SIMPLE STRESSES ND STRINS Mechanics of Materials Mechanics of materials deals with the elastic behavior of materials and the stability of members. Mechanics of materials concepts are used to determine the stress and deformation of axially loaded members, connections, torsional members, thin-walled pressure vessels, beams, eccentrically loaded members, and columns. Simple Stresses & Strains Stress is the internal resistance offered by the body per unit area. Stress is represented as force per unit area. Typical units of stress are N/m 2, ksi and MPa. There are two primary types of stresses: normal stress and shear stress. Normal stress,, is calculated when the force is normal to the surface area; where as the shear stress,, is calculated when the force is parallel to the surface area. P P normal _ to _ area parallel _ to _ area Linear strain (normal strain, longitudinal strain, axial strain),, is a change in length per unit length. Linear strain has no units. Shear strain,, is an angular deformation resulting from shear stress. Shear strain may be presented in units of radians, percent, or no units at all. L parallel _ to _ area Height tan [ in radians] Hooke s law is a simple mathematical relationship between elastic stress and strain: stress is proportional to strain. For normal stress, the constant of proportionality is the modulus of elasticity (Young s Modulus), E. E Poisson s ratio,, is a constant that relates the lateral strain to the axial strain for axially loaded members. lateral axial Theoretically, Poisson s ratio could vary from zero to 0.5, but typical values are 0.35 for aluminum and 0.3 for steel and maximum value of 0.5 for rubber. Hooke s law may also be applied to a plane element in pure shear. For such an element, the shear stress is linearly related to the shear strain, by the shear modulus (also known as the modulus of rigidity), G. G
For an elastic isotropic material, the modulus of elasticity E, shear modulus G, and Poisson s ratio are related by G E 2 1 E 2G 1 The bulk modulus (K) describes volumetric elasticity, or the tendency of an object's volume to deform when under pressure; it is defined as volumetric stress over volumetric strain, and is the inverse of compressibility. The bulk modulus is an extension of Young's modulus to three dimensions. For an elastic, isotropic material, the modulus of elasticity E, bulk modulus K, and Poisson s ratio by are related E 3K 1 2 Uniaxial Loading and Deformation The deformation,, of an axially loaded member of original length L can be derived from Hooke s law. Tension loading is considered to be positive; compressive loading is negative. The sign of the deformation will be the same as the sign of the loading. L L E PL E This expression for axial deformation assumes that the linear strain is proportional to the normal stress E and that the cross-sectional area is constant. When an axial member has distinct sections differing in cross-sectional area or composition, superposition is used to calculate the total deformation as the sum of individual deformations. PL E P L E When one of the variables (e.g., ), varies continuously along the length, PdL E P dl E The new length of the member including the deformation is given by L f L The algebraic deformation must be observed. Elastic Strain Energy In Uniaxial Loading
Strain energy, also known as internal energy per unit volume stored in a deformed material. The strain energy is equivalent to the work done by the applied force. Simple work is calculated as the product of a force moving through a distance. Work = force x distance = FdL Work per volume = FdL L = d Work per unit volume corresponds to the area under the stress-strain curve. For an axially loaded member below the proportionality limit, the total strain energy is given by, The strain energy per unit volume is Three Dimensional Strains U u 1 2 P U L 2 P L 2E Hooke s law, previously defined for axial loads and for pure shear, can be derived for three-dimensional 2 2E Stress-strain relationships and written in terms of the three elastic constants, E, G, and equations can be used to find the stresses and strains on the differential element in Fig. (1). x y z 1 E 1 E 1 E x y z y z x z x y. The following xy xy G yz yz G zx zx G Stress-Strain curve
Some basic definitions Elastic Material when materials return to their original state when unloaded Homogenous- a material of uniform structure without any flaws Inhomogeneous material varies from point to point in structure Isotropic exhibit uniform properties through out in all direction. nisotropic Does not exhibit uniform properties throughout in all direction. Orthotropic different properties in different planes. Proof Stress The stress which when removed produces a permanent strain of 0.1 % of original gauge length. Necking takes place when load reduces Repeated loading and unloading will produce a yield point approaching the ultimate stress value but strain to failure will be much reduced. Ductility The ability of a material to be drawn out plastically. Malleability- Represents the ability of a material to allow permanent extension in all lateral direction under compressive loading. s the carbon content increases the % elongation and reduction of area decreases while the tensile strength and elastic limit increases. FOS- ratio of max stress to allowable working stress. It is 2.5 for static and 10 for stock load. Change in length due to increase in temperature = L.α.t Thermal strain ϵ = = αt and Thermal stress σ = E.α.t Toughness- The ability of a material to withstand the shock Creep- gradual increase of plastic strain in a material with time at constant load. Equivalent dia(d) for tapered bar of dia d and d2 d =
Compound ars: 1. =. W 2. = 3. Common extension, x = 4. Difference in free length = Thin-Walled Pressure Vessels: Cylindrical shells
F 0: (2 t x) p(2 r x ) 0 z 1 Hoop stress or circumferential stress = 1 pr t F 0: (2 rt) p(2 r ) 0 x 2 2 Longitudinal stress = 2 pr 2t max 2 pr 2t
1 max( in plane) 2 2 pr 4t Spherical shells F 0: (2 rt) p(2 r ) 0 x 2 2 Hoop stress = longitudinal stress = 1 2 pr 2t 1 max 2 1 pr 4t 1. If a tension test bar is found to taper uniformly from (D a) to (D + a) diameter, prove that the error involved in using the mean diameter to calculate the Young s Modulus is Solution: Let the two diameters be (D + a) and (D a). Let E be the Young s Modulus of elasticity. Let the extension of the member be. Then, =.
Cu Steel Cu (D + a) (D + a) P P L If the mean diameter D is adopted, let Then = be the computed Young s modulus. = Hence, percentage error in computing Young s modulus = = 100 = 100 = percent. 2. weight of 45 kn is hanging from three wires of equal length. The middle one is of steel and the two other wires are of copper. If the cross section of each wire is 322 sq. mm, find the load shared by each wire. Take = 207 N/ and = 124.N/. Solution: Let = load carried by the copper wire = load carried by the steel wire = stress in copper wire = stress in steel wire Then, + + = 45000 N or 2 + = 45000 lso strain in steel wire = strain in copper wire Hence = or = = = 06 ut 45000 = 2 + = 2 = 322 (2 + ) = 322(2 0.6 + ) = = 63.5 N/ = a = 63.5 322 = 20.447 kn = = 12.277 kn 45 kn
3. cylindrical shell 900 mm long, 150 mm internal diameter, having a thickness of metal as 8 mm, is filled with a fluid at atmospheric pressure. If an additional 20000 of fluid is pumped into the cylinder, find (i) the pressure exerted by the fluid on the cylinder and (ii) the hoop stress induced. Take E = 200 kn/ and = 0.3 Solution: Let the internal pressure be p. N/. Hoop stress = = 9.375p N/ Longitudinal stress = =4.6875p N/ Circumferential = = (9.375 0.3 4.6875) = 7.96875 Longitudinal strain = = (4.6875 0.3 9.375) = 17.8125 Increase in volume = V = 20000 1708125 900 = 20000 (i) p = 14.12 N/ (ii) = = 132.4 N/ 4. copper rod of 15 mm diameter, 800 mm long is heated through 50 C. (a) What is its extension when free to expand? (b) Suppose the expansion is prevention by gripping it at both ends, find the stress, its nature and the force applied by the grips when one grip yields back by 0.5 mm. per C Solution: (a) Cross-sectional area of the rod = Extension when the rod is free to expand = = 18.5 = 0.74 mm (b) When the grip yields by 0.5 mm, extension prevented = = Temperature strain = Temperature stress = = 37.5 N/ (comp.) Force applied = 37.5 = 6627 N sq.mm - 0.5 = 0.24 mm 5. steel band or a ring is shrunk on a tank of 1 metre diameter by raising the temperature of the ring through 60 C. ssuming the tank to be rigid, what should be the original inside diameter of the ring before heating? lso, calculate the circumferential stress in the ring when it cools back to the normal temperature on the tank. = 200 kn/ Solution: Let d mm be the original diameter at normal temperature and D mm, after being heated through 60 C. D should be, of course, equal to the diameter of the tank for slipping the ring on to it. Circumference of the ring after heating = D mm. Circumference of the ring at normal temperature = d mm The ring after having been slipped on the tank cannot contract to d and it cools down resulting in tensile stress in it. Contraction prevented = (D d) Temperature strain = = or
from which d = 999.4 mm circumferential temperature stress or stress due to prevention of contraction of the ring = TE = 60 = 12 N/ 6. vertical steam boiler with 2 metres internal diameter and 4 metres high is constructed with 2 cm thick plates for a working pressure of 1 MPa. The end plates are flat and not stayed. Calculate (1) the stress in the circumferential plates due to the pressure on the end plates, (2) the stress in the circumferential plates due to resisting the bursting effect and (3) the increase in length, diameter and volume. ssume the Poisson s ratio as 0.3 and E = 200 GPa. Solution: The stress in the plate due to the pressure to end plates will obviously be the longitudinal stress. = 2 = 50 MPa = 25 MPa Net longitudinal strain = - = = since = 0.3 = = Increase in length, Circumferential stain = - = - = 0.02 cm = = = = Increase in diameter, = d = 0.0425 cm = Increase in volume, V = 5,970 cu.cm. pr 5.97 litres 7. thin spherical shell 1.5 diameter with its wall of 1.25 cm thickness is filled with a fluid at atmospheric pressure. What intensity of pressure will be developed in it if 160 cu.cm more of fluid is pumped into it? lso, calculate the hoop stress at that pressure and the increase in diameter. Take E = 200 GPa and m = 10/3. Solution: t atmospheric pressure of fluid in the shell, there will not be any increase in its volume since the outside pressure too is atmospheric. ut, when 160 cu. cm of fluid is admitted into it forcibly, the sphere shall have to increase its volume by 160 cu. cm. increase in volume = 160 cu. cm. V = (4 / 3) = (4 / 3) cu. cm. = = increase in diameter = = 0.00453 cm
8. compound tube is made by shrinking a thin steel tube on a thin brass tube. and brass tubes, and and are the corresponding values of the Young s Modulus. Show that for any tensile load, the extension of the compound tube is equal to that of a single tube of the same length and total cross-sectional area but having a Young s Modulus of. Solution: = = strain Where and are stresses in steel and brass tubes respectively. + = P where P total load on the compound tube. From equations (i) and (ii);. + = P = P Extension of the compound tube = dl = extension of steel or brass tube = From equations (iii) and (iv); = (i). (ii). (iii).. (iv).. (v) Let E be the Young s modulus of a tube of area ( + ) carrying the same load and undergoing the same extension = equation (v) = (vi) = or E =. (vi) 9. composite bar consists of an aluminum section rigidly fastened between a bronze section and a steel section as shown in Fig. 1 8a. xial loads are applied at the positions indicated. Determine the stress in each section. ronze luminum Steel 4000 lb 9000 lb 2000 lb 1.3 ft 1.6 ft 1.7 ft (a) 4000 lb 4000 lb 9000 lb 4000 lb 9000 lb 2000 lb Figure 1 8 (b) Solution: To calculate the stresses, we must first determine the axial load in each section. The appropriate free-body diagrams are shown in Fig. 1 8b, from which we determine (compression), and (compression). The free-body diagrams have been drawn by
isolating the portion of the bar lying to the left of imaginary cutting planes. Identical results would be obtained if portions lying to the right of the cutting planes had been considered. The stresses in each section are (tension) ns. (compression) ns. (compression) ns. Note that neither the lengths of the sections nor the materials from which the sections are made affect the calculations of the stresses. s you can see from this example, the first step in calculating the stress in a member is to determine the internal force carried by the member. This determination is accomplished by the analysis of correctly drawn free-body diagrams. Note that in this example, it would have been easier to determine the load in the steel portion by taking the section lying to the right of the exploratory section in the steel. 3 m (1) C (2) D E 30 kn 70 kn F G H 4 panels at 4 m = 16 (a) D 3 4 E 3 m 4 3 C C CE 30 kn E 4 m 4 m (b (c) Figure 1 9 10. large pipe, called a penstock in hydraulic work, is 1.5 m in diameter. Here it is composed of wooden staves bound together by steel hoops, each 300 in cross-sectional area, and is used to conduct water from a reservoir to a powerhouse. If the maximum tensile stress permitted in the hoops is 130 MPa, what is the maximum spacing between hoops under a head off water of 30 m? (The mass density of water is 1000 kg/.) Solution: The pressure corresponding to a head of water of 30 m is given by If the maximum spacing between hoops is denoted by L, then, as shown in Fig. 1 22, each hoop must resist the bursting force on the length L. Since the tensile force in a hoop is given by, we obtain from the free-body diagram
L F = pdl L L Figure 1 22 Spacing of hoops in a penstock. Which gives m = 177 mm ns. 11. Compute the total elongation caused by an axial load of 100 kn applied to a flat bar 20 mm thick, tapering from a width of 120 mm to 40 mm in a length of 10 m as shown in Fig. 2 6. ssume. 20 mm 20 mm thick 60 mm P = 100 kn P = 100 kn Figure 2 6 Solution: Since the cross-sectional area is not constant, Eq. (2 4) does not apply directly. However, it may be used to find the elongation in a differential length for which the cross sectional area is constant. Then the total elongation is the sum of these infinitesimal elongations. t section, the half width (mm) at a distance (m) from the left end is found from geometry to be or nd the area at that section is 10 m mm t section, in a differential length, the elongation may be found from Eq. (2 4): from which the total elongation is mm ns.
12. steel rod 2.5 m long is secured between two walls. If the load on the rod is zero at C, compute the stress when the temperature drops to. The cross-sectional area of the rod is 1200 and. Solve, assuming (a) that the walls are rigid and (b) that the walls spring together a total distance of 0.500 mm as the temperature drops. Solution: Part a. Imagine the rod is disconnected from the right wall. Temperature deformations can then freely occur. temperature drop causes the contraction represented by in Fig. 2 12. To reattach the rod to the wall will evidently require a pull to produce the load deformation. From the sketch of deformations, we see that, or, in equivalent terms from which we have ns. Note that cancels out of the preceding equation, indicating that the stress is independent of the length of the rod. P Figure 2 12 Rigid walls. P Figure 2 13 Nonrigid walls. Yield Part b. When the walls spring together, Fig. 2 13 shows that the free temperature contraction is equal to the sum of the load deformation and the yield of the walls. Hence yield Replacing the deformations by equivalent terms, we obtain or yield from which we obtain ns. Notice that the yield of the walls reduces the stress considerably, and also that the length of the rod does not cancel out as in Part a. 13. 5 m high column is made of two 6 6 cm angle-iron sections, connected back-to-back. It carries an axial load of 50 kn. Estimate the maximum stress in the column, taking into account, the weight of the column. Specific weight of the iron is 8 gm/. If the crushing stress for the material of the column is 250 MPa, what factor of safety has been provided? Solution: Sectional plan of the column is shown below. = sq. cm. = 1.6875
6 6 3/4 6 Volume of the column = 5 = 8437.5 cu.cm Its weight = 8437.5 8gm = 67.5 kg = 662 N = 0.662 kn Total load on the bottom section W = (50 + 0.662) = 50.662 kn Maximum stress produced, = Pa This is the stress to which the column will be actually subjected and could be termed as the working stress. Factor of safety = = = 8.327 3/4 3/4 14. Two length of a tie bar, each of diameter D are connected by a pin joint. The end of one part is forked, in which is fitted the eye of the other and both are secured by a pin of diameter d as shown in the Figure. If and are tensile and shear stresses in the bars and the pin respectively, establish a relation between their diameters and stresses. ssume that both offer equal resistance. P D d Solution: Tensile strength of bar = tensile force acting. = F = The pin can fail in double shear and hence the shearing strength of pin = 2. (i).....(ii) (i) = (ii) (given) = 15. spherical pressure vessel of 600 mm internal diameter is made of 3 mm thick cold drawn sheet steel with static strength properties of = 440 MPa and = 370 MPa. Determine the maximum pressure for (a) static yielding, and (b) eventual fatigue failure, when the pressure fluctuates between 0 and yielding is also not permitted, (Do not apply any factor of safety) (GTE 1997) Solution: = where P = internal pressure, when
D = diameter, T = thickness. = = = Pa = 7.4 MPa When pressure fluctuates between 0 and = s no yielding is also allowed. Soderberg s equation is to be used. i.e., + = Solving, P = 5051 + = 1 Pa = 5.51 Mpa Example 22.1. The diameters of the brass and steel segments of the axially loaded bar shown in Fig. 22.1 are 30 mm and 12 mm respectively. The diameter of the hollow section of the brass segment is 20 mm. Determine: (i) The displacement of the free end; (ii) The maximum normal stress in the steel and brass Take and. Solution. Fig. 22.1 10 kn 12 mm 5 KN 30 mm 20 mm Steel rass C D 0.15 m 0.2 m Fig. 22.1 0.125 m (i) The maximum normal stress in steel and brass: (ii) The displacement of the free end: The displacement of the free end 178 m or 0.09178 mm (ns.) Example 22.2. beam hinged at is loaded at as shown in Fig. 22.2. It is supported from the roof by a 2.4 m long vertical steel bar CD which is 3.5 cm square for the first 1.8 m length and 2.5 m square for the remaining length. efore the load is applied the beam hangs horizontally. Determine:
(i) The maximum stress in the steel bar CD; (ii) The total elongation of the bar. Take. Solution. Given: Let P = The pull in the bar CD. Then, taking moments about, we get P = 90 kn (i) The maximum stress in the steel bar CD, : The stress shall be maximum in the portion DE of the steel bar CD. 3.5 cm C 1 1.8 cm Roof E (ii) The total elongation of the bar CD, 2.5 cm 2 0.6 cm D 0.3 m 60 kn Fig. 22.2 Example 22.6. composite bar made up of a aluminium bar and steel bar is firmly held between two unyielding supports as shown in Fig. 22.7 10 luminum bar 200 kn 15 Steel bar C 0.6 m 10 15 Fig. 22.7 n axial load of 20 kn is applied at at 50. Find the stresses in each material when the temperature is 100. Take E for aluminium and steel as 70 GN/ and 210 GN/ respectively. Coefficient of expansion for aluminium and steel are 24 per and 11.8 per respectively. Solution. Given: = 15 cm = 0.15 m; = 50, = 100 ; Load, P = 20kN = 70 GN/ ; = 210 GN/ ; = 24 per, = 11.8 per. Stresses; ; Out of the load of 200 kn (P) applied at, let kn be taken up by and (20 - ) kn by C. Since the supports are rigid, Elongation of = contraction of C.
Stress in aluminium, 50 MN/ (Tensile) Stress in steel, = 100 MN/ (Compressive) These are the stresses in the two materials (luminium and steel) at 50. Now let the temperature be raised to 100. In order to determine the stresses due to rise of temperature, assume that the support at C is removed and expansion is allowed free. Rise of temperature = Expansion of m... (i) Expansion of C m. Let a load be applied at C which causes a total contraction equal to the total expansion and let C be attached to rigid supports. If this load causes stress σ N/ in C, its value must be 15 σ and hence stress in must be = 1.5 σ N/ Total confraction cursed by the load From eqns. (i) ans (ii), we have 0.6 σ = 210 208500 = 43785000 σ = 72.97 N/ or 72.97 MN/ (Compressive) t 100, Stress in aluminium, = - + 1.5 72.97 59.45 MN/ (comp.) ns.) Stress in steel, = 172.97 MN/ (comp.) (ns.) QUESTIONS ND NSWERS 2. small element at the critical section of a component is in a bi-axial state of stress with the two principal stresses being 360 MPa and 140 MPa. The maximum working stress according to Distortion Energy Theory is () 220 MPa () 110 MPa (C) 314 MPa (D) 330 MPa (GTE 1997) ns: (C) ccording to distorsion energy theorem, + - Working stress = = 314.32 MPa
SHERING FORCE ND ENDING MOMENT Sign Conventions The shear force, V at a section of a beam is the sum of all vertical forces acting on the beam between that section and any one of its ends. It has units of Newtons, pounds, kips, etc. Shear force is not the same as shear stress, since the area of the object is not considered. The direction (i.e., to the left or right of the section) in which the summation proceeds is not important. Since the values of shear will differ only in sign for summation to the left and right ends, the direction that results in the fewest calculations should be selected. V F i section _ to one _ end Shear is positive when there is a net upward force to the left of a section, and it is negative when there is a net downward force to the left of the section. Figure. Shear force sign conventions The bending moment, M, at a section of a beam is the algebraic sum of all moments and couples located between the section and any one of its ends. M F d i sec tion to one _ end _ i C i sec tion to one _ end _ ending moments in a beam are positive when the upper surface of the beam is in compression and the lower surface is in tension. Positive moments cause lengthening of the lower surface and shortening of the upper surface. useful image with which to remember this convention is to imagine the beam smiling when the moment is positive. Figure. ending moment sign conventions Shear force and ending moment Relationships The change in magnitude of the shear at any point is equal to the integral of the load function, w(x), or the area under the load diagram up to that point.
V 2 V 1 x2 x1 w x dx w x dv x dx The change in magnitude of the moment at any point is equal to the integral of the shear function, or the area under the shear diagram up to that point. M 2 M 1 x2 x1 V x dx V x dm x dx Shear Force and ending Moment Diagrams oth shear force and bending moment can be described mathematically for simple loadings by the preceding equations, but the formulas become discontinuous as the loadings become more complex. It is more convenient to describe complex shear and moment functions graphically. Graphs of shear and moment as functions of position along the beam are known as shear force and bending moment diagrams. The following guidelines and conventions should be observed when constructing a shear diagram. The shear at any section is equal to the sum of the loads and reactions from the section to the left end. The magnitude of the shear at any section is equal to the slope of the moment function at that section. Loads and reactions acting upward are positive The shear diagram is straight and sloping for uniformly distributed loads. The shear diagram is straight and horizontal between concentrated loads. The shear is undefined at points of concentrated loads. The following guidelines and conventions should be observed when constructing a bending moment diagram. y convention, the moment diagram is drawn on the compression side of the beam. The moment at any section is equal to the sum of the moments and couples from the section to the left end. The change in magnitude of the moment at any section is the integral of the shear diagram, or the area under the shear diagram. concentrated moment will produce a jump or discontinuity in the moment diagram. The maximum or minimum moment occurs when the shear is either zero or passes through zero. The moment diagram is parabolic and is curved downward for downward uniformly distributed loads. PROPPED EMS & FIXED EMS 1. Static Indeterminacy: If the equations of static equilibrium are insufficient to determine the internal forces, the beams are called statically indeterminate. Equations of static equilibrium are: and
NOTE: shall be considered only if inclined horizontal loads exist. 2. Important cases of static indeterminacy: (a). Propped cantilever: Unknowns are: and Nos. Equations of: Nos. Point of contraflexure static indeterminacy = 3.2 = 1 To completely analyses, additional equation is considered by equating upward deflection due to prop and downward deflection due to external load at prop. (b). Fixed eams: Static Indeterminacy Redundancy = 4 2 = 2. (Or) Point of contraflexure (c). Continuous eams: Static Indeterminacy = 1 C 3. nalysis of propped beams: (i) Propped cantilever with through out. C = deflection due to Point to contraflexure 3L/4 3L/4 deflection = Maximum ve.m = Support moment = deflection (hogging).m.d (ii) Propped Cantilever Carrying Central Point Load: (Hogging) Max = M = 5wL/32 (sagging) Point of contra flexure: at 3L/11 from fixed support. 11W/16 8L/11 3L/11 5W/16 (iii) Simply supported beam with prop at centre with : or w/m (Hogging) L/2 L/2 4. nalysis of fixed beams: (a) Support Moments (by Moment rea Method): (i) rea of free and fixed.m.d s are numerically equal. s = area of free MD C
f = area of fixed MD (ii) Moment of area of M/EI diagram about any support is zero. (iii) For beam of constant EI, the C.E. of free.m.d and C.G of fixed.m.d will be equidistant from the same support. (b) Standard Cases: (eams of uniform section) (i) Fixed beam with central point load. W L/2 L/2 Ymax = Yc = ¼ max ve.m = (sag) max ve.m = (hog) free.m.d fixed WL/4 (hog).m.d No. of point of contra flexure = 2 Location of point of contra flexure (or) Inflection = L/4 from supports WL/8 (ii) Fixed beam with : W/2 S.F.D Ymax = Yc = max ve.m = max ve.m = Points of contra flexure: @ 0.212 from either supports. W/2.M.D 0.212 L S.F.D (iii) Eccentric Point Load: (hog) (hog) a b.m.d
al/(3a + (iv) couple {hog} {Sag} a L (v) (hog) (hog) 5. Sinking of Supports: Let sinking of = It induces support moments (hog) (sag) L 6. Rotation of supports: Support anti clockwise rotation θ radians. (hogging) (sagging) Shear Force and ending Moment 1. Types of supports: a) Movable Hinged support (or) Simple support (or) Roller support: - Support of the Fig. Shown a side - No bending moment - Free to rotate - Horizontal Displacement possible - Only vertical reaction or reaction normal to plane of rolling - Number of independent reaction components 1. b) Immovable hinged support: - Support of the Fig. shown above. - Free to rotate No translation - oth vertical and horizontal reactions. - No bending moment - Number of independent reaction components 2.
c) Fixed support: - Support C of the Fig. shown a side. C - Neither rotation nor translation - Vertical and horizontal reactions and.m. - Number of independent reaction components 3. D 2. a) Shear force (F): lgebraic sum of all transverse forces either to the left or right hand side of a section b) ending moment (M): lgebraic sum of moments of all transverse forces either to the left or right of a section. Sign convention S.F Left Position Right S.F Left Negative Right.M Hogging -Ve Sagging +Ve W/m 3. Relation between load, S.F. and.m.: a) or b) or F + F F For maximum.m., M x (M+ M) 4. Conclusions: i) Rate of change of S.F. (and hence slope of S.F.D.) is equal to intensity of loading ii) The rate of change of.m (and hence slope of.m.d) is equal to shear force. iii) The change of.m. from O to x is proportional to the area of S.F.D. from O to x iv).m is maximum; when S.F., is zero or changes sign v) Point of contra flexure: Point where.m. changes sign and is equal to zero. 5. Variation of S.F and.m for different loadings on spans of beams: S. No Type of loading Variation of S.F. Variation of.m 1 Point load Rectangle Inclined line for linear 2 U.D.I. Linear Square Parabola 3 U.V.I. or Triangular Parabolic Cubic Parabola 4 Parabolic Cubic Fourth degree polynomial 5 ending couple No shear variation vertical step at the point of application 6. Inclined loading: If the external load is not at right angles to the axis of the beam, the loading can be resolved axially and transversely to the beam Transverse: Components xial: Component produces.m. and S.F. produces pull or push 7. Horizontal thrust diagram: Diagram showing the variation of axial Thrust over a span 2t
8. Internal hinge: ending moment is zero H.T.D 9. Link: Only vertical force: Can not resist horizontal force H and.m. Point of contra flexures: The point or the points on the beam where the bending moment (.M) changes sign is defined as the point of contra flexure or point of inflexion. 1. Draw the shear and bending moment diagrams for the beam. The distributed load of 40 lb/in. extends over 12 in. of the beam, from to C, and the 400 lb load is applied at E. 40 lb/in. C D E 400 lb 12 in. 6 in. 4 in. 10 in. 32 in. Solution: Taking entire beam as free body, calculate reactions at and.
Determine equivalent internal force-couple systems at sections cut with in segments C, CD, and D. Plot results. 40 lb/in. C D E 400 lb 12 in. 6 in. 4 in. 10 in. 32 in. 480 lb C D E 32 in. 400 lb 6 in. 16 in 10 in. Solution: Taking entire beam as a free- body, calculate reactions at and.
= 515 lb Note: The 400 lb load at E may be a replaced by 400 lb force and 1600 lb-in. couple at D. 12 in 40 lb/in. 6 in. 160 lb.in. 14 in 1 C 2 D 515 lb 400 lb 365 lb 40 x M 515 lb V 480 lb M V 515 lb x Evaluate equivalent internal force-couple systems at sections cut within segments C, CD, and D. Form to C : V = 515 40 x From C to D:
V = 35 lb M = 12 in 40 lb/in. 6 in. 1600 lb.in. 14 in 515 lb 1 C 2 D 400 lb 3 365 lb 480 lb 1600 lb.in. M 515 lb 400 lb V Evaluate equivalent internal force-couple systems at sections cut within segments C, CD and D. From D to V = -365 lb M =
12 in 6 in. 14 in 40 lb/in. 515 lb 1600 lb.in. 1 C 2 D 3 400 lb 365lb V 10 in 515 lb 35lb 18 in. 32 in. 12 in. - 365lb M 3300 lb.in 3510 lb.in 5110 lb.in Plot results. From to C : V = 515 40 x M = From C to D : V = 35 lb From D to : V =
- - - - The change in moment between and is equal to area under shear curve between the points. The parabolic shear curve results in a cubic moment curve. at, The change in moment between and C is equal to area under shear curve between points. The constant shear curve results in a linear moment curve. 401. Write shear and moment equations for the beam loaded as shown in Fig. 4 10a and sketch the shear and moment diagrams. Solution:
egin by computing the reactions. pplying gives kn, and yields kn. check of these values is given by. The sections in the beam at which the loading conditions change are called change of load points and are designated by the letters,, C, and D. If a section is taken through the beam anywhere between and, the external loads on it appears as in Fig. 4 8. pplying the definitions of vertical shear and bending moment, and noting that they only to external loads, we obtain kn (a) kn.m (b) These equations are valid only for values of between 0 and 5 m, that is, between points and. To obtain shear and moment equations between and C, take another exploratory section anywhere between and C. Note that the location of section is still defined in terms of as measured from the left end of the beam, although now ranges between the limits of 5 m and 10 m. The effects of the external forces on this section are determined by applying the definitions of shear and moment to Fig. 4 9. kn (c) kn.m (d) The shear and moment equations for segment are obtained similarly by passing a section anywhere between and. The external loads acting on the beam to the left of this section are shown in Fig. 4 11, from which we obtain kn (e) kn.m (f) simpler method of obtaining is to consider the forces lying to the right of section as shown in Fig. 4 12. Noting that downward forces produce negative bending moment, we also obtain kn.m Figure 4 8 Figure 4
(a) Load diagram V (kn) (m) (b) Shear diagram M (kn) 99.23 65 (m) (c) Moment diagram (d) Elastic curve Figure 4 10 2.5 m Figure 4 12 Figure 4 11 Summarizing, we have computed by considering only the external forces lying to the left of any exploratory section, whereas may be computed by taking moments about the exploratory section caused by the external loads that lie either to the left or to the right of the section. We have been careful to assign plus signs to and caused by upward acting loads, and minus signs to and caused by downward acting loads. We shall be consistent in assigning a plus sign to any upward quantity and a minus sign to any quantity associated with the word down or its equivalent.
Note further that Figs. 4 8, 4 9, 4 11, and 4 12 have been used only for explanation; you will soon learn to visualize such diagrams directly from the original beam loading. summary of the principles presented here and in rt. 4 2 suggests the following procedure for the construction of shear and moment diagrams: 1. Compute the reactions. 2. Compute values of shear at the change of load points, using either or. 3. Sketch the shear diagram, determining the shape from Eq. (4 5); that is, the intensity of the load ordinate equals the slope at the corresponding ordinate of the shear diagram. 4. Locate the points of zero shear. 5. Compute values of bending moment at the change of load points and at the point of zero shear, using either or, whichever is more convenient. 6. Sketch the moment diagram through the ordinates of the bending moments computed in step 5. The shape of the diagram is determined from Eq. (4 6); that is, the intensity of the shear ordinate equals the slope at the corresponding ordinate of the moment diagram. V 14 2 Shear diagram M 40 36 Moment diagram Figure 4 12 Load, shear, and moment diagrams.
Example 22.15. Draw the shear force and the bending moment diagrams for the simply supported beam loaded with a concentrated force, P = 100 kn and a concentrated moment, = 200 kn/m as shown in Fig. 22.14 P = 100 kn C 200 knm 2 m Fig. 22.14 2 m Solution. Determination of reactions and : Taking moments about the point, we get C 100 kn = 0 lso, S. F. calculation: 200 knm 100 kn 2 m (a) Loaded beam 2 m S.F. Diagram is shown in Fig. 22.15 (b).m. calculations: 100 kn 100 kn (b) S.F. Diagram 200 kn m.m. Diagram is shown in Fig. 22.15. (c) (c).m. Diagram Fig. 22.15
TORSION OF SHFTS 1. Torsion: If moment is applied in a plane perpendicular to the longitudinal axis of the beam (or) shaft, it will be subjected to Torsion. E.g: 1. Shaft Transmitting Torque or power. 2. L beams 3. Portico beams 4. Curved beams 5. Close coiled springs. 2. Torsion formula: T Where T = Torque applied Θ = Twist of cross section = Maximum shear stress due to torsion R = Radius of shaft L = Length of shaft T J = Polar moment of inertia = = for Hollow circular shaft for solid circular shaft ssumptions: 1. Plane normal sections of shaft remain plane after twisting. 2. Torsion is uniform along the shaft 3. Material of the shaft is homogeneous, and isotropic. 4. Radii remain straight after torsion. 5. Stress is proportional to strain i.e., all the stresses are with in elastic limit. Note: 1. The stress setup at any point in a cross section is one of pure shear or simple shear. 2. The longitudinal axis is neutral axis. 3. The shear stress will vary linearly from zero at the centre to maximum at the outer surface (any point on periphery) Distribution along vertical 3. (a) Torsional Section Modulus:
s the value of Torsional modulus increases, the Torsional strength increases. For E.g.: hollow circular shaft compared to that of a solid shaft of same area, will have more Torsional strength. For a solid circular shaft, For a hollow circular shaft, = Outer diameter, = inner diameter. (b) Torsional Rigidity: CJ Unit: kg. or The torsional which produces unit twist per unit length. (c) ngle of Twist 4. Power Transmitted by a Shaft : In SI system : Power (P) is measured in watts (W) Metric System where T = verage Torque in kn.m N = rpm 1 watt = 1 Joule / sec = 1 N. m / sec 1 metric H.P = 746 watts 0.75 kw H.P Where T = average torque in kg.m 5. Design of Shaft: To be safe against maximum permissible shear stress. Diameter of shaft 6. (a) Composite Shafts: When two dissimilar shafts are connected together to form one shaft, the shaft is known as composite shaft. (b) Shafts in Series: If the driving torque is applied at one end, and the resisting torque at the other end, the shafts are said to have been connected in series. T For such shaft, i) both the parts carry some Torque i.e., ii) Total angle of twist at fixed and is sum of separate angles of twist of two shafts. (c) Shaft in Parallel: OR
If the Torque T is applied at the junction of two shafts and resisting Torque at their remote ends, the shafts are said to be connected in parallel. For such a case, ; T = If both the shafts are of same material 7. Combined bending and Torsion: a) Let a shaft be subjected to a bending moment of M and twisting moment T at a sector. Now bending stress, Shear stress, Principal stresses are, b) Equivalent Torque: It is the twisting moment, which acting along produce the maximum shear stress due to combined bending and Torsion. c) Equivalent ending Moment: The bending moment to produce the maximum bending stress equal to greater principle stress. 8. Comparison of Hollow and Solid Shafts: (a) When the areas of solid and hollow sections are equal, For E.g: If K (b) When radius of solid shaft is equal to external radius of hollow shaft, (c) The ratio of the weight of a hollow shaft, and solid shaft of equally strong is 9. Strain Energy due to Torsion: a) In a solid shaft :- b) For a hollow shaft D = ext dia, d = internal dia c) The ratio of strain energy stored in a solid shaft and hollow shaft of external dia D and internal dia d having equal volumes is = )
16. shaft is transmitting 97.5 kw at 180 rpm. If the allowable shear stress in the material is 60 N/, find the suitable diameter for the shaft. The shaft is not to twist more than 1 in a length of 3 meters. Take c = 80 N/ Solution: Let T = torque transmitted Power = 97.5 T = 5172 Nm = 5712 N- mm Let d be the diameter of the shaft. Then, T = 5172 = = 439 d = 76 mm Considering stiffness of the shaft i.e., =
= 113191150 d = 103(say 105 mm) We adopt the greater value. i.e., 105 mm. 302. Two solid shafts of different materials are rigidly fastened together and attached to rigid supports as shown in Fig. 3 4. The aluminum segment is 3 in. in diameter, and. The steel segment has a diameter of 2 in. and. The torque, T = 10 kip.in., is applied at the junction of two segments. Compute the maximum shearing stress developed in the assembly. 6 ft 3 ft luminum 3 in. dia. Steel T = 10 kip. in. 2 in. dia. Figure 3 4 Statically indeterminate composite shaft. Solution: This problem is statically indeterminate in that we do not know how the applied torque is apportioned to each segment. The procedure we follow is exactly the same as that discussed in rt. 2 5 for statically indeterminate axially loaded members. pplying the conditions of static equilibrium and of geometric compatibility, we obtain first: (a) nother relation between and is obtained from the condition that each segment has the same angular deformation, so that. pplying Eq. (3 1) gives from which Solving Eqs. (a) and (b), we obtain: and pplying the torsion formula, we find the stresses to be ns. (b) ns. 342. load is supported by two steel springs arranged in series as shown in Fig. 3 16. The upper spring has 20 turns of 20-mm-diameter wire on a mean diameter of 150 mm. The lower spring consists of 15 turns of 10-mm-diameter wire on a mean diameter of 130 mm. Determine the maximum shearing stress in each spring if the total deflection is 80 mm and.
P Figure 3 16 Solution: The total deflection is the sum of the deflection in each spring. y applying Eq. (3 the load to be 11), we find Knowing, we can now find the stresses. For the upper spring,. pplying Wahl s formula, Eq. (3 10), we obtain Max. Max. MPa ns. Similarly, for the lower spring where and we find Max. ns. If we had used Eq. (3 9) to compute these maximum shearing stresses, the results would have been 11.4 MPa in the upper spring and 76.7 MPa in the lower spring. Thus the approximate formula gives results that are 10.2% and 6.35% lower than the more precise Wahl formula. Example 22.37. solid shaft in a rolling mill transmits 20 kw at 2 Hz. Determine the diameter of the shaft if the shearing stress is not to exceed 40 MPa and angle of twist is limited to 6 in a length of 3 m. Use C = 83 GPa. Solution. Given: Diameter of the shaft D: Let ω be the angular velocity of the shaft is rad/sec. We know,. lso, power transmitted nu the shaft, Consider shear stress:
Consider angle of twist: Hence D = 58.7 mm (larger of the two values) (ns.) Example 22.38. While transmitting power of a steam turbine, the angle of twist of a rotating shaft was measured and was found to be 1.2 degrees over a length of 6 metres. The external and internal diameters of the shaft are 250 mm and 170 mm respectively. The rotational speed of the shaft is 250 revolutions per minute. The shear modulus, C is 80 GPa. Determine the power being transmitted by the shaft and the maximum shear stress developed in it. Solution. Given: (i) Maximum shear stress developed : Using the relation, (ii) Power transmitted by the shaft; P: Let T be the torque transmitted by the shaft. We know that, N/mm or 84178.4 Nm Now, Power transmitted by the shaft, Example 22.39. hollow shaft is 50 mm outside diameter and 30 mm internal diameter. n applied torque of 1.5 Nm is found to produce an angular twist of 0.4, measured on a length of 0.2 m of the shaft. Calculate the value of the modulus of rigidity. Calculate also the maximum power which could be transmitted by the shaft at 2000 r.p.m. if the maximum allowable shearing stress is 70 MN/. Solution. Given: Modulus of rigidity, C: We know that, Maximum power transmitted P:
Now, Hence, maximum power transmitted under the given conditions, Example 22.40. hollow steel shaft; 80 mm internal diameter and 120 mm external diameter is to be replaced by a solid alloy shaft. If the polar modulus has the same value for both, calculate the diameter of the latter and the ratio of torsional rigidities. The shear modulus of elasticity for steel is 2.4 times that of the alloy. Solution. Let, ( = polar modulus of the steel shaft section = polar modulus of alloy shaft section, = outer dia of steel shaft, = inner dia of steel shaft, and = dia. of the alloy shaft. Diameter of the alloy shaft : Now, (Given) Substituting the values, we get Ratio of the torsional rigidities: Example 22.41. Determine the shaft diameter and bolt size for a marine flange-coupling transmitting 3.75 MW at 150 r.p.m. The allowable shear stress in the shaft and bolts may be taken as 50 MPa. The number of bolts may be taken as 10 and bolt pitch circle diameter as 1.6 times the shaft diameter. Solution. Given: Shaft diameter, D: lso,
olt size, : olt pitch circle diameter, Now,
DEFLECTION OF EMS The curvature of a beam caused by a bending moment is given by Eq. (1), where is the radius of curvature, c is the largest distance from the neutral axis of the beam, and max is the maximum longitudinal normal strain in the beam. 2 1 M d y d ------- (1) mac 2 c EI dx dx max c ------- (2) Using the preceding relationships, the deflection and slope of a loaded beam are related to the moment M(x), shear V(x), and load w(x) by Eqs. (3) through (7). y deflection ------- (3) dy ------- (4) y ' slope dx 2 d y dx '' y 2 3 d y dx ''' y 3 4 d y dx '''' y 4 M x EI V x EI w x EI ------- (5) ------- (6) ------- (7) If the moment function, M(x), is known for a section of the beam, the deflection at any point on that section can be found from Eq. (8). The constants of integration are determined from the beam boundary conditions in the table shown below. EIy M x dx ------- (8) Table. eam oundary Conditions End condition y y y V M Simple Support 0 0 uilt-in Support 0 0 Free end 0 0 0 Hinge 0 When multiple loads act simultaneously on a beam, all of the loads contribute to deflection. The principle of superposition permits the deflections at a point to be calculated as the sum of the deflections from each individual load acting singly. Superposition can also be used to calculate the shear and moment at a point and to draw the shear and moment diagrams. This principle is valid as long as the normal stress and strain are related by the modulus of elasticity, E. Generally this is true when the deflections are not excessive and all stresses are kept less than the yield point of the beam material. Points to be remembered
1. Relation between curvature, slope and deflection: a) Curvature - - - - - for ve.m or EI / = M - - - - - for.m is + ve b) Slope θ = dy / dx radians EIθ = EI. dy / dx = c) Deflection = y, EIy = d) EI y / = dm / dx = + F e) EI y / = df / dx = + W f) If a beam is subjected to pure bending, it bends into an arc of a circle. δ (2R δ) = L/2. L/2 = /4 E radians D 2. Methods of determining slope and deflection: a) Double integration method *Not suitable for objective type questions. b) rea moment method: - for cantilever, slopes and deflections can be determined very quickly. c) Conjugate beam method: - very much suitable for beams of varying sections, subjected to couples, for cantilevers of S. S. eams. Student is advised to understand this method properly d) Macaulay s method: - lso successive integration method. Note: In double integration of Macaulay s method two constants of integration and will be obtained. These are determined using end conditions. 3. Mohr s Theorem s : Moment rea Method: Theorem 1: The angle between tangents drawn at any two points on the deflected curve, is equal to the area of M / EI diagram between the two points. i.e., θ = area of M / EI diagram.. = area of.m.d. Theorem 2 : The intercept on a vertical line made by two tangents drawn at the two points on the deflected curve, is equal to the moment of M / EI diagram between the two points about the vertical line. = distance of C.G. of.m.d. E.G: (Suitable for cantilevers) from objective point of view. a) Step 1: To determine slope and deflection at any point say. Step 2 : Draw (MD) / (EI) i.e., M / EI Step 3 : Slope = area of (M / EI) diagram between fixed end point under consideration. L Step 4 : Deflection / EI, =.M.D area between fixed end and point under consideration. = distance of C.G. of M/EI from point under consideration. L M
4. Maxwell s Law of Reciprocal Deflections: Consider cantilever beam. Let C be an intermediate point. Then the deflection at C due to a point load P at say, is equal to deflection at due to a point load P at C i.e., C Slope and deflection of beams SL No. 1 Type of Loading
W Maximum ending Moment Slope Maximum Deflection SL No. 2 Type of Loading W a C b Maximum ending Moment Slope Maximum Deflection SL No. 3 Type of Loading
w/unit run Maximum ending Moment where W = ( total load on the Slope cantilever) Maximum Deflection SL No. 4 Type of Loading w/unit run C - a Maximum ending Moment Slope where W = Maximum Deflection SL No. 5 Type of Loading
w/unit run a Maximum ending Moment Slope Maximum Deflection SL No. 6 Type of Loading Maximum ending Moment Slope Maximum Deflection SL No.7 Type of Loading
run Maximum ending Moment Slope Maximum Deflection SL No. 8 Type of Loading W C Maximum ending Moment Slope Maximum Deflection SL No.9 Type of Loading
W C Maximum ending Moment Slope Maximum Deflection SL No. 10 Type of Loading w/unit run Maximum ending Moment where W = ( total load on the beam ) Slope Maximum Deflection SL No.11 Type of Loading
w/unit run C Maximum ending Moment Slope Maximum Deflection ( at x = 0 519 and ) SL No. 12 Type of Loading w/unit run Maximum ending Moment Slope Maximum Deflection Sign conventions used : Slope: Clockwise Counter- clockwise Deflection : upwards
Downward 49. strut of length has its ends built into a material which exerts a constraining couple equal to k times the angular rotation in radians. Show that the buckling load is given by the equation tan Solution: Y + )y = where = - M - Py y = sin mx + cos mx -
P y P M - k x t x = 0, y = 0; 0 = = = m cos mx - m sin mx t x = 0, = - ; - = m = - t x =, = 0 0 = m cos - m sin = tan (i.e.,) = tan tan = -. 602. Find the value of at the position midway between the supports and at the overhanging end for the beam shown in Fig. 6 5. y 1 m 3 m 2 m 2 m 600 N C D E Figure 6 5 Solution: This is the same beam for which we determined the general moment equation on page 185. pplying the differential equation of the elastic curve, and integrating twice, we obtain To determine, we note that at, which gives. Note that we ignore the negative terms in the pointed brackets. Next, we use the condition that at the right support N.m where. This gives or N. N. N.
Finally, to obtain the midspan deflection, we substitute in the deflection equations for segment obtained by ignoring negative values of the bracketed terms and. We obtain N. ns. Example 22.27. Determine the slope and deflection at the end of the prismatic cantilever beam when it is loaded as shown in Fig. 22.33, knowing that the flexural rigidity of the beam is. 50kN 3 m Fig. 22.33 90kN Solution. Refer Fig. 22.33. Consider a section XX at a distance x from the end. ending moment at XX, X 50kN 3 m 90kN Fig. 22.34 Integrating, we get when Hence slope equation becomes, Slope at (at x = 3 m) (where = constant of integration) (i) (Since EI = 104 kn/ ) Deflection at, : Integrating eqn. (i), we get When Hence
Example 22.30. tube 40 mm outside diameter, 5 mm thick and 1.5 m long simply supported at 125 mm from each end carries a concentrated load of 1 kn at each extreme end. (i) Neglecting the weight of the tube, sketch the shearing force and bending moment diagrams; (ii) Calculate the radius of curvature and deflection at mid-span. Take the modulus of elasticity of the material as 208 GN/. Solution. Given GN/ N/ (i) S.F. diagram and.m. diagram are shown respectively in Fig. 22.38 (b, c) (ii) Radius of coordinate R. s per bending equation: or Here, Substituting the values in eqn. (i), we get Deflection at mid-span: W W Integrating, we get D a C a = W = W (a) eam W (b) S.F. Diagram W Integrating again, we get When (c).m. Diagram Fig. 22.38 t mid-span, i.e., Hence deflection at mid-span = 1.366 mm (Upwards) (ns.)
Example 22.31. beam 4m long is freely supported at the ends. It carries concentrated loads of 20 kn each at points 1m from the ends. Calculate the maximum slope and deflection of the beam and slope and deflection under each load. Take EI = 13000. Solution. Refer Fig. 22.39. Let and be the reactions at the left and right hand supports respectively. Since the loads are placed symmetrically on the beam, 20 kn 20 kn lm lm 2m lm E D Elastic curve = 20 kn 4m Fig. 22.39 = 20 kn pply Macaulay s method in this problem. Consider an imaginary section XX at a distance x from in the segment D. Integrating the above expression, we have Integrating again, we get When When From eqn. (iv), we have (i) (ii) Substituting for constants and, the slope and deflection equations reduce to Maximum slope: Maximum slope will occur at the supporting ends and. Consider the slope equation, Similarly slope at Maximum deflection, :
For obtaining maximum deflection, we should first determine the position where slope is zero. Considering the slope equation we have: [Since the term 10 will be negative for the segment CD and hence neglected] or from. Now in order to determine maximum deflection, put x = 2m in the deflection equation Hence = 2.82 mm (downwards) Slope under the load 20 kn at C, : Substituting x = 1 m in the slope equation, we get Slope under the load 20 kn at D, : Substituting x = 3 m in the slope equation, we get Deflection under the load 20 kn at C : Substituting x = 1, m in the deflection equation, we get Hence 2.05 mm (downwards) (ns.) Deflection under the load 20 kn t D, : Hence = 2.05 mm (downwards) (ns.) Example 22.50. simply supported beam is subjected to a single force P at a distance b from one of the supports. Obtain the expression for the deflection under the load using castigliano s theorem. How do you calculate deflection at the mid-point of the beam? Solution. Let load P acts at a distance b from the support, and l be the total length of the beam. Reaction at, Reaction at,, and
P C Fig. 22.49 Strain energy stored by beam, U = Strain energy stored by C ( + strain energy stored by C ( ) Deflection under the load Deflection at the mid-span of the beam can be found by Macaulay s method. y Macaulay s method, deflection at any section is given by (ns.) Where y is deflection at any distance x from the support. t at mid-span,
MOHR S CIRCLE Thin-Walled Pressure Vessels: Cylindrical shells 1 Hoop stress or circumferential stress = pr t 2 Longitudinal stress = pr 2t max 2 pr 2t 1 max( in plane) 2 2 pr 4t Spherical shells
1 2 Hoop stress = longitudinal stress = pr 2t 1 max 2 1 pr 4t 44. t a certain point a material is subjected to the following strains: ; ;.