MAHALAKSHMI ENGINEERING COLLEGE

MAHALAKSHMI ENGINEERING COLLEGE TIRUCHIRAALLI - 6113. QUESTION WITH ANSWERS DEARTMENT : CIVIL SEMESTER: V SUB.CODE/ NAME: CE 5 / Strength of Materials UNIT 3 COULMNS ART - A ( marks) 1. Define columns If the member of the structure is vertical and both of its ends are fixed rigidly while subjected to axial compressive load, the member is known as column. Example: A vertical pillar between the roof and floor.. Define struts. If the member of the structure is not vertical and one (or) both of its ends is Linged (or) pin jointed, the bar is known as strut. Example: Connecting rods, piston rods etc, 3. Mention the stresses which are responsible for column failure. i. Direct compressive stresses ii. Buckling stresses iii. Combined of direct compressive and buckling stresses. State the assumptions made in the Euler s column theory.. (AUC Nov/Dec 011) 1. The column is initially perfectly straight and the load is applied axially.. The cross-section of the column is uniform throughout its length. 3. The column material is perfectly elastic, homogeneous and isotropic and obeys Hooke s law.. The self weight of column is negligible. 5. What are the important end conditions of columns? 1. Both the ends of the column are hinged (or pinned). One end is fixed and the other end is free. 3. Both the ends of the column are fixed.. One end is fixed and the other is pinned. 6. Write the expression for crippling load when the both ends of the column are hinged. l = Crippling load E = Young s Modulus I = Moment of inertia l = Length of column MUTHUKUMAR./Asst.rof./CIVIL-III age 1

7. Write the expression for buckling load (or) Crippling load when both ends of the column are fixed? L = Crippling load E = Young s Modulus I = Moment of inertia l = Length of column 8. Write the expression for crippling load when column with one end fixed and other end linged. l = Crippling load E = Young s Modulus I = Moment of inertia l = Length of column 9. Write the expression for buckling load for the column with one fixed and other end free. l = Crippling load E = Young s Modulus I = Moment of inertia l = Length of column 10. Explain equivalent length (or) Effective length. If l is actual length of a column, then its equivalent length (or) effective length L may be obtained by multiplying it with some constant factor C, which depends on the end fixation of the column (ie) L = C x l. 11. Write the Equivalent length (L) of the column in which both ends hinged and write the crippling load. Crippling Load L Equivalent length (L) = Actual length (l) = Crippling load E = Young s Modulus I = Moment of inertia L= Length of column MUTHUKUMAR./Asst.rof./CIVIL-III age

1. Write the relation between Equivalent length and actual length for all end conditions of column. Both ends linged L = l Constant = 1 Both ends fixed l L 1 Constant = One end fixed and other l 1 L Constant = end hinged One end fixed and other end free L l Constant = 13. Define core (or) Kernel of a section.. (ACU Nov/Dec 011) (ACU May/June 01) When a load acts in such a way on a region around the CG of the section So that in that region stress everywhere is compressive and no tension is developed anywhere, then that area is called the core (or) Kernal of a section. The kernel of the section is the area within which the line of action of the eccentric load must cut the cross-section if the stress is not to become tensile. 1. Derive the expression for core of a rectangular section. (ACU Nov/Dec 003) The limit of eccentricity of a rectangular section b x d on either side of XX axis (or) YY axis is d/6 to avoid tension at the base core of the rectangular section. Core of the rectangular section = Area of the shaded portion 1 b d 3 6 bd 18 15. Derive the expression for core of a solid circular section of diameter D. (ACU April /May 010) (AUC Nov/Dec 011) base. The limit of eccentricity on either side of both XX (or) YY axis = D/8 to avoid tension of the Core of the circular section = Area of the shaded portion D /8 D 6 16. A steel column is of length 8m and diameter 600 mm with both ends hinged. 5 Determine the crippling load by Euler s formula. Take E.110 N/mm. I 6 6 9 d 600 6.36 10 mm Since the column is hinged at the both ends, Equivalent length L = l MUTHUKUMAR./Asst.rof./CIVIL-III age 3

cr L 5.110 6.36 10.06 10 8000 8 N 9 17. Define Slenderness ratio. (AUC Nov/Dec 013) It is defined as the ratio of the effective length of the column (L) to the least radius of gyration of its cross section (K) (i.e) the ratio of K L is known as slenderness ratio. Slenderness ratio = K L 18. State the Limitations of Euler s formula. (AUC April /May 005) a. Euler s formula is applicable when the slenderness ratio is greater than or equal to 80 b. Euler s formula is applicable only for long column c. Euler s formula is thus unsuitable when the slenderness ratio is less than a certain value. 19. Write the Rankine s formula for columns. (AUC April /May 010) fc A L 1 K K = Least radius of gyration I A = Crippling load A = Area of the column f c = Constant value depends upon the material. = Rankine s constant f c E 0. Write the Rankine s formula for eccentric column. ey 1 c k fc A L 1 k K = Least radius of gyration I A = Crippling load A = Area of the column f c = Constant value depends upon the material. = Rankine s constant f c E MUTHUKUMAR./Asst.rof./CIVIL-III age

1. Define thick cylinder.. (ACU April /May 011) If the ratio of thickness of the internal diameter of a cylindrical or spherical shell exceeds 1/0, it is termed as a thick shell. The hoop stress developed in a thick shell varies from a maximum value at the inner circumference to a minimum value at the outer circumference. Thickness > 1/0. State the assumptions involved in Lame s Theory (AUC Nov/Dec 013) i. The material of the shell is Homogeneous and isotropic. ii. lane section normal to the longitudinal axis of the cylinder remains plane after the application of internal pressure. iii. All the fibers of the material expand (or) contact independently without being constrained by there adjacent fibers. 3. What is the middle third rule? (ACU Nov/Dec 003) In rectangular sections, the eccentricity e must be less than or equal to b/6. Hence the greatest eccentricity of the load is b/6 form the axis Y-Y and with respect to axis X X 1 the eccentricity does not exceed d/6. Hence the load may be applied with in the middle third of the base (or) Middle d/3.. What are the limitations of the Euler s formula? 1. It is not valid for mild steel column. The slenderness ratio of mild steel column is less than 80.. It does not take the direct stress. But in excess of load it can withstand under direct compression only. 5. Write the Euler s formula for different end conditions. (ACU Nov/Dec 01) 1. Both ends fixed. E = л / ( 0.5L). Both ends hinged E = л / (L) 3. One end fixed,other end hinged. E = л / ( 0.7L). One end fixed, other end free. E = л / ( L) L = Length of the column 6. How many types will you determine the hoop stress in a thick compound cylinder? (AUC May/June 01) 1. Radial stress. Circumferential or hoop stress 3. Maximum at the inner circumference. Minimum at the outer circumference of a thick cylinder 7. How columns are differentiate the type of column depending upon the Slenderness ratio.. (ACU April /May 011) Slenderness ratio is used to differentiate the type of column. Strength of the column depends upon the slenderness ratio, it is increased the compressive strength of the column decrease as the tendency to buckle is increased. MUTHUKUMAR./Asst.rof./CIVIL-III age 5

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11. A hollow cylindrical cast iron column whose external diameter in 00mm and has a thickness of 0mm is.5m long and is fixed at both ends. Calculate the safe load by Rankin s formula using a factor safety of.5. Take crushing strength of material as 580N/mm and Rankin s constant as 1/1600. Find also the ratio of Rulers to Rankin s load E= 150Gpa. (AUC May/June 01) MUTHUKUMAR./Asst.rof./CIVIL-III age 19

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1. A pipe of 00mm internal diameter and 50mm thickness carries a fluid at a pressure of a 10Mpa. Calculate the maximum and minimum intensities of circumferential stress distribution and circumferential stress distribution across the section. (AUC Apr011) MUTHUKUMAR./Asst.rof./CIVIL-III age 1

13. Explain the failure of long column. Solution: A long column of uniform cross-sectional area A and of length l, subjected to an axial compressive load, as shown in fig. A column is known as long column if the length of the column in comparison to its lateral dimensions is very large. Such columns do not fail y crushing alone, but also by bending (also known buckling) The load, at which the column just buckles, is known as buckling load and it is less than the crushing load is less than the crushing load for a long column. Buckling load is also known as critical just (or) crippling load. The value of buckling load for long columns are long columns is low whereas for short columns the value of buckling load is high. Let l = length of the long column p = Load (compressive) at which the column has jus buckled. A = Cross-sectional area of he column e = Maximum bending of the column at the centre. 0 = Stress due to direct load A Where = Stress due to bending at the centre of the column b = e Z Z = Section modulus about the axis of bending. The extreme stresses on the mid-section are given by Maximum stress = 0 + Minimum stress = 0 - b b The column will fail when maximum stress (i.e) 0 + b is more the crushing stress f c. In case of long column, the direct compressive stresses are negligible as compared to buckling stresses. Hence very long columns are subjected to buckling stresses. MUTHUKUMAR./Asst.rof./CIVIL-III age

1. State the assumptions made in the Euler s column Theory. And explain the sign conventions considered in columns. (AUC April/May003) (AUC May/June01) (AUC Nov/Dec 010) The following are the assumptions made in the Euler s column theory: 1. The column is initially perfectly straight and the load is applied axially. The cross-section of the column is uniform throughout its length. 3. The column material is perfectly elastic, homogeneous and isotropic and obeys Hooke s law.. The length of the column is very large as compared to its lateral dimensions 5. The direct stress is very small as compared to the bending stress 6. The column will fail by buckling alone. 7. The self-weight of column is negligible. The following are the sign conventions considered in columns: 1. A moment which will tend to bend the column with its convexity towards its initial centre line is taken as positive.. A moment which will tend to bend the column with its concavity towards its initial center line is taken as negative. 15. Derive the expression for crippling load when the both ends of the column are hinged. (AUC Nov/Dec 011) Solution: Consider a column AB of length L hinged at both its ends A and B carries an axial crippling load at A. Consider any section X-X at a distance of x from B. Let the deflection at X-X is y. The bending moment at X-X due to the load, M =. y d y y k y dx Where k p d y ` k y 0 dx Solution of this differential equation is MUTHUKUMAR./Asst.rof./CIVIL-III age 3

y Acos kx B sin kx y Acos x p Bsin x p By using Boundary conditions, At B, x = 0, y = 0 A = 0 At A, x = l, y = 0 0 B sin l p Sinl p 0 p l 0,,,3... Now taking the lest significant value (i.e) p l ; p l l p `The Euler s crippling load for long column with both ends hinged. p l 16. Derive the expression for buckling load (or) crippling load when both ends of the column are fixed. (AUC April/May010) Solution: Consider a column AB of length l fixed at both the ends A and B and caries an axial crippling load at A due to which buckling occurs. Under the action of the load the column will deflect as shown in fig. Consider any section X-X at a distance x from B.Let the deflection at X-X is y. Due to fixity at the ends, let the moment at A or B is M. MUTHUKUMAR./Asst.rof./CIVIL-III age

Total moment at XX = M.y Differential equation of the elastic curve is d dx y M y d dx d dx y y py py M IE M IE p p d dx y py M The general solution of the above differential equation is M y Acos x / B sin x / (i) Where A and B are the integration constant From (i) At, N. x = 0 and y = 0 0 A 1 B 0 M p M A p Differentiating the equation (i) with respect to x, dy A dx Sin x dy At the fixed end B, x = 0 and 0 dx B 0 Either B = 0 (or) 0. / B Cos x. 0 MUTHUKUMAR./Asst.rof./CIVIL-III age 5

Since 0 as p 0 B = 0 Subs M A and B = 0 in equation (i) p M y cos x. M M y 1 cos x.. Again at the fixed end A, x = l, y = 0 M 0 1 Cos l. / l. / 0,,,6... Now take the least significant value l. l. l The crippling load for long column when both the ends of the column are fixed L 17. Derive the expression for crippling load when column with one end fixed and other end hinged. (ACU April/May 003) (AUC Nov/Dec 013) Solution: Consider a column AB of length l fixed at B and hinged at A. It carries an axial crippling load at A for which the column just buckles. As here the column AB is fixed at B, there will be some fixed end moment at B. Let it be M. To balance this fixing moment M, a horizontal push H will be exerted at A. MUTHUKUMAR./Asst.rof./CIVIL-III age 6

Consider any section X-X at a distance x from the fixed end B. Let the deflection at xx is y. Bending moment at xx = H (l-x) - y Differential equation of the elastic curve is, d dx y H l x y d dx y 1 l x y d dx y y H l x p d y H l x p y dx The general solution of the above different equation is p p H l x y Acos x B x. sin. Where A and B are the constants of integration. (i) At B, x = 0, y = 0 From (i) A Hl B B H H p Again at the end A, x = l, y=0. substitute these values of x, y, A and B in equation (i) Hl 0 Cos H l. / Sinl. / H p Hl l. / Cosl. / Sin. MUTHUKUMAR./Asst.rof./CIVIL-III age 7

l. /. l /. l tan The value of /. l tan in radians has to be such that its tangent is equal to itself. The only angle whose tangent is equal to itself, is about.9 radians. /. l.9 l l.9 (approx) l The crippling load (or) buckling load for the column with one end fixed and one end hinged. l 18. Derive the expression for buckling load for the column with one end fixed and other end free. (AUC April/May 003) (AUC May/June01) Solution: Consider a column AB of length l, fixed at B and free at A, carrying an axial rippling load at D de to which it just buckles. The deflected form of the column AB is shown in fig. Let the new position of A is A 1. Let a be the deflection at the free end. Consider any section X-X at a distance x from B. Let the deflection at xx is y. Bending moment due to critical load at xx, d y M a y dx d y a py dx MUTHUKUMAR./Asst.rof./CIVIL-III age 8

d y py pq dx The solution of the above differential equation is, y Acos x B x. a sin. Where A and B are constants of integration. At B, x = 0, y = 0 From (i), A = 0 Differentiating the equation (I w.r. to x dy A Sin x. B dx dy At the fixed end B, x = 0 and 0 dx 0 B As 0 p 0 Cos x. Substitute A = -a and B = 0 in equation (i) we get, y a cos x. a y a cos x.. 1 (ii) At the free end A, x = l, y = a, substitute these values in equation (ii) a a 1 cos 1.. cos1.. 0 MUTHUKUMAR./Asst.rof./CIVIL-III age 9

3 5 1,, Now taking the least significant value, 1 1 l The crippling load for the columns with one end fixed and other end free. l 19. A steel column is of length 8 m and diameter 600 mm with both ends hinged. Determine the crippling load by Euler s formula. Take E =.1 x 10 5 N/mm Solution: Given, Actual length of the column, l = 8m = 8000 mm Diameter of the column d= 600 mm E =.1 x 10 5 N/mm I 6 d 6 600 I 6.36 10 mm Since the column is hinged at the both ends, Equivalent length L =l Euler s crippling load, 9 MUTHUKUMAR./Asst.rof./CIVIL-III age 30

cr L 5.110 6.36 10 8000 =.06 x 10 8 N 9 0. A mild steel tube m long, 3cm internal diameter and mm thick is used as a strut with both ends hinged. Find the collapsing load, what will be the crippling load if Solution: i. Both ends are built in? ii. One end is built in and one end is free? Given: Actual length of the mild steel tube, l = m = 00 cm Internal diameter of the tube, d = 3 cm Thickness of the tube, t = mm = 0.cm. External diameter of the tube, D = d + t = 3+(0.) = 3.8 cm. Assuming E for steel = x 10 6 Kg/cm M.O.I of the column section, I D 6 6 d 3.8 3 I = 6.6 cm i. Since the both ends of the tube are hinged, the effective length of the column when both ends are hinged. L = l = 00 cm Euler s crippling load cr L 6 10 6.6 00 MUTHUKUMAR./Asst.rof./CIVIL-III age 31

cr 77.30Kg. The required collapsed load = 77.30 Kg. ii. When both ends of the column are built in, then effective length of the column, l 00 L 00cm Euler s crippling load, cr L cr 6 10 00 = 3089.19 Kg. 6.6 iii. When one end of the column is built in and the other end is free, effective length of the column, L = l = x 00 = 800 cm Euler s crippling load, cr L 6 10 6.6 800 cr = 193.07 Kg. 1. A column having a T section with a flange 10 mm x 16 mm and web 150 mm x 16 mm is 3m long. Assuming the column to be hinged at both ends, find the crippling load by using Euler s formula. E = x 10 6 Kg/cm. Solution: Given: Flange width = 10 mm = 1 cm Flange thickness = 16 mm = 1.6 cm Length of the web = 150 mm = 15cm Width of the web = 16mm = 1.6cm MUTHUKUMAR./Asst.rof./CIVIL-III age 3

E = 10 6 Kg/cm Length of the column, l = 3m = 300 cm. Since the column is hinged at both ends, effective length of the column. L = l = 300 cm. axis. From the fig. Y-Y is the axis of symmetry. The C.G of the whole section lies on Y-Y Let the distance of the C.G from the 16 mm topmost fiber of the section = Y Y 1.6 15 1 1.6 15 1.61.6 1 1.6 15 1.6 Y 5. 1cm Distance of C.G from bottom fibre = (15+1.6) - 5.1 = 11.19cm Now M.O.I of the whole section about X-X axis. I XX 1 1.6 1 3 1 1.6 5.1 1.6 1.6 15 1 3 1.6 1511.19 15 I XX 1188.9cm M.I of the whole section about Y-Y axis I yy 1.6 1 3 3 1 15 106 1 35.5cm I 35 cm min. 5 Euler s Crippling load, cr L 6 10 35.5 300 ; cr 51655.3Kg. MUTHUKUMAR./Asst.rof./CIVIL-III age 33

. A steel bar of solid circular cross-section is 50 mm in diameter. The bar is pinned at both ends and subjected to axial compression. If the limit of proportionality of the material is 10 Ma and E = 00 Ga, determine the m minimum length to which Euler s formula is valid. Also determine the value of Euler s buckling load if the column has this minimum length. Solution: Given, Dia of solid circular cross-section, d = 50 mm Stress at proportional limit, f = 10 Mpa = 10 N/mm Young s Modulus, E = 00 Ga = 00 x 10 3 N/mm Least moment of inertia of the column section, Area of cross section, A 50 1963.9mm I 6 Least radius of gyration, 3 50 3.6.79 10 mm I 306.79 10 k A 1963.9 The bar is pinned at both ends, Effective length, L = Actual length, l Euler s buckling load, cr cr A L E L / K 3 50 156.5mm For Euler s formula to be valid, value of its minimum effective length L may be found out by equating the buckling stress to f L E L K k 10 5 E 10 L 10 10 156.5 MUTHUKUMAR./Asst.rof./CIVIL-III age 3

L = 111.89 mm = 11 mm = 1.1 m The required minimum actual length l =L = 1.1 m For this value of minimum length, L Euler s buckling load Result: 5 10 306.75 11 = 15 N = 1.5 KN 10 3 Minimum actual length l = L = 1.1 m Euler s buckling Load =1.5 KN 3. Explain Rankine s Formula and Derive the Rankine s formula for both short and long column. Solution: Rankine s Formula: Euler s formula gives correct results only for long columns, which fail mainly due to buckling. Whereas Rankine s devised an empirical formula base don practical experiments for determining the crippling or critical load which is applicable to all columns irrespective of whether they a short or long. If is the crippling load by Rankine s formula. c is the crushing load of the column material E is the crippling load by Euler s formula. Then the Empirical formula devised by Rankine known as Rankine s formula stand as: 1 1 1 e E For a short column, if the effective length is small, the value of E will be very high and the 1 1 value of will be very small as compared to and is negligible. C E MUTHUKUMAR./Asst.rof./CIVIL-III age 35

For the short column, (i.e) 1 1 c = C Thus for the short column, value of crippling load by Rankine is more or less equal to the value of crushing load: For long column having higher effective length, the value of E is small and large enough in comparison to 1 1. So is ignored. C C 1 1 E will be 1 For the long column, (i.e) p C E E Thus for the long column the value of crippling load by Rankine is more or less equal to the value of crippling load by Euler. 1 1 1 c E 1 E E c c E p ; c c E p c c 1 Substitute the value of c = f c A and E in the above equation, E fc A p fc A 1 / L Where, f c = Ultimate crushing stress of the column material. A = Cross-sectional are of the column L = Effective length of the column I = Ak Where k = Least radius of gyration. L MUTHUKUMAR./Asst.rof./CIVIL-III age 36

p p fc A f A 1 / L fc A L 1 K c 1 fc A fc A L EAk where = Rankine s constant f c E Crushing Load = 1 L / k When Rankine s constant is not given then find f c E The following table shows the value of f c and for different materials. Material f c N/mm E Wrought iron 50 1 9000 Cast iron 550 1 1600 Mild steel 30 1 7500 Timber 50 1 750 f c. A rolled steel joist ISMB 300 is to be used a column of 3 meters length with both ends fixed. Find the safe axial load on the column. Take factor of safety 3, f c = 30 N/mm 1 and. roperties of the column section. 7500 Area = 566 mm, I XX = 8.603 x 10 7 mm I yy =.539 x 10 7 mm Solution: Given: Length of the column, l = 3m = 3000 mm Factor of safety = 3 MUTHUKUMAR./Asst.rof./CIVIL-III age 37

f c = 30 N/mm, 1 7500 Area, A = 566 mm I XX = 8.603 x 10 7 mm I yy =.539 x 10 7 mm The column is fixed at both the ends, Effective length, l 3000 L 1500mm Since I yy is less then I xx, The column section, I I yy mm 7 min I.539 10 Least radius of gyration of the column section, 7 K I.539 10 89. mm A 566 8 Crippling load as given by Rakine s formula, p cr fc A 30566 L 1 1500 1 1 K 7500 89.8 cr = 1335.38 N Allowing factor of safety 3, Safe load = Crippling Factor of Load safety 1335.38 780.79N 3 Result: i. Crippling Load ( cr ) = 1335.38 N ii. Safe load =780.79N MUTHUKUMAR./Asst.rof./CIVIL-III age 38

5. A built up column consisting of rolled steel beam ISWB 300 with two plates 00 mm x 10 mm connected at the top and bottom flanges. Calculate the safe load the column carry, if the length is 3m and both ends are fixed. Take factor of safety 3 f c = 30 N/mm 1 and 7500 Take properties of joist: A = 6133 mm Solution: Given: I XX = 981.6 x 10 mm ; I yy = 990.1 x 10 mm Length of the built up column, l = 3m = 3000 mm Factor of safety = 3 Sectional area of the built up column, 00 10 10133 A 6133 mm f c =30 N/mm 1 7500 Moment of inertia of the built up column section abut xx axis, 3 00 10 I XX 981.6 10 1 = 1.9 x 10 8 mm Moment of inertia of the built up column section abut YY axis, 00 10155 I YY 10 00 990.110 1 = 0.3 x 10 8 mm 3 Since I yy is less than I xx, The column will tend to buckle about Y-Y axis. Least moment of inertia of the column section, I I YY 0 mm 8 min I.3 10 The column is fixed at both ends. Effective length, MUTHUKUMAR./Asst.rof./CIVIL-III age 39

l 3000 L 1500mm Least radius of gyration o the column section, 8 K J 0.310 7. mm A 10133 6 Crippling load as given by Rankine s formula, p cr fc A 3010133 L 1 1500 1 1 K 7500 7.6 = 8603.3 N Safe load = Crippling load 8603.3 9567.3N 3 Factor of safety Result: i. Crippling load = 8603.3 N ii. Safe load = 9567.3 N 6. Derive Rankine s and Euler formula for long columns under long columns under Eccentric Loading? (AUC Nov/Dec 010) i. Rankine s formula: Consider a short column subjected to an eccentric load with an eccentricity e form the axis. Maximum stress = Direct Stress + Bending stress f c A M Z Z I y A p. e. y Ak c I Ak where k I A MUTHUKUMAR./Asst.rof./CIVIL-III age 0

A = Sectional are of the column Z = Sectional modulus of the column y c = Distance of extreme fibre from N.A k = Least radius of gyration. f c 1 A ey k c Eccentric load, fc A eyc 1 k Where ey 1 c k is the reduction factor for eccentricity of loading. For long column, loaded with axial loading, the crippling load, Where fc A L 1 K L 1 K is the reduction factor for buckling of long column. Hence for a long column loaded with eccentric loading, the safe load, ey 1 K f c c 1 A L K ii. Euler s formula Maximum stress n the column = Direct stress + Bending stress l e sec / A Z MUTHUKUMAR./Asst.rof./CIVIL-III age 1

Hence, the maximum stress induced in the column having both ends hinged and an e l eccentricity of e is sec / A Z The maximum stress induced in the column with other end conditions are determined by changing the length in terms of effective length. 7. A column of circular section has 150 mm dia and 3m length. Both ends of the column are fixed. The column carries a load of 100 KN at an eccentricity of 15 mm from the geometrical axis of the column. Find the maximum compressive stress in the column section. Find also the maximum permissible eccentricity to avoid tension in the column section. E = 1 x 10 5 N/mm Solution: Given, Diameter of the column, D = 150 mm Actual length of the column, l = 3m = 3000 mm Load on the column, = 100 KN = 1000 x 10 3 N E = 1 x 10 5 N/mm Eccentricity, e = 15 mm Area of the column section A D 150 = 17671 mm Moment of inertia of the column section N.A., I D 150 6 6 =.85 x 10 6 mm Section modulus, Z I y I D / 6.85 10 = 331339mm 150 Both the ends of the column are fixed. 3 Effective length of the column, l 3000 L 1500mm MUTHUKUMAR./Asst.rof./CIVIL-III age

Now, the angle 3 100 10 L 5 110.85 10 = 0.150 rad = 8.61 o / 6 1500 Maximum compressive stress, A e sec Z / L 100 10 17671 = 10. N/mm To avoid tension we know, M A Z 3 100 10 15 sec8.61 331339 3 o A p esec.8.61 Z o 100 10 17671 3 3 o 100 10 esec.8.61 331339 e = 18.50 mm Result: i. Maximum compressive stress = 10. N/mm ii. Maximum eccentricity = 18.50 mm 8. State the assumptions and derive Lame s Theory? 1. The assumptions involved in Lame s Theory. i. The material of the shell is homogenous and isotropic ii. lane sections normal to the longitudinal axis of the cylinder remain plane after the application of internal pressure. iii. All the fibres of the material expand (or) contract independently without being constrained by their adjacent fibres. MUTHUKUMAR./Asst.rof./CIVIL-III age 3

Derivation of Lame s Theory Consider a thick cylinder Let r c = Inner radius of the cylinder r 0 = Outer radius of the cylinder i = Internal radial pressure o = External radial pressure L = Length of the cylinder f = Longitudinal stress. Lame s Equation: f x p x a where x b a x b f x a a x b f x a x f x = hoop stress induced in the ring. p x = Internal radial pressure in the fig. x + d x = External radial pressure in the ring. The values of the two constants a and to b are found out using the following boundary conditions: i. Since the internal radial pressure is i, At x = r i, x = i ii. Since the external radial pressure is 0, At x = r 0, x = 0 9. A thick steel cylinder having an internal diameter of 100 mm an external diameter of 00 mm is subjected to an internal pressure of 55 M pa and an external pressure of 7 Mpa. Find the maximum hoop stress. Solution: Given, MUTHUKUMAR./Asst.rof./CIVIL-III age

100 Inner radius of the cylinder, r i 50mm 00 Outer radius of the cylinder, r o 100mm Internal pressure, i = 55 Mpa External pressure, 0 = 7 Mpa In the hoop stress and radial stress in the cylinder at a distance of x from the centre is f x and p x respectively, using Lame s equations, f x b a x (i) b x a x (ii) where a and b are constants, Now by equation, at x = 50 mm, x = 55 Ma (Boundary condition) Using these boundary condition in equation (ii) b x a x 55 b a 50 (iii) Then x = 100 mm, p x = 7 Mpa Using these boundary condition is equation (ii) b a 100 7 (iv) Solving (iii) & (iv) b / 100 a 7 / 50 a 55 b (- ) (+) 3b = - 8 10000 MUTHUKUMAR./Asst.rof./CIVIL-III age 5

b = 160000 a = 9 Substitute a & b in equation (i) 160000 f x 9 x The value of f x is maximum when x is minimum Thus f x is maximum for x = r i = 50 mm 160000 Maximum hoop stress 9 50 = 73 Mpa (tensile) Result: Maximum hoop stress = 73 Ma (tensile) 30. A cast iron pipe has 00 mm internal diameter and 50 mm metal thickness. It carries water under a pressure of 5 N/mm. Find the maximum and minimum intensities of circumferential stress. Also sketch the distribution of circumferential stress and radial stress across the section. Solution: Given: Internal diameter, d i = 00 mm Wall thickness, t = 50 mm Internal pressure, i = 5 N/mm External pressure, 0 = 0. di 00 r ri t 100 50 150mm Internal radius r i 100mm External radius 0 Let f x and x be the circumferential stress and radial stress at a distance of x from the centre of the pipe respectively. Using Lame s equations, f x p x b a x (i) b a x (ii) MUTHUKUMAR./Asst.rof./CIVIL-III age 6

where, a & b are arbitrary constants. Now at x = 100 mm, x = 5 N/mm At x = 150 mm, x = 0 Using boundary condition is (ii) b 5 100 a (ii) b 0 150 a (iv) By solving (iii) & (iv) a = ; b = 90000 90000 90000 f x,, x x x utting x = 100 mm, maxi circumferential stress. 90000 f x 13N / mm 100 tensile utting x = 150 mm, mini circumferential stress. 90000 f x 8N / mm 150 tensile 31. Explain the stresses in compound thick cylinders. Solution: Consider a compound thick cylinder as shown in fig. Let, r 1 = Inner radius of the compound cylinder r = Radius at the junction of the two cylinders r 3 = Outer radius of the compound cylinder When one cylinder is shrunk over the other, thinner cylinder is under compression and the outer cylinder is under tension. Due to fluid pressure inside the cylinder, hoop stress will develop. The resultant hoop stress in the compound stress is that algebraic sum of the hoop stress due to initial shrinkage and that due to fluid pressure. a. Stresses due to initial shrinkage: Applying Lame s Equations for the outer cylinder, MUTHUKUMAR./Asst.rof./CIVIL-III age 7

x b x 1 a 1 f x b x At x = r 3, x = 0 1 a 1 and at x = r, p x = p Applying Lame s Equations for the inner cylinder x f x b x b x a a At x = r, x = p and at x = r 3, p x = 0 b. Stresses due to Internal fluid pressure. To find the stress in the compound cylinder due to internal fluid pressure alone, the inner and outer cylinders will be considered together as one thick shell. Now applying Lame s Equation, x f x B x B x At x = r 1, x = p f At x = r 3, p x = 0 A A ( f being the internal fluid pressure) The resultant hoop stress is the algebraic sum of the hoop stress due to shrinking and due internal fluid pressure. 3. A compound cylinder is composed of a tube of 50 mm internal diameter at 5 mm wall thickness. It is shrunk on to a tube of 00 mm internal diameter. The radial pressure at the junction is 8 N/mm. Find the variation of hoop stress across the wall of the compound cylinder, if it is under an internal fluid pressure of 60 N/mm Solution: Given: Internal diameter of the outer tube, d 1 = 50 mm Wall thickness of the outer tuber, t = 5 mm Internal diameter of the inner tube, d = 00 mm Radial pressure at the junction = 8 N/mm MUTHUKUMAR./Asst.rof./CIVIL-III age 8

Internal fluid pressure within the cylinder f = 60 N/mm External radius of the compound cylinder, d1 t r 1 50 5 150mm Internal radius of the compound cylinder, d 00 r1 100 mm d1 50 Radius at the junction, r1 15mm Let the radial stress and hoop stress at a distance of x from the centre of the cylinder be p x and f x respectively. i. Hoop stresses due to shrinking of the outer and inner cylinders before fluid pressure is admitted. a. Four outer cylinder: Applying Lame s Equation b1 x a 1 (i) x b1 f x a 1 (ii) x Where a 1 and b 1 are arbitrary constants for the outer cylinder. Now at x = 150 mm, x = 0 X = 15 mm, x = 8 N/mm b1 o a 1 (iii) 150 b1 8 a 1 (iv) 15 Solving equation (iii) & (iv) a 1 = 18 ; b 1 = 09091 09091 f x 18 (v) x utting x = 150 mm in the above equation stress at the outer surface, MUTHUKUMAR./Asst.rof./CIVIL-III age 9

09091 f x 18 36N / mm (Tensile) 150 Again putting x = 15 mm in equation (v), stress at junction, f x 09091 15 18 N / mm (Tensile) b). For inner cylinder: Applying Lame s Equation with usual Notations. x f x b a x (iv) b a x (v) Now at x = 15 mm, x = 8 N/mm x =100 mm, x = 0 b 8 a (vi) 15 o b a (vii) 100 By solving (vi) & (vii) a = - b = - f x.n / mm (Comp) 100 f x 36.N / mm (Comp) 15 iii. Hoop stresses due to internal fluid pressure alone for the compound cylinder: In this case, the two tubes will be taken as a single thick cylinder. Applying Lame s equations with usual notations. x B A x (viii) MUTHUKUMAR./Asst.rof./CIVIL-III age 50

B f x A x (ix) At x = 150 mm, x = 0 x = 100 mm, x = p f = 60 N/mm From Equation (viii) B O 150 A (x) B 60 A (xi) 100 By solving (x) & (xi) A = 133, B = 3 x 10 6 6 310 f x 133 x utting x = 150 mm, hoop stress at the outer surface 6 310 f x 133 66 N / mm (Tensile) 150 Again putting x = 15 mm, hoop stress at the junction 310 f x 133 35 N / mm 15 6 Tensile utting x = 100 mm, hoop stress at the inner surface 310 f x 133 33 N / mm 100 6 Tensile iii. Resultant hoop stress (shrinkage +Fluid pressure): a. Outer cylinder Resultant hoop stress at the outer surface = 36 + 66 = 30 N/ mm (Tensile) Resultant hoop stress at the junction = + 35 = 369 N/mm (tensile) b. Inner cylinder; MUTHUKUMAR./Asst.rof./CIVIL-III age 51

Resultant hoop stress at the inner face = -. + 33 = 388.8 N/mm (Tensile) Resultant hoop stress at the junction = - 36. + 35 = 88.8 N/mm (Tensile) 33. A column with alone end hinged and the other end fixed has a length of 5m and a hollow circular cross section of outer diameter 100 mm and wall thickness 10 mm. If E = 1.60 x 10 5 N/mm and crushing strength 0 350N / mm, Find the load that the column may carry with a factor of safety of.5 according to Euler theory and Rankine Gordon theory. If the column is hinged on both ends, find the safe load according to the two theories. (ACU April/May 003) Solution: Given: L = 5 m = 5000 mm Outer diameter D = 100 mm Inner diameter d = D-t = 100 (10) = 80 mm Thickness = 10 mm I = 1.60 x 10 5 N/mm i. Calculation of load by Euler s Theory: 0 350N / mm f =.5 Column with one end fixed and other end hinged. l 5000 L 3536.06 mm L 1.60 3536.06 5 3.1 10 I I D 6 d 100 80 6 100000000 0960000 6 I = 8.96 x 10 5 mm MUTHUKUMAR./Asst.rof./CIVIL-III age 5

5 3.1 1.60 10 8.96 10 1503716.1 p = 73.07 x 10 3 N 5 ii. Calculation of load by Rankine-Gordon Theory: 1 Rankine s Constant a (assume the column material is mild steel.) 7500 A c p f L 1 a K K = lest radius of Gyration A I 8.96 10 86 5 3.01 A 100 80 10000 600 f c = c = 86 mm 1 350 8.6 1 7500 3536.06 3.01 989100 1.33 10 103.036 60.9 10 N iii. Both ends are hinged Euler s theory L = l L 5 3.1 1.6010 8.96 5000 10 = 18.7 x 10 18.7 10 N ; Safe Load =.5 5 MUTHUKUMAR./Asst.rof./CIVIL-III age 53

Rankine s Theory f c A p L 1 a K = 73096 N 350 86 1 5000 1 7500 3.01 989100 1.33 10 398.81 Safe load = 30.80 x 10 30.80 10 = 1190 N.5 Result: i. Euler s Theory One end fixed & one end hinged = 73.07 x 10 3 N Both ends hinged = 18.7 x 10 N ii. Rankine s Theory One end fixed & one end hinged = 60.9 x 10 N Both ends hinged = 30.80 x 10 N iii. Safe Load Euler s Theory = 73096 N Rankine s theory = 1190 N 3. A column is made up of two channel ISJC 00 mm and two 5 cm x 1 cm flange plate as shown in fig. Determine by Rankine s formula the safe load, the column of 6m length, with both ends fixed, can carry with a factor of safety. The properties of one channel are A = 17.77 cm, I xx = 1,161. cm and I yy = 8. cm. Distance of centroid from back of web = 1.97 cm. Take f c = 0.3 KN/mm 1 and Rankine s Constant 7500 Solution: Given: Length of the column l = 6 m = 600 mm Factor of safety = Yield stress, f c = 0.3 KN/mm Rankine s constant, a 1 7500 Area of column, A = (17.77+5 x 1) A = 85.5 cm A = 855 mm Moment of inertia of the column about X-X axis (ACU April /May 003) MUTHUKUMAR./Asst.rof./CIVIL-III age 5

3 5 1 I 5 110.5 XX 1,161. = 7839.0 cm 1 3 1 5 I YY 8. 17.77 5 1.97 =,99.0 cm 1 I yy < I XX The column will tend to buckle in yy-direction I = I yy =99.0 cm Column is fixed at both the ends l 6000 L 3000mm I 99 10 K 7. 5mm A 855 f c. A 0.3 855. A K 1 3000 1 a 1 L 75000 7.5 Safe load of column F. O. S 8 =557 KN Result: Safe load = 557 KN = 8 KN MUTHUKUMAR./Asst.rof./CIVIL-III age 55