Direct and Shear Stress

Direct and Shear Stress 1

Direct & Shear Stress When a body is pulled by a tensile force or crushed by a compressive force, the loading is said to be direct. Direct stresses are also found to arise when bodies are heated or cooled under constraints and in vessels under pressure. Tension Compression Rigid Constraint Applied Heat Induced Compression Stress (σ) Direct stress is defined as the ratio of the applied load (F) to the cross-sectional area (A), normal to the load. Hence we have the familiar equation: Stress = Load Area or σ = F A Force has units of newton s (N) and area has units of m 2, hence stress has units of N/m 2. Since stresses are often large values we often quote them in terms of: kn/m 2 = 10 3 N/m 2 MN/m 2 = 10 6 N/m 2 GN/m 2 = 10 9 N/m 2 Note Some textbooks still use the name pascal (Pa) in place of N/m 2. Strain (ε) Whenever a body is subjected to stress, whether tensile or compressive, it will experience a change in size. Bodies subjected to tension will extend; bodies subjected to compression will shorten. F L = original length x = extension L x Body subjected to tension The ratio of the extension to the original length is called strain. Thus we have: 2

Strain = Change in length Original length or ε = x L Since x and L have the same units it follows that strain is unitless. Young s Modulus of lasticity or the Relationship between Stress and Strain If we are lucky, when we remove loading on a body it will return to its original size. This is because most materials exhibit a degree of elasticity. However, if we load beyond the materials elastic limit, permanent set or extension will occur. Young s Modulus of lasticity helps to define the point at which this elastic limit is exceeded. Hooke s Law This states that within the elastic limit that strain is directly proportional to stress. Stress Strain = A constant = Young s Modulus Since strain is unitless it follows that has the same units as σ. Values of may be found in tables and has quite large values. For example, for steel is in the range 196 210 GN/m 2. is also related to the stiffness or rigidity of a material since the higher its value, the greater the load required to produce a given stiffness. Bars of Varying Cross-Section F A B The load in each section will be the same however the stress will be different because of the difference in area σ A = F Area A and σ B = F Area B Provided section A and section B are made of the same material then: ε A = σ A and ε B = σ B Worked xamples 1. A rod of length 325 mm is subjected to a tensile load of 60 kn and found to extend 0.160 mm. If = 200 GN/m 2 determine the diameter of the rod. ε = x L = 0.160 325 = 4.923 10 4 = σ ε Therefore σ = ε = 200 4.923 10 4 = 0.09846 GN/m 2 = 98460 kn/m 2 σ = F A 3

Therefore A = F σ = 60 98460 = 6.0938 10 4 m 2 A = πd2 4 Therefore d = 4A π = 4 6.0938 10 4 π = 0.02785 m = 27.85 mm 2. A vertical steel hanger ABC is 2 m long and carries a load of 135 kn at the lower end, as shown in the diagram below. The upper length is 50 mm diameter and the lower length is 35 mm diameter. If = 200 GN/m 2 calculate the total extension of the bar A 1m 50 mm dia (a) σ (a) = F σ A (b) = F (a) A (b) C F is the same in both sections 1m 35 mm dia (b) ε (a) = σ (a) ε = x L ε (b) = σ (b) B 135 kn Area (a) = π o.o52 4 Area (b) = π 0.0352 4 σ (a) = σ (b) = 135 1.963 10 135 9.621 10 = 1.963 10 3 m 2 = 9.621 10 4 m 2 3 = 68772.3 kn/m2 4 = 140318 kn/m2 ε (a) = 68772.3 103 200 10 9 = 3.438 10 4 ε (b) = 140318 103 200 10 9 = 7.016 10 4 x (a) = 3.438 10 4 1 = 3.438 10 4 m = 0.3438 mm x (b) = 7.016 10 4 1 = 7.016 10 4 m Therefore, x = 3.438 10 4 + 7.016 10 4 = 1.0454 10 3 m = 1.0454 mm 4

Tutorial Problems From Applied Mechanics by Hannah & Hillier 1. A bar of 25 mm diameter is subjected to a tensile load of 50 kn. Calculate the extension on a 300 mm length. = 200 GN/m 2. (0.153 mm) 2. A steel strut, 40 mm diameter, is turned down to 20 mm diameter for one-half its length. Calculate the ratio of the extensions in the two parts due to axial loading. (4:1) 3. When a bolt is in tension, the load on the nut is transmitted though the root area of the bolt which is smaller than the shank area. A bolt 24 mm in diameter (root area = 353 mm 2 ) carries a tensile load. Find the percentage error in the calculated value of the stress if the shank area is used instead of the root area. (21.96%) 4. A light alloy bar is observed to increase in length by 0.35% when subjected to a tensile stress of 280 MN/m 2. Calculate Young s modulus for the material. (80 GN/m 2 ) 5. A duralumin tie, 600 mm long, 40 mm diameter, has a hole drilled out along its length. The hole is 30 mm diameter and 100 mm long. Calculate the total extension of the tie due to a load of 180 kn/m 2. Take as 84 GN/m 2. (1.24 mm) 6. A steel strut of rectangular section is made up of two lengths. The first, 150 mm long, has breadth 40 mm and depth 50 mm; the second, 100 mm long, is 25 mm square. If = 220 GN/m 2, calculate the compression of the strut under a load of 100 kn. (0.107 mm) 7. A solid cylindrical bar, of 20 mm diameter and 180 mm long, is welded to a hollow tube of 20 mm internal diameter, 120 mm long, to make a bar of total length 300 mm. Determine the external diameter of the tube if, when loaded axially by a 40 kn load, the stress in the solid bar and that of the tube are to be the same. Hence calculate the total change in length of the bar. = 210 GN/m 2. (28.3 mm; 0.184 mm) 5

Shear Stress (τ) Body subjected to shear When two equal and opposite parallel forces not in the same plane act on parallel faces of a body, as shown, then the body is said to be loaded in shear and we have the equation τ = F. F is the applied load and A is the A area parallel to the force. Shear stress has the same units as direct stress (N/m 2 ). xamples of shear stress include: Shearing metal on a guillotine Shearing a hole on a press Rivet in single or double shear Shear Strain (ϕ) B C Shear strain is defined as the angle of deformation BAC in radians. Unstressed Block A Stressed Block ϕ = BAC BC BA radians Analogous with the direct stress-strain relationship, we have: Shear stress = constant G or τ = G = Modulus of Rigidity Shear strain ϕ Factor of Safety Most engineering parts have a margin of safety built in to them by assuming a safe working stress based on the materials ultimate tensile strength (UTS). Factor of Safety = UTS Safe working stress Worked xamples 1. A load of 6 kn is applied to the member shown below, and is carried by a single rivet. The angle of the joint is 60 o to the axis of the load. Calculate the tensile and shear stress in the rivet which is 18 mm diameter. Fcos30 o Fsin30 o Area of rivet = πd2 = π 0.0182 = 2.545 10 4 m 2 4 4 6

Direct pull on rivet = Fcos30 o = 6cos30 o = 5.196 kn Therefore, direct stress on rivet = Direct pull Area Shear force on rivet = Fsin30 o = 6sin30 o = 3 kn = 5.196 2.545 10 4 = 20416.50 kn/m2 = 20.416 MN/m 2 Shear force 3 Therefore, shear stress on rivet = = = 11787.82 Area 2.545 10 4 kn/m2 = 11.787 MN/m 2 Note Since the direct stress is considerably larger than that of the shear stress it would probably be wise to change to a bolt. 2. A solid coupling is to transmit 80 kw at 180 rev/min through 8 equally spaced bolts (see diagram below). If the bolts are 10 mm diameter and are on a pitch circle diameter of 150 mm, calculate the average shear stress on each bolt. Area of bolt = πd2 = π0.0102 = 7.854 10 5 m 2 4 4 P = Tω where ω = 2πN 60 Therefore P = 2πNT 60 Hence T = 60P = 60 80 103 = 4244.13 Nm 2πN 2π 180 T = F r Therefore F = T r = 4244.13 0.075 = 56588.4 N Therefore, load/bolt = 56588.4 8 = 7073.55 N Hence shear stress per bolt = 7073.55 7.854 10 5 = 90063025.21N/m2 = 90.03 MN/m 2 7

Tutorial Problems From Applied Mechanics by Hannah & Hillier 1. Calculate the maximum thickness of plate which can be sheared on a guillotine if the ultimate shearing strength of the plate is 250 MN/m2 and the maximum force the guillotine can exert is 200 kn. The width of the plate is 1 m. (0.8 mm) 2. A rectangular hole 50 mm by 65 mm is punched in a steel plate 6 mm thick. The ultimate shearing strength of the plate is 200 N/mm 2. Calculate the load on the punch. (276 kn) 3. Calculate the maximum diameter of hole which can be punched in 1.5 mm plate if the punching force is limited to 40 kn. The plate is aluminium having ultimate shear strength of 90 MN/m 2. (94.5 mm) 4. A bar is cut at 45 to its axis and joined by two 12 mm diameter bolts (see diagram below). If the pull in the bar is 80 kn, calculate the direct and shear stress in each bolt. (250 MN/m 2 ; 250 MN/m 2 ) 5. A solid coupling is to transmit 225 kw at 10 rev/s. The coupling is fastened with six bolts on a pitch circle diameter of 200 mm. If the ultimate shear stress is 300 MN/m 2, calculate the bolt diameter required. The factor of safety is to be 4. (10.1 mm) 8

Compound Bars When two or more members are rigidly fixed together so they share the same load and extend the same amount, the members are said to be compound. Stresses in such members are calculated as follows: (1) Total load is the sum of loads taken by each member. (2) Load taken by each member is given by the product of its stress and its area. (3) xtension or contraction is the same for each member Hence we are able to write: F = σ AA A + σ BA B ε A = ε B or σ A A = σ B B Reinforced Bar Tutorial Problems From Applied Mechanics by Hannah & Hillier 1. A rectangular timber tie, 180 mm by 80 mm, is reinforced by a bar of aluminum of 25 mm diameter. Calculate the stresses in the timber and reinforcement when the tie carries an axial load of 300 kn. for timber = 15 GN /m 2 ; for aluminum = 90 GN/m 2. (17.8 MN/m 2 ; 106.8 MN/m 2 ) 2. A concrete column having modulus of elasticity 20 GN/m 2 is reinforced by two steel bars of 25 mm diameter having a modulus of 200 GN/m 2. Calculate the diameter of square section strut if the stress in the concrete is not to exceed 7 MN/m 2 and the load is to be 400 kn. (220 mm square) 3. A cast iron pipe is filled with concrete and used as a column to support a load W. If the outside diameter of the pipe is 200 mm diameter and the inside diameter is 150 mm, what is the maximum permissible value of W if the compressive stress in the concrete is limited to 5 MN/m 2. Take for concrete as 0ne tenth that of cast iron. (97 t) 4. A concrete column is reinforced with steel bars and carries a load of 20 t. The overall cross-sectional area of the column is 0.1 m 2 and the steel reinforcement accounts for 3% of this area. Find the stress taken by the concrete. If the length of the column is 4 m, how much does it shorten? Take e for steel as 200 GN/m 2 and for concrete, 20 GN/m 2. (1.54 MN/m 2 ; 0.31 mm) 5. A cylindrical mild steel bar of 40 mm diameter and 150 mm long, is enclosed by a bronze tube of the same length having an outside diameter 60 mm and an inside diameter of 40 mm. The compound strut is subjected to an axial compressive load of 9

200 kn. Find; (a) the stress in the steel rod; (b) the stress in the bronze tube; (c) the shortening of the strut. For steel = 200 kn/mm 2, For bronze = 100 kn/mm 2. (Steel 98 MN/m 2 ; Bronze 49 MN/m 2 ; 0.0735 mm) 6. A compound assembly is formed by brazing a brass sleeve on to a solid steel bar of 50 mm diameter. The assembly is to carry a tensile axial load of 250 kn. Find the crosssectional area of the brass sleeve so that the sleeve carries 30% of the load. Find, for this composite bar, the stresses in the brass and the steel. for brass = 84 GN/m 2 ; for steel = 210 GN/m 2. (2110 mm 2 ; steel 89.1 MN/m 2 ; brass 35.6 MN/m 2 ) 10

Thermal Strain A change in temperature of material gives rise to a thermal expansion such that x = Lαt where L = original length; α = coefficient of expansion; t = change in temperature. Hence thermal strain is given by ε = x L = Lαt L = αt There will be no stress associated with this unless the bar is restricted see below. Induced Stress in Constrained Bar Rigid Support When a material is heated and not allowed to expand freely, stresses are induced which are known as temperature stresses. Their value will depend upon the temperature change and the restraint, the extreme case being when the expansion is fully restrained. If the above bar is heated such that temperature increases by t o the resulting stress may be found by considering the process to take place in two stages. (1) The bar is allowed to expand freely such that x = Lαt. (2) A corresponding force is then applied so as to restore the bar to its original length. The previous expansion x now becomes the compression of the bar under load. Thus Compressive strain ε = x L = αt And compressive stress = ε = αt Worked xample A steel bar 280 mm long, 25 mm diameter is machined down to 20 mm diameter for 90 mm of its length. It is then heated 35 o above room temperature, clamped at its ends and allowed to cool back to room temperature. If the distance between the clamps is held rigid, determine the maximum stress in the bar. Assume α = 12.5 x 10-6 / o C and = 210 GN/m 2. 280 mm 90 mm σb = σmax since area is smallest 25 mm dia (A) 20 mm dia (B) If allowed to contract freely the contraction would be x = αlt = 12.5 x 10-6 x 0.280 x 35 = 1.225 x 10-4 m This contraction is in essence an expansion. 11

Now total expansion = extensiona + extensionb Or x = xa + xb = αlt quation (1) But ε = x L and = σ ε Therefore, x = εl = σl Therefore, So, from (1) we have x A = σ AL A and x B = σ BL B αlt = 1.225 x 10-4 = σ AL A + σ BL B quation (2) We need an expression for σa in terms of σb F = σaaa = σbab Therefore, σa = σ BA B A A Substituting this into (2) gives 1.225 x 10-4 = ( σ BA B L A A A + σ BL B ) = σ B ( A BL A + L B ) A A Or 1.225 x 10-4 = σb ( 0.0202 0.190 0.025 2 210 10 3 + 0.090 210 10 3) This gives σb = 121.57 MN/m 2 = σmax Tutorial Problems From Applied Mechanics by Hannah & Hillier 1. Steel railroads 10m long are laid with a clearance of 3mm at a temperature of 15 o C. At what temperature, will the rails just touch? What stress would be induced in the rails at that temperature if there were no initial clearance? Assuming α = 11.7 x 10-6 / o C and = 200GN/m 2. (40.64 o C, 60MN/m 2 ) 2. A brittle steel rod is heated to 150 o C and then suddenly clamped at both ends. It is then allowed to cool and breaks at a temperature of 90 o C. Calculate the breaking stress of the steel. = 210 GN/m 2 ; α =12 x 10-6 / o C. (151 MN/m 2 ) 3. A steel bar of 100 mm diameter is rigidly clamped at both ends so that all axial extension is prevented. A hole of 40 mm diameter is drilled out for one-third of the length. If the bar is raised in temperature by 30 o C above that of the clamp, calculate the maximum axial stress in the bar. = 210 GN/m 2 ; α = 0.000012/ o C. (84.7 MN/m 2 ) 12

Volumetric Strain As a result of an elastic tensile strain on a linear dimension x the new length is given by New length = x + x = x + εx = x(1 + ε) where ε = strain Similarly, after compression New length = x(1 ε) Now consider a block immersed in a liquid This will produce a volumetric strain Let the three strains be ε x, ε y, and ε z New volume = x(1 ε x )y(1 ε y )z(1 ε z ) = xyz(1 + ε x ε y + ε y ε z + ε z ε x ε x ε y ε z ε x ε y ε z ) Since the strains will be small, the product terms will be very small and can be neglected Therefore, new volume = xyz + xyz( ε x ε y ε z ) = xyz xyz(ε x + ε y + ε z ) But xyz = original volume Therefore, volume change = new volume - original volume = xyz(ε x + ε y + ε z ) Therefore, volumetric strain ε v = xyz(ε x+ε y +ε z ) xyz Poisson's Ratio (ν) = (ε x + ε y + ε z ) In addition to the change in length in the direction of a direct load there is a simultaneous dimensional change in the two directions normal to such a load. Thus a tensile load produces tensile (longitudinal) strain along its own axis and a compressive (lateral) strain along the other two axes resulting in a reduction in the corresponding dimensions. 13

Within the elastic limit the ratio of the lateral strain to the longitudinal strain is called Poisson's Ratio and denoted ν. ν = Lateral strain Longitudinal strain Worked xample A test piece 50 mm by 12.5 mm has a length of 100 mm. Young's modulus = 200 GN/m 2 and ν = 0.25, estimate for a load of 70 kn: (a) Longitudinal strain (b) Lateral strain (c) Volumetric strain (d) Change in volume Longitudinal strain: σ x = 70 (0.05 0.0125 = 1.12 105 kn/m 2 ε x = σ x = 1.12 105 200 10 9 = 5.6 10 4 Lateral strain: σ y = σ z = 0 Therefore ε y = ε z = νσ x = 0.25 5.6 10 4 = 1.4 10 4 Volumetric strain: ε v = ε x + ε y + ε z = ε x 2ε y ε v = 5.6 10 4 + ( 2 1.4 10 4 ) = 2.8 10 4 Volumetric change: Volume change = ε v V original = 2.8 10 4 (0.05 0.0125 0.1) = 1.75 10 3 (increase) Consider now a block subjected to stress in the x and y directions. σ y σ x ε y = σ y + νσ x 14

and ε x = σ x νσ y If the block is now subjected to stress in all three planes: σ y σ x σ z ε y = σ y + νσ x + νσ z ε x = σ x νσ y + νσ z You write the equation for ε z Tutorial Problem 1. A test piece 50 mm wide, 12.5 mm thick has a gauge length of 100 mm. Taking as 208 GN/m 2 and Poisson s ratio as 0.25 determine, for a load of 67.5kN: (a) The longitudinal strain (b) The lateral strain (c) Volumetric strain (d) The change in volume (5.15 10 4 ; 1.297 10 4 ; 2.596 10 4 ; 1.623 10 8 m 3 ) 2. If the test piece in question 1 is also subjected to and additional horizontal stress of 62MN/m 2, the stress in the remaining direction being zero, calculate the vertical and horizontal strains when the second stress is (a) compressive and (b) tensile. (5.94 10 4 ; 4.447 10 4 ; 4.278 10 4 ; 1.683 10 4 ; 5.53 10 5 ; 2.04 10 4 ) 3. A cube is stressed in three mutually perpendicular directions x, y and z. The stresses in these directions are σ x = 50 kn m 2, σ y = 80 kn m 2 and σ z = 100 kn m 2. 15

Determine the strains in the x, y and z directions. Poisson s ratio is 0.34 and = 71 GN/m 2. (8x10-7, 1.386x10-6, -2.06x10-6 ) 16

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