weight density, w 7.644 0-5 N/mm Length of bar, L,500 mm s 0.4 N/mm / 5 wl 7. 644 0, 500 0. 4 0.4775 e 0.4775.6 e 0.4775.6 π π d d. 6 4 4 d d. 6 d. 6 60. 6 60.696 76.8 mm Diameter at the top 76.8 mm. Exercise..0-m-long vertical circular bar is subjected to a uniform stress of 0. N/mm throughout its length. The diameter of the bottom is 50 mm. What is the diameter of the bar at the middle section if the weight density of the bar is 5. 0-5 N/mm?

Chapter Volumetric Stress and Volumetric Strain stress which acts on a body equally from all the directions is known as volumetric stress; as an example hydrostatic pressure acts equally in all the directions. Consider a sphere of diameter D at a depth h (from free surface of the liquid) in a liquid of weight density, w (Fig..6). p h D p p Volum. stress p e, elastic limit pressure p p O Vol. strain, e v D Figure.6 Volumetric stress and volumetric strain Pressure on sphere, p wh. s per Pascal s Law of Fluid Mechanics, a liquid transmits pressure equally in all the directions. Say V initial volume of sphere V final volume of sphere under pressure V - V dv, change in volume Volumetric strain, e V δv V s the depth of the body in the liquid increases, the pressure also increases on the body. Volumetric strain is proportional to pressure that is, p a e v. p Ke v, where K is the constant of proportionality or ulk modulus. ulk modulus, p Volumetric stress K ε Volumetric strain Until elastic limit p e, the pressure is linearly proportional to the volumetric strain. V Example. spherical ball of a material of a diameter of 60 mm goes down to a depth of 600 m in sea water. If the specific weight of sea water is 0. kn/m and the bulk modulus of the material of the ball is 60 kn/mm, determine the change in volume of ball. Specific weight, w 0. kn/m Depth, h 600 m Pressure, p wh 0. 600 6,0 kn/m 6. N/mm ulk modulus, K 60,000 N/mm

Simple Stresses and Strains p Volumetric strain, ε v K Initial volume, Now, e v.85 0-5 6.. 85 0 5 60, 000 πd π V 60.44 0 mm 6 6 Change in volume dv e v V.85 0-5.44 0 6 8 mm 6 Exercise. n aluminium ball of a diameter of 50 mm is immersed in sea water to a depth of h m. The specific weight of sea water is 0. kn/m. What is h if the pressure acting on ball is 4.8 N/ mm? If K for aluminium is 66 kn/mm, what is the change in volume of balls? Statically Indeterminate Problems In a statically determinate problem, the equations of equilibrium of forces are sufficient to determine the unknown forces as reactions in a structure. However, in certain problems, the equations of equilibrium of forces are not sufficient to determine the unknown forces. These are called statically indeterminate problems and an additional equation involving the geometry of deformed structure is necessary to determine unknown forces. This type of problem is a statically indeterminate problem. This equation provides condition for structural compatibility. In Fig..8, a bar held between two rigid supports is shown. force P is applied to the bar. Reactions at ends are P and P. Then, P P + P (.5) This equation is not sufficient to determine the values of P and P independently. dditional Equation bar is held between two rigid supports; as shown in Fig..7 therefore, net deformation in the bar along its axis is zero. Due to reactions P and P, portion I comes under tension and portion II comes under compression, as shown in Fig..8. If E is the Young s modulus of this material, then E P P P P P I P a I II b I P L P II Figure.7 ar field between two rigid supports Figure.8 Two portions under axial loads

4 Chapter P a δl I (extension in portion I) E P b δl II (contraction in portion II) E where is the cross-sectional area of the bar. Then, P a P b 0 E E (.6) or P a + P b b P a P (.7) Now, Eqs (.5) and (.6) can determine the reactions, P and P. The second equation i.e. Eq. (.6) is known as deformation compatibility equation. Example. n aluminium bar of a diameter of 0 mm and a length of 400 mm is rigidly held at its upper end. It is subjected to an axial force P kn at section as shown in Fig..9. E for aluminium 67 kn/mm. Gap between the lower end and the floor is 0. mm. Determine the stresses in portions and C of the bar. Under the action of the force P, the portion of the bar extends and the end C of the bar touches the floor. Then, reaction offered from floor provides a compressive force for portion C of bar. Say P tensile force in portion P compressive force in portion C P P + P (.8) Extension in, P δl I E 50 Contraction in C, P δl II E 50 where is the cross-sectional area of the bar and E is the Young s modulus of the bar. Then, δl δl 0. gap (.9) I II mm π rea, 0 78. 54 mm 4 P 50 P 50 0. 78. 54 E 78. 54 E 50 P - 50 P 0. 78. 54 67, 000 50 P - 50 P,05,46 P.6667 P + 706.4 (.0) 50 50 P C luminium bar dia 0 mm P kn Figure.9 d gap 0. mm

Simple Stresses and Strains 5 P + P kn,000 (.) or.6667 P + 706.4,000 - P or.6667 P 498.76 cm m P 868.9 N P,000-888.9 0. N Stresses in portions and C: cm P 0. 78. 54 I + + 9. 0 N/mm m 868. 9 II. 8 N/mm 78. 54 C 0.4 mm Figure.0 Exercise. -m-long stepped steel bar, shown in Fig..0, with a cross-sectional area of the portion is cm and that of the portion C, cm is subjected to a load of P 5 kn. Gap between the bar and the ground floor is 0.4 mm. What are the stresses developed in the portions and C of the bar, given that E 00 GPa? (Note that GPa 0 9 N/m 0 N/mm ) Strain Energy and Resilience When a body is subjected to an external force, stress and strain are produced in the body and the work done on the body is absorbed as strain energy in the body. If the external forces are removed then strain energy absorbed is fully recovered, and it is said to be resilience of the body. For the property of resilience, springs are used in mechanisms and machines. Figure. shows a graph between stress (s) and strain (e) on a body. s e is the elastic limit stress. Within the elastic limit, the strain energy is fully recoverable. Within the elastic limit, say, stress is s and strain is e ; then Strain energy per unit volume.. ε E E using Hooker s law ε E Shaded area ose shows resilience up to the elastic limit, s e. Strain energy absorbed e E V, where V is the volume of the body. Stress s o Elastic region s s e e Strain Plastic region Figure.

6 Chapter e V is known as proof resilience. Modulus of resilience e E E In case of t (shear stress) and f (shear strain), Modulus of resilience τ e G where t e is the shear stress at elastic limit and G is shear modulus. In case of volumetric stress, p e (at elastic limit), and volumetric strain, e v, e Modulus of resilience p K where p e is hydrostatic pressure at elastic limit and K is the bulk modulus. is proof resilience per unit volume. Example.4 40-mm cubical block is subjected to shear stress and it is observed that t e 40 N/mm. t e If shear modulus G 84 kn/mm, determine (i) the modulus of resilience, (ii) the shear strain at elastic limit and (iii) the total strain energy absorbed at elastic limit. 40 f Figure. shows a 40-mm cube subjected to a shear stress, t e, of 40 N/mm. Volume of the block 40 40 40 64 0 mm Shear modulus, G 84,000 N/mm Figure. Modulus of resistance τ e 40 4. 86 0 Nmm/mm G 84, 000 Shear strain at elastic limit, τ e G 40 84, 000. 857 0 Total strain energy absorbed at elastic limit, U e Proof resilience 4. 86 0 64 0 9. 5 Nmm.9 Nm 40 40 mm Exercise..4 copper bar of a diameter of 0 mm and a length of 600 mm is subjected to an axial tensile load. If e 5 N/mm, determine. the resilience at 00 N/mm,. the proof resilience and. the modulus of resilience, given E 0 kn/mm. Sudden Load Till now, we have considered the application of gradually applied loads, where the load is gradually increased starting from zero magnitude and the stress produced in a material is the load/cross-sectional

Simple Stresses and Strains 7 area. However, if the whole of the magnitude of the load is applied suddenly on a body, then the stress produced is more than the stress produced by the same load when applied slowly and gradually. Say a heavy load is being lifted by a crane hook, slings are tightened on the load and slings pass through the throat of the crane hook. When wire rope moves up, loose slings become taut and the whole of the load acts suddenly on the hook and the wire rope. Consider that a load W is applied suddenly on a bar and sudden deformation produced in bar is d as shown in Fig... Work done on the bar Wd strain energy absorbed by the bar However, the internal resistance of the body develops gradually. So that ( / ) Rδ Wδ where internal resistance, R W. Stress is developed as an internal resistance of the body per unit area. Stress developed in the bar R W, where is the cross-sectional area. W sudden gradual. This shows that the stress developed due to load applied suddenly is double the stress developed in the body due to the same load applied gradually. Example.5 Water under a pressure of 5 N/mm is suddenly admitted on a plunger of a diameter of 0 mm. Plunger CR P p is connected to a steel connecting rod of a length of 4 m and a diameter of 0 mm. Determine the amount of stress developed in the connecting rod and the deformation produced in p Plunger the connecting rod. E for steel 00 kn/mm (Fig..4). Water pressure, p 5 N/mm Plunger diameter, D 0 mm rea of plunger, π π D 4 4 0 09.7 mm Force on plunger, F p 5 09. 7 56548. 5 N Cross-sectional area of connecting rod π 0 706. 86 mm 4 Length of connecting rod, L 4 m 4,000 mm Stress produced in connecting rod, sudden F/ W O R D C d Figure. Graph between sudden load and deformation W CR Connecting rod Figure.4 d

8 Chapter 56, 548. 5 706. 86 60 N/mm s Deformation in connecting rod, δ L E 60 00 000 4,,, 000. mm Exercise.5 copper column of a diameter of 50 mm and a length of 500 mm is subjected to a load W 0 kn applied suddenly as shown in Fig..5. If E for copper 05 kn/mm, what is the stress developed in copper and what is the deformation produced? Impact Load The load which is applied with some velocity on a body is known as impact or shock load. The kinetic energy of the load is absorbed as strain energy in the body. This type of load is unsafe and must be avoided in normal applications. However, the impact loads are used in industry to produce forged parts and to drive piles in the ground for reinforcing the earth for heavy building structure. In other words, if a load is allowed to fall through a height, it will gain some velocity due to gravity and will strike the body with kinetic energy. Figure.6 shows a bar of a uniform cross-sectional area of and a length of L with a collar at lower end and upper end fixed in ceiling. weight W falling under gravity through height h strikes the collar producing an instantaneous elongation d in the bar. Loss of potential energy of weight W h + δ ( ) Gain in strain energy of the bar i E V where i instantaneous stress produced in the bar V volume L E Young s modulus d i i E L, where L is length of bar i W ( h + δ E L i ) Using the principle of conservation of energy, Wh W L i i + E E L Multiplying throughout by E, we get L i (.) W gap 0 500 mm d 50 mm Collar Figure.5 W h Figure.6 ar under impact load L d i d i

Simple Stresses and Strains 9 Wh E WL E + L E L i i W W Eh or i i 0 (.) L Solving the quadratic equation for instantaneous stress i, i W 4W 8W Eh ± + L W W Eh ± + W L W W Eh + + W L Note that negative sign is inadmissible, as the stress produced in the bar is tensile and the term in the under root is greater than. W Eh Instantaneous stress produced, i + + WL s the load strikes the collar, the bar starts vibrating with some amplitude, but the vibrations die down due to air and material damping and finally the load comes to rest W W If distance h 0, load will become sudden load, h 0, s ( + + 0) i Instantaneous elongation in the bar, δi E L, where L is the length of the bar Example.6 50-N load falls through a height of 50 mm onto a steel bar of a diameter of 5 mm and a length of 400 mm, supported on the ground. If E 00 GPa, what is the instantaneous stress developed in the bar and what is the change in its length? The load falls on the bar producing a compressive stress in the bar and its length has instantaneous contraction. W 50 N h 50 mm d 5 mm π π d 5 490. 87 mm 4 4 Length of the bar, L 400 mm E,00,000 N/mm Eh 00 000 490 87 50 WL,,. 50 400 4,90,870

0 Chapter Putting these values in the expression, W 50 490 87 0. 086. ( ) i 0. 086 + + 4, 90, 870 ( ) 0. 086 + 700. 6 7. 47 N/mm (compressive stress) δ i i E L 7. 47 00 0000 400 0. 4 mm (contraction),, Exercise.6 copper bar of a diameter of 0 mm and a length of 00 cm is stressed by a weight of 0 N dropping freely through a height of 40 mm onto a collar provided at the end of the bar before commencing to stretch the bar. Find the instantaneous stress and the instantaneous elongation in the bar. Given E 05 kn/mm, s ut of material 60 N/mm. Comment whether the bar will break. Tensile Test on Mild Steel Mild steel is the most commonly used engineering material. Majority of engineering components and structures are made of mild steel and the material is easily available in the market. In material testing labs of engineering colleges, majority of the tests are performed on mild steel samples. specimen of the shape shown in Fig..7(a), circular section, with collars at the ends is fitted in the grips of a testing machine. gradually and slowly increasing tensile load P is applied on the sample through a mechanism provided in the machine. Due to the increasing tensile load, the specimen is continuously stretched. Tensile load and extension in the specimen are continuously recorded by the operator. Nowadays, there are machines available in which an automatic graph between the load and the extension is obtained as shown in Fig..7(b). Many important mechanical properties are obtained from this load extension graph as follows:. O to is a straight line, P δ L (proportional), that is, the material obeys Hooke s law. Slope of this line provides information on the Young s modulus of the material.. Point, an elastic limit point, is very close to, and many a times it is hardly noticed. If the load is removed from the sample at this point, then the residual strain in sample will be zero.. Point C, upper yield point of the material. The material has yielded, meaning that plastic strain is developed in the sample. 4. Yield strength of the material is defined at this point, that is, load at upper yield point/crosssectional area. 5. Resistance of the material falls from C to D, lower yield point. This is due to the presence of extra rows of carbon and nitrogen atoms in the material. 6. D to E is called yield plateau. In this portion average stress remains constant. 7. t E, resistance of the material again starts increasing and point E is termed as point of strain hardening.

Simple Stresses and Strains P Collar d P L (a) Specimen F Ultimate load C at F P C D E G, reaking load t F Neck C - Upper yield point D - Lower yield point Cup Cone O Extension Recovered extension Shear fracture Tensile fracture (b) Load extension diagram (c) t necking Figure.7 8. t point F, in any ductile material as mild stress, necking takes place, that is, in some weaker section of the bar, the diameter is heavily reduced. Further extension takes place in the vicinity of this neck. Point F gives the ultimate load and the ultimate load/original area of sample gives the ultimate tensile strength of material 9. Resistance of the material falls through the portion F to G and finally specimen breaks into two pieces, showing the cup and cone type of fracture, marked by two regions: (i) shear fracture in outer ring, marked by shining lines and (ii) straight tensile fracture in inner core marked by dull granular structure. Join the two broken pieces to find out (i) the final gauge length, L (ii) the diameter at the neck, d. L Percentage elongation L 00 indicates ductility of the material L π π d d Percentage reduction in area 4 4 d d 00 00 π d d 4 Percentage reduction in area provides the information on degree of cold working of the material. Two important information on (i) yield strength and (ii) percentage elongation decide the suitability of steel for structural applications. Up to the point F, ultimate load extension in sample is proportional to its length and beyond F up to breaking load extension is proportional to lateral dimension, that is, the diameter of the bar.

Chapter Total extension bl + c dl bl + c, which is known as arba s Law, where b and c are arba s constants, L is gauge length and is the cross-sectional area of specimen. Example.7 round section of wrought iron of a diameter of.5 mm and a gauge length of 00 mm was tested in tension up to fracture and the following observations were made. (i) Yield load 9.6 kn (ii) Maximum load 44.8 kn (iii) Load at fracture 7 kn (iv) Diameter at neck 9. mm (v) Total extension in sample 7.6 mm. Determine (i) the yield strength, (ii) the ultimate tensile strength, (iii) the actual breaking strength and (iv) the percentage elongation. Diameter, d.5 mm Gauge length, L 00 mm Diameter at neck, d 9. mm Original cross-sectional area, π π d 4 4 5..7 mm rea at neck, π π d 9. 4 4 65.04 mm (i) Yield strength, s 9. 6 00 yp 4. N/mm. 7 (ii) Ultimate tensile strength, s ut 44. 8, 000 65.06 N/mm. 7 (iii) ctual breaking strength, s 7 00 bk 568.9 N/mm 65. 04 (iv) Percentage elongation, 7. 6 00 7.6% 00 Exercise.7 specimen of aluminium alloy of a diameter of mm was tested under tension. Linear strain parallel to the applied load P was measured accurately with the help of strain gauges as follows: Pin kn 4 8 6 0 4 8 6 8 9.5 e a 0-0.4 0.69.0.9.7.07.40.74 4.0 5.65 8.40

Simple Stresses and Strains Plot graph (P ) versus axial strain (e a ) and determine Young s modulus of elasticity and 0. per cent proof stress from the graph. To determine 0. per cent proof stress, a line from 0.00 length is drawn parallel to initial slope of load-extension graph. This line intersects the graph providing load. Stress Concentration We have studied stresses and strains in bars of uniform section throughout the axial length and we find that the stress is equal to the load/cross-sectional area, a uniform stress. However, if there is an abrupt change in the geometry of a component, then the stress distribution becomes non-uniform with larger values of stress at corners, fillets and holes. Consider a stepped bar of diameters D and d as shown in Fig..8, subjected to axial load P, stress in portion I, that is, 4 P suddenly increases to 4 P in portion II at the corner section bb πd π d of the stepped bar. Increase in stress is concentrated along the circumferential edge bb, and stress distribution becomes non-uniform with s max at corner b, as shown in Fig..8(b). However, at section cc, away from discontinuity stress distribution again becomes uniform. Due to stress concentration the stepped bar under load will try to break along corner bb. To avoid failure due to stress concentration, the cross-sectional area should be gradually reduced by providing a fillet radius as shown in Fig..9. Effect of fillet radius R on stress concentration factor (SCF) in a bar shown in Fig..9 is obtained through photoelasticity (an experimental technique). P I D b b a a II c c d D P a b b a s max s max Figure.8 Stepped bar under axial load c c s 4 P pd Moreover, one can consider the effect of a hole in a plate as shown in Fig..40(a). Stress distribution in the section abba becomes non-uniform, with maximum stress max at the edge of the hole. SCF max av Maximum stress verage stress, with average stress over a section with no effect of abrupt change. s max P a b b a R Fillet radius P P R Figure.9 P Figure.40 (a)

4 Chapter Detailed discussion on SCFs is beyond the scope of this book. P n empirical relation of SCF for an elliptical hole in a plate [Fig..40(b)] subjected to an axial load is SCF+ b a where a semi-minor axis and b semi-major axis. a Example.8 In a rectangular section bar, shown in Fig..9, breadth 5 mm. What should be the minimum fillet radius R so b that SCF does not exceed.65? Table. provides values of SCF for various ratios of R/ i.e. fillet radius/breadth From Table. for SCF of.65, R P 0.4 Fillet radius, R 0.4 0.4 5.575 mm Figure.40(b) Plate with elliptical hole Exercise.8 In a sample made from a 40-mm-broad thin sheet, a fillet radius of 7 mm is provided. Find SCF for this flat. [Hint: Plot a graph between R and SCF using values given in Table. and find SCF.] Table. R/, Ratio 0. 0. 0.4 0.08 SCF.5.50.65.8 Factor of Safety material is tested for its ultimate strength. While designing a machine component or any engineering component, we do not want that it should fail in service; therefore, it is designed based for a safe allowable stress. ut allowable FOS where ut is the ultimate strength and FOS is the factor of safety. The necessity for an FOS or safe design depends upon the application of an engineering component such as (i) a wire rope in a crane carrying the load through crane hook and the wire rope should not break, as the breaking of the wire rope may cause the casualty of human beings those who are working with the crane; (ii) a structural bridge should not collapse during service, as the collapse of a bridge may cause extensive damage to the property and human lives; and (iii) a bed of machine damage may cause less problems; therefore, a smaller FOS can be taken in this type of applications. higher FOS is desired for bridges and wire ropes. Then, we assume that the martial is homogeneous (having same physical properties throughout the volume used) and isotropic (having same elastic properties throughout); however, in reality, the material may contain internal defects such as holes, cracks and slag inclusions and as such no material is homogeneous.

Simple Stresses and Strains 5 Therefore, to account for all these factors, we take some suitable value of FOS depending upon (i) the type of load on the component such as gradual, sudden or shock load; (ii) the type of service conditions and (iii) the mechanical properties of the material such as strength, ductility and hardness. Example.9 Two 0-mm-thick plates are joined by a single rivet of a diameter of 0 mm. The ultimate shear strength of the material of the rivet 460 N/mm. Take FOS as and determine the load which can be safely applied on the plates, as shown in Fig..4. The plates are 50 mm wide. Ultimate shear strength 460 MPa Factor of safety 460 llowable shear stress, τ all 5. MPa Rivet diameter, d 0 mm Cross-sectional area of rivet, π π d 0 00π mm 4 4 The rivet can break in shear along the section aa, and the rivet is in single shear. π Permissible load, P d τ all 4 00p 5. 48,70 N 48.7 kn P a a 0 mm P b 50 mm Riveted joint Figure.4 Exercise.9 The ultimate tensile strength of mild steel is 660 N/mm. tie bar of circular section is subjected to an axial load of 80 kn. Take FOS as 4 and determine the diameter of the tie bar. Problem. hole of a diameter of 50 mm is to be punched in a.6-mm-thick mild steel sheet. If the ultimate shear stress of mild steel is 480 N/mm, determine the force required to punch the hole.

6 Chapter Figure.4 shows a thin sheet of thickness t and punch of diameter d. When the hole is punched, shearing area pdt. Ultimate shearing strength t us Thus shear force. P pdt t us p 50.6 480 0,67 N 0.67 kn, force required to punch the hole. Problem. Two parts of a certain machine component are joined by a rivet of a diameter of 0 mm. Determine the shear and normal stresses in the rivet if the axial force P 5 kn and the angle of joint is 60 to the axis of the load as shown in Fig..4. Rivet diameter, d 0 mm Cross-sectional area π 0 4 00p mm P P 5,000 N Normal component, P n P cos 0 5,000 0.866,990 N Tangential component, P t P sin 0 5,000 0.5 7,500 N Normal stress, Pn, 990 4. 5 N/mm 00π Shear stress, G Pt 7, 500 τ. 87 N/mm 0. 00π Problem. -m rigid square plate is supported over four legs,, C and D of equal length. load of,00 N is applied on a square plate at point K, such that F 0.4 m and G 0. m as shown in Fig..44. Determine the reactions in each leg. Say the reactions in legs are R, R, R C and R D, respectively. Then, R + R + R C + R D,00 N (.4) 0.4 F m P d Punch Hole t t Figure.4 60 P 0 Figure.4 P 00 N D F 0.4 m m G 0. m K E C m Figure.44