UNIT I ENERGY PRINCIPLES Strain energy and strain energy density- strain energy in traction, shear in flexure and torsion- Castigliano s theorem Principle of virtual work application of energy theorems for computing deflections in beams and trusses Maxwell s reciprocal theorem. 1. Define strain energy and Proof stress. Two Marks Questions and Answers Strain energy Whenever a body is strained, the energy is absorbed in the body. The energy which is absorbed in the body due to straining effect is known as strain energy. The strain energy stored in the body is equal to the work done by the applied load in stretching the body Proof stress The stress induced in an elastic body when it possesses maximum strain energy is termed as its proof stress. 3. Define Resilience, Proof Resilience and Modulus of Resilience. Resilience The resilience is defined as the capacity of a strained body for doing work on the removal of the straining force. The total strain energy stored in a body is commonly known as resilience. Proof Resilience The proof resilience is defined as the quantity of strain energy stored in a body when strained up to elastic limit. The maximum strain energy stored in a body is known as proof resilience. Modulus of Resilience It is defined as the proof resilience of a material per unit volume. Proof resilience Modulus of resilience = ------------------- Volume of the body 4. State the two methods for analyzing the statically indeterminate structures. a.displacement method (equilibrium method (or) stiffness coefficient method b. Force method (compatibility method (or) flexibility coefficient method) 5. Define Castigliano s first theorem second Theorem.
First Theorem. It states that the deflection caused by any external force is equal to the partial derivative of the strain energy with respect to that force. Second Theorem It states that If U is the total strain energy stored up in a frame work in equilibrium under an external force; its magnitude is always a minimum. 6. State the Principle of Virtual work. It states that the workdone on a structure by external loads is equal to the internal energy stored in a structure (Ue = U i ) Work of external loads = work of internal loads 7. What is the strain energy stored in a rod of length l and axial rigidity AE to an axial force P? Strain energy stored P L U= -------- AE 8. State the various methods for computing the joint deflection of a perfect frame. 1. The Unit Load method. Deflection by Castigliano s First Theorem 3. Graphical method : Willot Mohr Diagram 9. State the deflection of the joint due to linear deformation. n δ v = Σ U x 1 n δ H = Σ U x 1 PL = --------- Ae U= vertical deflection U = horizontal deflection 10. State the deflection of joint due to temperature variation. n δ = Σ U X A 1 = U 1 1 + U + + U n n If the change in length ( ) of certain member is zero, the product U. for those members will be substituted as zero in the above equation. 11. State the deflection of a joint due to lack of fit.
n δ = Σ U 1 = U 1 1 + U + + U n n If there is only one member having lack of fit 1, the deflection of a particular joint will be equal to U 1 1. 1. What is the effect of change in temperature in a particular member of a redundant frame? When any member of the redundant frame is subjected to a change in temperature, it will cause a change in length of that particular member, which in turn will cause lack of fit stresses in all other members of the redundant frame. 13. State the difference between unit load and strain energy method in the determination of structures. In strain energy method, an imaginary load P is applied at the point where the deflection is desired to be determined. P is equated to zero in the final step and the deflection is obtained. In the Unit Load method, a unit load (instead of P) is applied at the point where the deflection is desired. 14. State the assumptions made in the Unit Load method. 1. The external and internal forces are in equilibrium. Supports are rigid and no movement is possible 3. The material is strained well within the elastic limit. 15. State the comparison of Castigliano s first theorem and unit load method. The deflection by the unit load method is given by n PUL δ = Σ ------- 1 AE δ = n PL Σ ------- x U 1 AE n = Σ x U ----- (i) 1 The deflection by castigliano s theorem is given by n δ= 1 PL AE P W --------- (ii) By comparing (i) & (ii) P W =U
16. State Maxwell s Reciprocal Theorem. The Maxwell s Reciprocal theorem states as The work done by the first system of loads due to displacements caused by a second system of loads equals the work done by the second system of loads due to displacements caused by the first system of loads. 17. Define degree of redundancy. A frame is said to be statically indeterminate when the no of unknown reactions or stress components exceed the total number of condition equations of equilibrium. 0. Define Perfect Frame. If the number of unknowns is equal to the number of conditions equations available, the frame is said to be a perfect frame. 1. State the two types of strain energies. a.strain energy of distortion (shear strain energy) b. strain energy of uniform compression (or) tension (volumetric strain energy). State in which cases, Castigliano s theorem can be used. 1. To determine the displacements of complicated structures.. To find the deflection of beams due to shearing (or) bending forces (or) bending moments are unknown. 3. To find the deflections of curved beams springs etc. 3. Define Proof stress. The stress induced in an elastic body when it possesses maximum strain energy is termed as its proof stress. 16 Marks Questions And Answers 1. Derive the expression for strain energy in Linear Elastic Systems for the following cases. (i) Axial loading (ii) Flexural Loading (moment (or) couple) (i)axial Loading Let us consider a straight bar of Length L, having uniform cross- sectional area A. If an axial load P is applied gradually, and if the bar undergoes a deformation, the work done, stored as strain energy (U) in the body, will be equal to average force (1/ P) multiplied by the deformation. Thus U = ½ P. But = PL / AE U = ½ P. PL/AE = P L / AE ---------- (i) If, however the bar has variable area of cross section, consider a small of length dx and area of cross section Ax. The strain energy du stored in this small element of length dx will be, from equation (i) P dx
du = --------- A x E The total strain energy U can be obtained by integrating the above expression over the length of the bar. L U = 0 P dx A x E (ii) Flexural Loading (Moment or couple ) Let us now consider a member of length L subjected to uniform bending moment M. Consider an element of length dx and let d i be the change in the slope of the element due to applied moment M. If M is applied gradually, the strain energy stored in the small element will be du = ½ Md i But d i d ------ = ----- (dy/dx) = d y/d x = M/EI d x d x M d i = ------- dx EI Hence du = ½ M (M/EI) dx Integrating = (M /EI) dx U = M dx EI L 0. State and prove the expression for castigliano s first theorem. Castigliano s first theorem: It states that the deflection caused by any external force is equal to the partial derivative of the strain energy with respect to that force. A generalized statement of the theorem is as follows: If there is any elastic system in equilibrium under the action of a set of a forces W 1, W, W 3.W n and corresponding displacements δ 1, δ, δ 3. δ n and a set of moments M 1, M, M 3 M n and corresponding rotations Φ 1, Φ, Φ 3,.. Φ n, then the partial derivative of the total strain energy U with respect to any one of the
forces or moments taken individually would yield its corresponding displacements in its direction of actions. Proof: Expressed mathematically, U =δ W 1 1 ------------- (i) U M 1 =φ 1 ------------- (ii) Consider an elastic body as show in fig subjected to loads W 1, W, W 3 etc. each applied independently. Let the body be supported at A, B etc. The reactions R A,R B etc do not work while the body deforms because the hinge reaction is fixed and cannot move (and therefore the work done is zero) and the roller reaction is perpendicular to the displacements of the roller. Assuming that the material follows the Hooke s law, the displacements of the points of loading will be linear functions of the loads and the principles of superposition will hold. Let δ 1, δ, δ3 etc be the deflections of points 1,, 3, etc in the direction of the loads at these points. The total strain energy U is then given by U = ½ (W 1 δ 1 + W δ +.) --------- (iii) Let the load W 1 be increased by an amount dw 1, after the loads have been applied. Due to this, there will be small changes in the deformation of the body, and the strain energy will be increased slightly by an amount du. expressing this small increase as the rate of change of U with respect to W 1 times dw 1, the new strain energy will be U + U W 1 xdw 1 --------- (iv) On the assumption that the principle of superposition applies, the final strain energy does not depend upon the order in which the forces are applied. Hence assuming that dw 1 is acting on the body, prior to the application of W1, W, W 3 etc, the deflections will be infinitely small and the corresponding strain energy of the second order can be neglected. Now when W1, W, W 3 etc, are applied (with dw 1 still acting initially), the points 1,, 3 etc will move through δ 1, δ, δ3 etc. in the direction of these forces and the strain energy will be given as above. Due to the application of W 1, rides through a distance δ 1 and produces the external work increment du = dw 1. δ 1. Hence the strain energy, when the loads are applied is U+dW 1.δ 1 ----------- (v)
Since the final strain energy is by equating (iv) & (v). U xdw W 1 U+dW 1.δ 1 = U + δ 1 = 1 1 U Walignl Which proves the proportion. Similarly it can be proved that Φ 1 = U M 1. Deflection of beams by castigliano s first theorem: If a member carries an axial force the energies stored is given by L U = 0 P dx A x E In the above expression, P is the axial force in the member and is the function of external load W 1, W,W 3 etc. To compute the deflection δ 1 in the direction of W 1 L P p δ 1 = U = dx 0 AE W 1 Walignl If the strain energy is due to bending and not due to axial load L U = 0 1 1 M dx EI L δ 1 = U = M M dx 0 W 1 EI Walignl If no load is acting at the point where deflection is desired, fictitious load W is applied at the point in the direction where the deflection is required. Then after differentiating but before integrating the fictitious load is set to zero. This method is sometimes known as the fictitious load method. If the rotation Φ 1 is required in the direction of M 1. L Φ 1 = U M = 1 0 M M dx M 1 EI 3. Calculate the central deflection and the slope at ends of a simply supported beam carrying a UDL w/ unit length over the whole span. Solution: a) Central deflection: Since no point load is acting at the center where the deflection is required, apply the fictitious load W, then the reaction at A and B will (WL/ + W/) each.
L δ c = U W = 0 M W dx EI Consider a section at a distance x from A. Bending moment at x, M= ( wl + W wx ) x Putting W=0, b) Slope at ends M x = x l δ c = EI 0 l δ c = EI 0 = δ c = 5 384 (( wl + W wx ) x ) x dx (( wl x ) wx ) x dx EI (( wlx 3 wx4 1 wl 4 EI l 16 ))0 To obtain the slope at the end A, say apply a frictions moment A as shown in fig. The reactions at A and B will be ( wl m l ) and ( wl + m l ) Measuring x from b, we get φ A = u m = 1 l EI Mx 0 Mx M. Dx -------------------------------- Where Mx is the moment at a point distant x from the origin (ie, B) is a function of M. Mx = ( wl + m l ) x - Wx Mx m = x l in
φ A = 1 l EI 0 ( wl + m l ) x - Wx X/ Dx Putting M=0 φa= 1 Ei 0 l wl x WX x l dx φ A = 1 EI [ wx 3 wx 4 L 6 8L ]0 φ A = wl3 4 EI 4. State and prove the Castigliano s second Theorem. Castigliano s second theorem: It states that the strain energy of a linearly elastic system that is initially unstrained will have less strain energy stored in it when subjected to a total load system than it would have if it were self-strained. u = 0 t For example, if λ is small strain (or) displacement, within the elastic limit in the direction of the redundant force T, u = λ t λ =0 when the redundant supports do not yield (or) when there is no initial lack of fit in the redundant members. Proof: Consider a redundant frame as shown in fig.in which Fc is a redundant member of geometrical length L.Let the actual length of the member Fc be (L- λ ), λ being the initial lack of fit.f C represents thus the actual length (L- λ ) of the member. When it is fitted to the truss, the member will have to be pulled such that F and F coincide.
According to Hooke s law F F 1 = Deformation = T (l λ ) AE = TL AE (approx ) Where T is the force (tensile) induced in the member. Hence FF 1 =FF -F 1 F λ = TL AE ------------------------------------ ( i ) Let the member Fc be removed and consider a tensile force T applied at the corners F and C as shown in fig. FF 1 = relative deflection of F and C = u1 ------------------------------------------ ( ii ) T According to castigliano s first theorem where U 1 is the strain energy of the whole frame except that of the member Fc. (or) Equating (i) and (ii) we get u1 TL = λ -- T AE u1 T + TL AE = λ ----------------------- ( iii ) To strain energy stored in the member Fc due to a force T is U FC = ½ T. TL AE = TL AE U FC T = TL AE Substitute the value of TL AE in (iii) we get
u' T + U FC U =λ (or) T T =λ When U= U 1 + U Fc.If there is no initial lack of fit, λ =0 and hence U T =0 Note: i) Castigliano s theorem of minimum strain energy is used for the for analysis of statically indeterminate beam ands portal tranes,if the degree of redundancy is not more than two. ii) If the degree of redundancy is more than two, the slope deflection method or the moment distribution method is more convenient. 5) A beam AB of span 3mis fixed at both the ends and carries a point load of 9 KN at C distant 1m from A. The M.O.I. of the portion AC of the beam is I and that of portion CB is I. calculate the fixed end moments and reactions. Solution: There are four unknowns M a, Ra, M b and R b.only two equations of static are available (ie) v=0 and M =0 This problem is of second degree indeterminacy. First choose M A and M B as redundant. δ A = Mx EI M x R A dx U AB R A =0= -----------(1) θ A = U AB =0= M A A 1) For portion AC: B M x EI Taking A as the origin M x M A dx -------------() M x = -M A + R A x M x R A = x; M x M A = 1
M.O. I=I Limits of x: 0 to 1m Hence C M x M 1 x (-M dx= A + R A x ) x dx A EI R A 0 EI = 1 EI ( M A (1 ) 1 EI ( R A 3 M A ) + R A (1 )3 3 ) C M x And A EI M x R A 1 (-M dx= A + R A x) ( 1 ) dx 0 EI = 1 EI ( M A (1 ) R A (1 ) ) =1 EI ( M A R A ) For portion CB, Taking A as the origin we have M x = M A +R A X 9( X 1) M x R A = x; M x M A = 1 M.O.I = I Limits of x : 1 to 3 m Hence B C M x EI M 3 x (-M A + R A x-9(x-1)) x dx= dx R A 1 EI = 1 EI [ 4M A +6 3 R A 4 ] And B M x M 3 x (-M A + R A x-9(x-1))-1 dx= C EI M A 1 EI = 1 EI [ M A 4R A +18 ] dx Subs these values in (1) & () we get
U AB R A =0 1 EI [ R A 3 M A ] +1 EI [ 4M A +6 3 R A 4 ] =0.08 M A = 9.88 (3) U AB M A =0 1 EI [ M A 1 Solving (3) & (4) R A ] +1 EI [ M A 4R A +18 ]=0 M A 1.7R A = -7. -------------- (4) M A = 4.8 KN M (assumed direction is correct) R A = 7.05 KN To find M B, take moments at B, and apply the condition M=0 there. Taking clockwise moment as positive and anticlockwise moment as negative. Taking M B clockwise, we have M B M A =R A (3) 9x = 0 M B 4.8 + (7.05x 3) -18 = 0 M B = 1.65 KN m (assumed direction is correct) To find R B Apply V =0 for the whole frame. R B = 9 R A = 9-7.05 = 1.95 KN 6.Using Castigliano s First Theorem, determine the deflection and rotation of the overhanging end A of the beam loaded as shown in Fig. Sol: Rotation of A: R B x L = -M R B = -M/L R B = M/L ( )
& R C = M/L ( ) θ A = U M = 1 B EI A M x. M x M dx+ 1 B EI C M x. M x.dx (1) M For any point distant x from A, between A and B (i.e.) x = 0 to x = L/3 M x = M ; and M x =1 () M For any point distant x from C, between C and B (i.e.) x = 0 to x = L Subs () & (3) in (1) M x = (M/L) x ; and M x M = x L (3) θ A = U M = 1 L/3 M (1).dx+ 1 L EI 0 EI ( M 0 L x ) x L dx = ML 3 EI = ML 3 EI + ML 3 EI ( clockwise) b) Deflection of A: To find the deflection at A, apply a fictitious load W at A, in upward direction as shown in fig. R B xl= ( M + 4 3 WL) 1 L 1 L R B = ( M+ 4 3 WL) R B =( M + 4 3 WL) ( ) R C =(M + 1 3 WL) 1 L ( ) δ A = U W = 1 B EI A For the portion AB, x = 0 at A and x = L/3 at B M M x x W + 1 B EI C M x M x W.dx M x = M + W x
M x W = x For the portion CB, x = 0 at C and x = L at B M x = ( M+ 1 8 WL ) 1 L. x M x W = x 3 Putting W = 0 L/3 δ A = 1 EI 0 L/3 δ A = 1 EI 0 (M +Wx ) x+ 1 EI 0 L (Mx )dx+ 1 EI 0 L ( Mx 3L ) dx ( M + 1 3 WL ) x L. x 3 dx δ A = M EI ( x ) L/3 0 + M 3 EI ( x3 3 ) L 0 δ A = ML 18 EI + ML 9 EI δ A = ML 6 EI 7. Determine the vertical and horizontal displacements of the point C of the pin-jointed frame shown in fig. The cross sectional area of AB is 100 sqmm and of AC and BC 150 mm each. E= x 10 5 N/mm. (By unit load method) Sol: The vertical and horizontal deflections of the joint C are given by PuL AE δ H = Pu ' L AE δ V = A) Stresses due to External Loading: AC = 3 +4 =5m Reaction: R A = -3/4 R B = 3/4 Sin θ = 3/5 = 0.6; Cos θ = 4/5 = 0.8
Resolving vertically at the joint C, we get 6 = P AC cos θ + P BC sin θ Resolving horizontally at the joint C, we get P AC cos θ = P BC sin θ; P AC = P BC P AC sin θ + P BC sin θ = 6 P AC sin θ = 6 P AC = 6/sin θ = 6/ x 0.6 = 5 KN (tension) P AC = P BC = 5 KN (tension) Resolving horizontally at the joint C, we get P AB = P AC cos θ P AB = 5 cos θ ; P AB = 5 x 0.8 P AB = 4 KN (comp) B) Stresses due to unit vertical load at C: Apply unit vertical load at C. The Stresses in each member will be 1/6 than of those obtained due to external load. u AC =u BC =5 /6 u AB = 4/6= /3 C) Stresses due to unit horizontal load at C: Assume the horizontal load towards left as shown in fig. Resolving vertically at the joint C, we get (u CA )'sinθ=(u CB )'sin θ (u CA )'=(u CB )' Resolving horizontally at the joint C, we get (u CB )'cosθ+ (u CA )'cosθ=1 (u CB )'cosθ+ (u CB )'cosθ=1 u CB 'cosθ=1 u CB '= 1 cosθ =1 =5/8 KN (tension) x0.8 u CA '= 5/8 KN u CA '=5/8 KN (comp) Resolving horizontally at the joint B, we get u AB '= u BC 'cosθ u AB '= 5 /8x0.8= 0.5 KN u AB '=0.5 KN (comp )
P(KN) U (kn) PUL/A U (KN) PU L/A Member Length(L) Area mm (mm) AB 8000 100-4 -/3 640/3-1/ 160 BC 5000 150 5 5/6 500/18 5/8 500/4 CA 5000 150 5 5/6 500/18-5/8 500/4 E = X 10 5 n/mm = 00 KN/m δv= Pul AE =491 00 =.45mm δh= pu' l AE =160 00 =0.8mm 8) The frame shown in fig. Consists of four panels each 5m wide, and the cross sectional areas of the member are such that, when the frame carries equal loads at the panel points of the lower chord, the stress in all the tension members is f n/mm and the stress in all the comparison members of 0.8 f N/mm.Determine the values of f if the ratio of the maximum deflection to span is 1/900 Take E=.0 x 10 5 N/mm. Sol: The top chord members will be in compression and the bottom chord members, verticals, and diagonals will be in tension. Due to symmetrical loading, the maximum deflection occurs at C. Apply unit load at C to find u in all the members. All the members have been numbered 1,, 3.. etc., by the rule u 8 = u10 = u 1 = 0. u 7 = R A sinθ = ( comp) Reaction R A = R B = 1/ θ = 45º ; cos θ = sin θ = u 3 =u 7 cos θ=.1 =1 =u 4 (tension) 1 u 9 = u 4 cosθ = (tension) Also, u 7 cosθ +u 9 cosθ=u 1
u 1 = x 1 + x 1 =1.0( comp) Member Length (L) mm P (N/mm ) U PUL 1 500-0.8 F -1.0 +000F 3 500 +F +1/ +150F 4 500 +F +1/ +150F 7 500 () 0.5-0.8F -() 0.5 / +000F 8 500 +F 0 0 9 500() 0.5 +F +() 0.5 / +500F n δ C = 1 PUL E =9000+ =0.09 F mm 5 x10 Sum: +9000F δ C = 1 900 xspan= 1 900 x10000=100 mm 9 Hence 0.09 F = 100/9 (or) F = 100/(9 x 0.09) = 13.5 N/mm. 9. Determine the vertical deflection of the joint C of the frame shown in fig. due to temperature rise of 60º F in the upper chords only. The coefficient of expansion = 6.0 x 10-6 per 1º F and E = x 10 6 kg /cm. Sol: Increase in length of each member of the upper chord = L α t = 400 x 6x 10-6 x 60 = 0.144 cm The vertical deflection of C is given by δ= uδ To find u, apply unit vertical load at C. Since the change in length ( ) occurs only in the three top chord members, stresses in these members only need be found out. Reaction at A = 4/1 = 1/3 Reaction at B = 8/1 = /3 Passing a section cutting members 1 and 4, and taking moments at D, we get U 1 = (1/3 x 4) 1/3 = 4/9 (comp)
get Also Similarly, passing a section cutting members 3 and 9 and taking moments at C, we u 3 = ( 3 x4 ) 1 3 =8 9 (comp) u =u 1 = 4 9 (comp ) δ C =u 1 Δ 1 +u Δ +u 3 Δ 3 δ C ={( 4 9 ) + ( 4 9 ) + ( 8 δ C = 0.56 cm 9 )} x (+0. 144 ) 10) Using the principle of least work, analyze the portal frame shown in Fig. Also plot the B.M.D. Sol: The support is hinged. Since there are two equations at each supports. They are H A, V A, H D, and V D. The available equilibrium equation is three. (i.e.) M=0, H =0, V =0. The structure is statically indeterminate to first degree. Let us treat the horizontal H ( ) at A as redundant. The horizontal reaction at D will evidently be = (3-H) ( ). By taking moments at D, we get (V A x 3) + H (3-) + (3 x 1) ( 1.5) (6 x ) = 0 V A = 3.5 H/3 V D = 6 V A =.5 + H/3 By the theorem of minimum strain energy, U H =0 U AB H + U BE H + U CE H + U DC H =0 (1)For member AB: Taking A as the origin.
1. x M= M H =x U AB H =1 EI 0 +H.x 3 M M H dx 3 3 = 1 EI 0 1 EI [ Hx3 3 ( x + Hx) x dx = 1 [9H 10.1 ] EI x4 8 ]0 () For the member BE: Taking B as the origin. M=( Hx3 ) (3x11.5 )+ ( 3.5 H 3 ) x M=3H 4.5+3.5x Hx 3 M H =3 x 3 U BE 1 H =1 EI 0 M M H dx 1 = 1 EI 0 1 = 1 EI 0 1 = 1 EI 0 ( 3H 4.5+3.5x Hx 3 )( 3 x 3 ) dx ( 9H 13.5+10.5x Hx Hx+1.5x 1.67 x Hx + 9 ) dx ( 9H 13. 5+1x Hx 1. 67 x Hx + 9 ) dx = 1 EI ( 9 Hx 13.5x +6x Hx 0.389x 3 + Hx3 1 7 ) 0 = 1 EI ( 9H 13.5+6 H 0.389+ H 7 ) = 1 [9H 7.9] EI (3) For the member CE: Taking C as the origin
M= (3 H ) x+(.5+ H 3 )x M= 6+H+.5x+ Hx 3 U CE H =1 EI 0 = 1 EI 0 = 1 EI 0 = 1 EI 0 M M H 3 Hx 6+H+.5x+ [( 3 )( + x 3 )] [ 1+4H+5x+6.67 Hx x+6.67 Hx+0.833 x Hx + 9 ] dx [ 1+4H+3x+13.34 Hx x+0.833 x Hx + 9 ] dx 1 = (10.96H - 15.78) EI (4) For the member DC: Taking D as the origin M= (3 H ) x= 3x+Hx M x =x U DC H =1 EI 0 M M H dx = 1 EI 0 ( 3x+Hx ) ( x ) dx = 1 EI 0 ( 3x +Hx ) dx = 1 EI ( 3x3 3 + Hx 3 3 ) dx = 1 0 EI ( x3 + Hx3 3 ) dx 0 1 = (.67H -8) EI Subs the values U H =0 1/EI (9-10.) + (8.04H-7.9) + (10.96H-15.78) + (-8+.67H) = 0 30.67H = 41.80 H = 1.36 KN Hence V A = 3.5 - H/3 = 3.5-1.36/3 = 3.05 KN V D =.5 + H/3 =.5 + 1.36/3 =.95 KN
M A = M D =0 M B = (-1 x 3 )/ + (1.36 x 3) = -0.4 KN m M C = - (3-H) = - (3-1.36) =-3.8KNm Bending moment Diagram: 11) A simply supported beam of span 6m is subjected to a concentrated load of 45 KN at m from the left support. Calculate the deflection under the load point. Take E = 00 x 10 6 KN/m and I = 14 x 10-6 m 4. Solution: Taking moments about B. V A x 6 45 x 4=0 V A x 6-180 = 0 V A = 30 KN V B = Total Load V A = 15 KN Virtual work equation: L (δ c ) V = 0 mmdx EI Apply unit vertical load at c instead of 45 KN R A x 6-1 x 4 =0 R A = /3 KN R B = Total load R A = 1/3 KN Virtual Moment: Consider section between AC M 1 = /3 X 1 [limit 0 to ] Section between CB M = /3 X -1 (X - ) [limit to 6 ] Real Moment:
The internal moment due to given loading M 1 = 30 x X 1 M = 30 x X -45 (X -) (δ c ) V = m 1 M 1 dx 1 0 EI + 6 m M dx EI ( x 1 3 ) (30 x 1 ) = 0 EI 6 ( 3 dx 1 + x (x )) (30 x 45 ( x ) ) dx EI ( 3 x x + ) ( 30x 45 x +90)dx 1 EI 0 6 0 x 1 + = 1 EI 0 = 1 EI 0 = 1 EI [ 0 x 1 6 0 x 1 + 6 0 x 1 + 3 3 ]0 ( x 3 + ) ( 15 x +90)dx 5x 30 x 30x +180dx +[ 5x 3 3 6 3 60x +180 x ] = 0 EI ( 8 3 ) + 1 EI ( 5 3 (63 3 ) 30 (6 )+180 (6 1 ) ) = 1 [53.33+346.67 960+70 ] EI = 160 = 160 =0.0571 m(or)57.1mm EI 00 x10 6 6 x 14 x10 The deflection under the load = 57.1 mm 1) Define and prove the Maxwell s reciprocal theorem. The Maxwell s reciprocal theorem stated as The work done by the first system loads due to displacements caused by a second system of loads equals the work done by the second system of loads due to displacements caused by the first system of loads. Maxwell s theorem of reciprocal deflections has the following three versions:
1. The deflection at A due to unit force at B is equal to deflection at B due to unit force at A. δ AB = δ BA. The slope at A due to unit couple at B is equal to the slope at B due to unit couple A Φ AB = Φ BA 3. The slope at A due to unit load at B is equal to deflection at B due to unit couple. ' φ ' AB =δ AB Proof: By unit load method, Where, δ= Mmdx EI M= bending moment at any point x due to external load. m= bending moment at any point x due to unit load applied at the point where deflection is required. Let m XA= bending moment at any point x due to unit load at A Let m XB = bending moment at any point x due to unit load at B. When unit load (external load) is applied at A,
M=m XA To find deflection at B due to unit load at A, apply unit load at B.Then m= m XB Hence, Similarly, δ BA = Mmdx = m XA.m XB dx (i) EI EI When unit load (external load) is applied at B, M=m XB To find the deflection at A due to unit load at B, apply unit load at A.then m= m XA δ AB = Mmdx = mb.m XA dx (ii) EI EI Comparing (i) & (ii) we get δ AB = δ BA 13. Using Castigliano s theorem, determine the deflection of the free end of the cantilever beam shown in the fig. Take EI = 4.9 MN/m. (NOV / DEC 003) Solution: Apply dummy load W at B. Since we have to determine the deflection of the free end. Consider a section xx at a distance x from B. Then M x =Wx+30 ( x 1)+0 1 ( x 1.5 )+16 (x ) δ= M M EI W dx [ 1 1 EI 0 3 x 1 Wx xdx + Wx x+30( x 1)x+0( x 1)( {{ 1 ) x } dx+ Wx x+30( x 1)x {0 1( x 1.5) x+1
= 1 EI [ W ( x3 +[ Wx 3 3 3 ) 0 +[ 1 Wx 3 3 3 +30( x3 3 x ) ( ] +10 x4 4 x3 + x 3 )]1 +30( x3 x ) ( +0 x3 3 ) ( 0.75 x +16 x3 3 3 ) ] x Putting W =0 δ= 1 EI [ 30 ( 7 3 3 ) +10 ( 15 4 14 3 +3 ) +30 ( 19 3 5 ) 0 ( 19 3 3.75 ) +16 ( 19 3 5 )] δ= 1 EI [ 305 6 +10 7 +30 3 6 +0.58+16 4 3 ] δ= 1x103 4.9x 10 6 (5+5.83+115+51.6+1.33) δ=0.446m(or )44.64 mm 14. Fig shows a cantilever, 8m long, carrying a point loads 5 KN at the center and an udl of KN/m for a length 4m from the end B. If EI is the flexural rigidity of the cantilever find the reaction at the prop. (NOV/DEC 004) Solution: To find Reaction at the prop, R (in KN) Portion AC: ( origin at A ) 4 ( Rx ) dx EI U 1 = 0 =[ R x 3 64 R R 4= =3 6 EI ]0 6 EI 3 EI Portion CB: ( origin at C ) Bending moment M x = R (x+4) 5x x / = R (x+4) 5x x U = 0 Total strain energy = U 1 +U 4 (M x ) dx EI
At the propped end U R =0 4 U R =64 R 3EI + 0 ( M x EI x dm x dr ) dx = 64 R 3 EI + 1 4 EI [ R ( x 4) 5x x ]( x+4)dx 0 = 64 R 3 EI + 1 4 EI [ R (x 4 ) 5x ( x+4 ) x ( x+4)] dx 0 = 64 R 3 EI + 1 EI 0 4 [ R (x +8x+16) 5( x +4x) ( x 3 +4x )] dx 0 = 64 R 3 EI + 1 [ 4 EI ( R x3 3 +4x x3 +16 x) 5 ( 3 +x ) ( x4 4 4x3 3 ) ]0 = 64 R 3 + [ R ( 64 3 +64+64 ) 5(64 3 +3) (56 4 +56 3 ) ] = 1.33 R + (149.33R 66.67 149.33) = 1.33 R + (149.33 R 416) 1.33 R +149.33 R 416 =0 R =.347 KN 15. A simply supported beam of span L is carrying a concentrated load W at the centre and a uniformly distributed load of intensity of w per unit length. Show that Maxwell s reciprocal theorem holds good at the centre of the beam. Solution: Let the load W is applied first and then the uniformly distributed load w. Deflection due to load W at the centre of the beam is given by δ W = 5Wl4 384 EI Hence work done by W due to w is given by: 5 wl 4 384 EI U A, B =Wx Deflection at a distance x from the left end due to W is given by
δ W (x ) = W 48 EI (3l x 4x ) Work done by w per unit length due to W, l/ U B,A = 0 W wx 48EI (3l x 4x )dx U B,A = Ww 4 EI [ 3l l ( ) 4] l Hence proved. U B,A = Ww 4 EI [ 3l4 8 ( l4 U A, B = 5 384 Wwl 4 EI 16 )]
Strength of Materials (FOR IV SEMESTER) Question bank UNIT II INDETERMINATE BEAMS Compiled by, K.DIVYA ASSISTANT PROFESSOR DEPARTMENT OF CIVIL ENGINEERING FATIMA MICHAEL COLLEGE OF ENGINEERING AND TECHNOLOGY MADURAI - 0
UNIT II INDETERMINATE BEAMS Propped Cantilever and fixed end moments and reactions for concentrated load (central, non central), uniformly distributed load, triangular load (maximum at centre and maximum at end) Theorem of three moments analysis of continuous beams shear force and bending moment diagrams for continuous beams (qualitative study only) Two Marks Questions and Answers 1. Define statically indeterminate beams. If the numbers of reaction components are more than the conditions equations, the structure is defined as statically indeterminate beams. E = R r E = Degree of external redundancy R = Total number of reaction components r = Total number of condition equations available. A continuous beam is a typical example of externally indeterminate structure.. State the degree of indeterminacy in propped cantilever. For a general loading, the total reaction components (R) are equal to (3+) =5, While the total number of condition equations (r) are equal to 3. The beam is statically indeterminate, externally to second degree. For vertical loading, the beam is statically determinate to single degree. E = R r = 5 3 = 3. State the degree of indeterminacy in a fixed beam. For a general system of loading, a fixed beam is statically indeterminate to third degree. For vertical loading, a fixed beam is statically indeterminate to second degree. E = R r For general system of loading: R = 3 + 3 and r = 3 E = 6-3 = 3
For vertical loading: R = + and r = E = 4 = 4. State the degree of indeterminacy in the given beam. The beam is statically indeterminate to third degree of general system of loading. R = 3+1+1+1 = 6 E = R-r = 6-3 = 3 5. State the degree of indeterminacy in the given beam. The beam is statically determinate. The total numbers of condition equations are equal to 3+ = 5. Since, there is a link at B. The two additional condition equations are at link. E = R-r = +1+-5 = 5-5 E = 0 6. State the methods available for analyzing statically indeterminate structures. i. Compatibility method ii. Equilibrium method 7. Write the expression fixed end moments and deflection for a fixed beam carrying point load at centre. M A =M B = WL 8 y max = WL3 19 EI 8. Write the expression fixed end moments and deflection for a fixed beam carrying eccentric point load.
M A = Wab L M B = Wa b L y max = Wa3 b 3 (under theload ) 3 3 EIL 9. Write the expression fixed end moments for a fixed due to sinking of support. M A =M B = 6 EI δ L 10. State the Theorem of three moments. Theorem of three moments: It states that If BC and CD are only two consecutive span of a continuous beam subjected to an external loading, then the moments M B, M C and M D at the supports B, C and D are given by M B L 1 +M C ( L 1 +L )=M D. L = 6a 1 x 1 + 6a x L 1 L Where M B = Bending Moment at B due to external loading M C = Bending Moment at C due to external loading M D = Bending Moment at D due to external loading L 1 = length of span AB L = length of span BC a 1 = area of B.M.D due to vertical loads on span BC = area of B.M.D due to vertical loads on span CD a x 1 x = Distance of C.G of the B.M.D due to vertical loads on BC from B = Distance of C.G of the B.M.D due to vertical loads on CD from D. 11. Draw the shape of the BMD for a fixed beam having end moments M in one support and +M in the other. (NOV/DEC 003)
1. What are the fixed end moments for a fixed beam of length L subjected to a concentrated load w at a distance a from left end? (Nov/Dec 004) Fixed End Moment: M A = Wab L M B = Wab L 13. Explain the effect of settlement of supports in a continuous beam. (Nov/Dec 003) Due to the settlement of supports in a continuous beam, the bending stresses will alters appreciably. The maximum bending moment in case of continuous beam is less when compare to the simply supported beam. 14. What are the advantages of Continuous beams over Simply Supported beams? (i)the maximum bending moment in case of a continuous beam is much less than in case of a simply supported beam of same span carrying same loads. (ii) In case of a continuous beam, the average B.M is lesser and hence lighter materials of construction can be used it resist the bending moment. 15. A fixed beam of length 5m carries a uniformly distributed load of 9 kn/m run over the entire span. If I = 4.5x10-4 m 4 and E = 1x10 7 kn/m, find the fixing moments at the ends and deflection at the centre. Solution: Given: L = 5m W = 9 kn/m, I = 4.5x10-4 m 4 and E = 1x10 7 kn/m (i) The fixed end moment for the beam carrying udl: M A = M B = WL 1 = 9x(5) =18.75 KNm 1 (ii) The deflection at the centre due to udl:
y c = WL4 384 EI y c = 9x(5)4 =3.54mm 384 x1x10 7 4 x4.5x 10 Deflection is in downward direction. 16. A fixed beam AB, 6m long is carrying a point load of 40 kn at its center. The M.O.I of the beam is 78 x 10 6 mm 4 and value of E for beam material is.1x10 5 N/mm. Determine (i) Fixed end moments at A and B. Solution: Fixed end moments: M A =M B = WL 8 M A =M B = 50x6 8 =37.5 knm 17. A fixed beam AB of length 3m is having M.O.I I = 3 x 10 6 mm 4 and value of E for beam material is x10 5 N/mm. The support B sinks down by 3mm. Determine (i) fixed end moments at A and B. Solution: Given: L = 3m = 3000mm I = 3 x 10 6 mm 4 E = x10 5 N/mm δ = 3mm M A =M B = 6 EI δ L = 6xx 105 x3x10 6 x3 (3000 ) =1x10 5 N mm = 1 kn m. 18. A fixed beam AB, 3m long is carrying a point load of 45 kn at a distance of m from A. If the flexural rigidity (i.e) EI of the beam is 1x10 4 knm. Determine (i) Deflection under the Load. Solution: Given:
L = 3m W = 45 kn EI = 1x10 4 knm Deflection under the load: In fixed beam, deflection under the load due to eccentric load Wa 3 b 3 3 EIL 3 y C = y C = 45 x ()3 x(1 ) 3 3x1x 10 4 x(3) y C =0.000444m y C =0.444mm The deflection is in downward direction. 19. A fixed beam of 5m span carries a gradually varying load from zero at end A to 10 kn/m at end B. Find the fixing moment and reaction at the fixed ends. Solution: Given: L = 5m W = 10 kn/m (i) Fixing Moment: M A = WL 30 and M B = WL 0 M A = 10(5) = 50 =8.33 knm 30 30 M B = 10(5) 0 = 50 0 =1.5kNm (ii) Reaction at support: R A = 3WL 0 R A = 3 10 5 0 R B = 7 10 5 0 and R B = 7WL 0 = 150 =7. 5 kn 0 = 350 0 =17.5kN
0. A cantilever beam AB of span 6m is fixed at A and propped at B. The beam carries a udl of kn/m over its whole length. Find the reaction at propped end. Solution: Given: L=6m, w = kn/m Downward deflection at B due to the udl neglecting prop reaction P, y B = wl4 8 EI Upward deflection at B due to the prop reaction P at B neglecting the udl, y B = Pl3 3 EI Upward deflection = Downward deflection Pl 3 3 EI = wl 4 8EI P = 3WL/8 = 3**6/8 =4.5 kn 16 Marks Questions And Answers
1. A fixed beam AB of length 6m carries point load of 160 kn and 10 kn at a distance of m and 4m from the left end A. Find the fixed end moments and the reactions at the supports. Draw B.M and S.F diagrams. Solution: Given: L = 6m Load at C, W C = 160 kn Load at D, W C = 10 kn Distance AC = m Distance AD =4m First calculate the fixed end moments due to loads at C and D separately and then add up the moments. Fixed End Moments: For the load at C, a=m and b=4m M B1 = W C a b L M A1 = W C ab L 160 xx(4) M A1 = =14.kNm (6) M B1 = 160 x x( 4) (6) =71.11kNm For the load at D, a = 4m and b = m M A = W D ab L M A = 10 x x( 4) (6) =53.33kNm M B = W D a b L 160xx( 4) M B = =106.66kNm (6) Total fixing moment at A, M A = M A1 + M A = 14. + 53.33 = 195.55 knm M A
Total fixing moment at B, M B =M B1 + M B = 71.11 + 106.66 = 177.77 kn m B.M diagram due to vertical loads: Consider the beam AB as simply supported. Let R A * and R B * are the reactions at A and B due to simply supported beam. Taking moments about A, we get R B x6 =160 x +10 x4 R B = 800 =133.33 kn 6 * R A = Total load - R B* =(160 +10) 133.33 = 146.67 kn B.M at A = 0 * B.M at C = R A x = 146.67 x = 93.34 kn m B.M at D = 133.33 x = 66.66 kn m B.M at B= 0 S.F Diagram: Let R A = Resultant reaction at A due to fixed end moments and vertical loads R B = Resultant reaction at B Equating the clockwise moments and anti-clockwise moments about A, R B x 6 + M A = 160 x + 10 x 4 + M B R B = 130.37 kn R A = total load R B = 149.63 kn S.F at A = R A = 149.63 kn S.F at C = 149.63-160 = -10.37 kn S.F at D = -10.37 10 = -130.37 kn S.F at B= 130.37 KN
. A fixed beam AB of length 6m carries two point loads of 30 kn each at a distance of m from the both ends. Determine the fixed end moments and draw the B.M diagram. Sloution: Given: Length L = 6m Point load at C = W 1 = 30 kn Point load at D = W = 30 kn Fixed end moments: M A = Fixing moment due to load at C + Fixing moment due to load at D W 1 a 1 b 1 W a b = + L L 30 xx4 30 x4x + =40kN m 6 6 Since the beam is symmetrical, M A = M B = 40 knm B.M Diagram: To draw the B.M diagram due to vertical loads, consider the beam AB as simply supported. The reactions at A and B is equal to 30kN. B.M at A and B = 0 B.M at C =30 x = 60 knm B.M at D = 30 x = 60 knm
3. Find the fixing moments and support reactions of a fixed beam AB of length 6m, carrying a uniformly distributed load of 4kN/m over the left half of the span. Solution: Macaulay s method can be used and directly the fixing moments and end reactions can be calculated. This method is used where the areas of B.M diagrams cannot be determined conveniently. For this method it is necessary that UDL should be extended up to B and then compensated for upward UDL for length BC as shown in fig. The bending at any section at a distance x from A is given by, EI d y dx =R A x M A wx x +w*(x-3) ( x 3) =R A x M A - ( 4x x 3) ) +4( ) = R A x M A - x +(x-3) Integrating, we get EI dy dx =R A x -M Ax - x3 3 +C 1 + ( x 3)3 3 When x=0, dy dx =0. Substituting this value in the above equation up to dotted line, C 1 = 0 Therefore equation (1) becomes EI dy dx =R x A -M Ax - x3 ( x 3)3 + 3 3 Integrating we get x EI y=r 3 A 6 M A x x4 1 +C ( x 3)4 + 1 When x = 0, y = 0 By substituting these boundary conditions upto the dotted line, C = 0 EI y= R A x3 M A x x 1( x 3 )4 4+ 6 6 6 By subs x =6 & y = 0 in equation (ii) -------(1) (ii)
0= R A 63 6 M A 6 6 4 1(6 3 ) 4 + 6 6 =36 R A 18 M A 16+13. 5 18R A 9 M A = 101.5 At x =6, dy dx =0 in equation (i) ------------- (iii) 0=R A x 6 M A x6 3 x (6 )3 + (6 3 )3 3 18R A M A x6 144+18=0 18R A 6M A =16 By solving (iii) & (iv) M A = 8.5 knm By substituting M A in (iv) 16 = 18 R A 6 (8.5) R A = 9.75 kn R B = Total load R A R B =.5 kn By equating the clockwise moments and anticlockwise moments about B M B + R A x 6 = M A + 4x3 (4.5) M B = 3.75 knm Result: M A = 8.5 knm M B = 3.75 knm R A = 9.75 kn R B =.5 KN 4. A continuous beam ABC covers two consecutive span AB and BC of lengths 4m and 6m, carrying uniformly distributed loads of 6kN/m and 10kN/m respectively. If the ends A and C are simply supported, find the support moments at A,B and C. draw also B.M.D and S.F.D. Solution: Given Data: Length AB, L 1 =4m. Length BC, L =6m UDL on AB, w 1 =6kN/m UDL on BC, w =10kN/m
(i) Support Moments: Since the ends A and C are simply supported, the support moments at A and C will be zero. By using cleyperon s equation of three moments, to find the support moments at B (ie) M B. M A L 1 + M B (L 1 +L ) + M C L = 6a 1 x 1 6a x 4 6 0 + M B (4+6) + 0 = 6a 1 x 1 6a x 4 6 0M B = 3a 1 x 1 a x The B.M.D on a simply supported beam is carrying UDL is a parabola having an attitude of wl 8. Area of B.M.D = 3 *L*h = wl * Span * 3 8 The distance of C.G of this area from one end, = span. a 1 =Area of B.M.D due to UDL on AB, = 3 *4* 6(4 ) 8 =3 x 1 = L 1 = 4/ = m. a = Area of B.M.D due to UDL on BC, = 3 *6* 10(6 ) 8 = 180m. x =L / = 6 / =3m Substitute these values in equation(i). We get, 0M B = 3 3 +(180 3) = 96+540 M B =31.8 knm. (ii) B.M.D
(iii) The B.M.D due to vertical loads (UDL) on span AB and span BC. Span AB: = w 1 L 1 8 = 6 4 8 =1kNm Span BC: = w L 8 = 10 6 8 =45kNm S.F.D: To calculate Reactions, For span AB, taking moments about B, we get (R A *4)-(6*4*) M B =0 4R A 48 = 31.8 (M B =31.8, -ve sign is due to hogging moment. R A =4.05kN Similarly, For span BC, taking moment about B, (R c *6)-(6*10*3) M B =0 6R C 180=-31.8 R C =4.7kN. R B =Total load on ABC (R A +R B ) =(6*4*(10*6))-(4.05+4.7) =55.5kN. RESULT: M A =M C =0 M B =31.8kNm R A =4.05kN R B =55.5kN R C =4.7kN
5. A continuous beam ABCD of length 15m rests on four supports covering 3 equal spans and carries a uniformly distributed load of 1.5 kn/m length.calculate the moments and reactions at the supports. Draw The S.F.D and B.M.D. Solution: Given: Length AB = L 1 = 5m Length BC = L = 5m Length CD = L 3 = 5m u.d.l w 1 = w = w 3 = 1.5 kn/m Since the ends A and D are simply supported, the support moments at A and D will be Zero. M A =0 and M D =0 For symmetry M B =0
(i)to calculate support moments: To find the support moments at B and C, by using claperon s equations of three moments for ABC and BCD. For ABC, M A L 1 +[M B (L 1 +L )]+M C L = 6a 1 x 1 L 1 6a x L 0+[M B (5+5)]+[M C (5)]= 6a 1 x 1 6a x 5 5 0M B +5M C = 6 5 ( a 1 x 1 +a x ) --------------------------------------(i) a 1 =Area of BMD due to UDL on AB when AB is considered as simply supported beam. = 3 AB Altitude of parabola (Altitude of parabola= w 1 L 1 8 ) = 3 5 1.5 (5) 8 =15.65 x 1 =L 1 / =5/=.5m Due to symmetry.a =a 1 =15.65 x =x 1 =.5 subs these values in eqn(i) 0M B +5M C = 6 [(15.65.5)+(15.65.5)] 5 =93.75 Due to symmetry M B =M C 0M B +5M B =93.75 M B =3.75kNm. M B =M C =3.75kNm. (ii) To calculate BM due to vertical loads: The BMD due to vertical loads(here UDL) on span AB, BC and CD (considering each span as simply supported ) are shown by parabolas of altitude w 1 L 1 1.5 1.5 = =4. 6875 knm each. 8 8 (iii)to calculate support Reactions: Let R A,R B,R C and R D are the support reactions at A,B,C and D. Due to symmetry R A =R D R B =R C For span AB, Taking moments about B, We get M B =(R A *5)-(1.5*5*.5) -3.75=(R A *5)-18.75
R A =3.0kN. Due to symmetry R A =R D =3.0kN R B =R C R A +R B +R C +R D =Total load on ABCD 3+R B +R B +3=1.5*15 R B =8.5kN R C =8.5kN. Result: M A = M D = 0 M B =M C =3.75kNm. R A =R D =3.0kN R B =8.5kN R C =8.5kN. 6. a continuous beam ABCD, simply supported at A,B, C and D is loaded as shown in fig. Find the moments over the beam and draw B.M.D and S.F.D. (Nov/ Dec 003)
Solution: Given: Length AB = L 1 = 6m Length BC = L = 5m Length CD = L 3 = 4m Point load W 1 = 9kN Point load W = 8kN u.d.l on CD, w = 3 kn/m (i) B.M.D due to vertical loads taking each span as simply supported: Consider beam AB, B.M at point load at E = W 1 ab = 9 4 =1 knm L 1 6 W ab Similarly B.M at F = = 8 3 =9.6kNm L 6 B.M at the centre of a simply supported beam CD, carrying U.D.L = wl 3 8 = 3 4 8 =6 knm (ii) B.M.D due to support moments: Since the beam is simply supported M A =M D = 0 By using Clapeyron s Equation of Three Moments: a) For spans AB and BC M A L 1 + M B (L 1 +L ) + M C L = 6a 1 x 1 6a x 4 6 0+M B (6+5 )+M c (5 )= 6a 1 x 1 6a x 6 5 6 M B +5M C =a 1 x 1 5 a x ------------ (i) a 1 x 1 = ½*6*1*L+a/3 = ½*6*1*(6+)/3 = 96 a x = ½*5*9.6*L+b/3 = ½*5*9.6*(6+4)/3 = 64 Substitute the values in equation (i) M B + 5M C = 96+6/5*64 M B + 5M C = 17.8 ------------ (ii) b) For spans BC and CD M B L + M C (L +L 3 ) + M D L 3 = 6a x L 6a 3 x 3 L 3 M B *5 + M C (5+4) +0 = 6a x 6a 3 x 3 5 4
5M B +18M C = 6 ax 5 +6a 3 x 3 ----------- (iii) 4 a x = ½ * 5 * 9.6 *(L+a)/3 =1/ * 5 * 9.6 *(5+)/3 = 56 a 3 x 3 = /3 * 4*6*4/ =3 Substitute these values in equation (iii) 5M B +18M C = 6 56 + 6 3 5 4 5M B +18M C =115. By solving equations (ii) &(iv) M B = 6.84 knm and M C = 4.48 knm (iii) Support Reactions: For the span AB, Taking moment about B, M B = R A * 6 9*4 = 6R A 36 R A = 36 6.84 =4.86 KN 6 For the span CD, taking moments about C M C =R D 4 3 4 4 (M C = 4.48) R D = 4.88KN For ABC taking moment about C M c = R A (6+5) 9 (5+4 )+R B 5 8 3 5R B =81+4 4.86 11 R B = 9.41 kn R C = Total load on ABCD (R A +R B +R D ) R C = (9+8+4*3) (4.86+9.41+4.88) R C = 9.85 kn Result: M A = M D = 0 M B = 6.84 knm and M C = 4.48 knm R A = 4.86kN R B = 9.41kN R C = 9.85 kn R D = 4.88KN
7. Using the theorem of three moments draw the shear force and bending moment diagrams for the following continuous beam. (April / May 003) Solution: Given: Length AB, L 1 =4m. Length BC, L =3m. Length CD, L 3 =4m. UDL on AB, w=4 kn/m Point load in BC, W 1 =4kN/m Point load in CD, W 1 =6kN (i) Bending Moment to Vertical Loads: Consider beam AB, B.M= wl 8 = 4 4 8 =8kNm.
Similarly for beam BC, B.M= W 1 ab = 6 1 L 3 =4kNm Similarly for beam CD, B.M= W ab = 8 1 3 L 3 4 =6kNm (ii) Bending Moment to support moments: Let M A,M B,M C And M D be the support moments at A,B,C and D. Since the ends is simply supported, M A =M D =0. By using Clayperon s equation of three moments for span AB and BC, M A L 1 +[M B (L 1 +L ) ]+ M C L = 6a 1 x 1 L 1 + 6a x L 0+[M B (4+3)] M C (3) = 6a 1 x 1 + 6a x 4 3 14M B + 3M C = 1.5a 1 x 1 + a x ----------------------------(i) a 1 x 1 = Moment of area BMD due to UDL = 3 Base (Base Altitude) = 3 4 (4 8) =4.33 a x = Moment of area BMD due to point load about point B = 1 3 ( 4) =5.33 Using these values in eqn (i), 14M B + 3M C =1.5(4.33) +(*5.33) 14M B + 3M C =63.495+10.66 -------------------------(ii) For span BC and CD, M B L 1 +[M C (L +L 3 ) ]+ M D L 3 = 6a x L + 6a 3 x 3 L 3 M B (3)+[M C (3+3) ]+ M D L 3 = 6a x + 6a 3 x 3 3 3 3M B +1M C = a x + a 3 x 3 ------------------------(iii) a x = Moment of area BMD due to point load about point C =(1/)**4* 1 3
=.66 a 3 x 3 = Moment of area BMD due to point load about point D 1 = 1 6 3 3 =6 Using these values in Eqn(iii), 3M B + 1M C =(.66) + (*6) 3M B + 1M C = 17.3 -------------------(iv) Using eqn (ii) and (iii), M B = 5.69 kn m M C = 0.19 kn m (iii) Support Reaction: For span AB, taking moment about B M B =R A 4 4 4-5.69 = R A *4 3 R A *4=6.731 R A = 6.68 kn For span CD, taking moment about C M C =R D 4 8 1-0.19 = R D *4-8 R D = 1.967 kn Now taking moment about C for ABC M C =R A (7) 4 (4 5 )+R B 3 6 1 M C =7R A 4(0)+3R B 6 0.19=7(6.68) 80+3R B 6 R B = 13.037 kn R C = Total load (R A +R B + R C ) = [ (4 4 )+6 +8] (6. 68+1. 967+13. 037 ) R C = 8.316 kn Result: M A = M D = 0 M B = 5.69 kn m M C = 0.19 kn m R A = 6.68 kn R B = 13.037 kn R C = 8.316 kn R D = 1.967 kn
8. A beam AB of 4m span is simply supported at the ends and is loaded as shown in fig. Determine (i) Deflection at C (ii) Maximum deflection (iii) Slope at the end A. E= 00 x 10 6 kn/m and I = 0 x 10-6 m 4 Solution: Given: L = 4m E= 00 x 10 6 kn/m and I = 0 x 10-6 m 4 To calculate Reaction: Taking moment about A R B 4=0 1+10 ( +1+1) R B *4 = 0 + 0(3) R B = 80/4 = 0 kn R A = Total load - R B = (10*+0) -0 R A = 0 kn
By using Macaulay s method: M X =EI d y x ) =0 x 0( x 1) 10( dx Integrating we get EI dy dx =10x +C 1 10( x 1 ) 5( x ) 3 3 Integrating we get 10 x3 EIy= 3 +C 1 x+c x 1)3 5( x )4 10( ---------- (ii) 3 1 When x = 0, y = 0 in equation (ii) we get C = 0 When x = 4m, y = 0 in equation (ii) 0= 10 3 (4)3 +4C 1 10 3 (4 1)3 5 1 ( 4 )4 = 13.33 +4C 1 90-6.67 C 1 = -9.16 Hence the slope and deflection equations are Slope Equation: EI dy dx =10x 9.16 10( x 1) 5( x ) 3 3 Deflection Equation: EIy= 10 x3 3 9.16 x 10( x 1 )3 3 (i) Deflection at C, y C : 5( x )4 1 Putting x = m in the deflection equation, we get EIy= 10()3 9.16() 10( 1)3 3 3 = 6.67-58.3-3.33 = -34.98 y c = 8.74 (downward) (ii) Maximum Deflection, y max : The maximum deflection will be very near to mid-point C. Let us assume that it occurs in the sections between D and C. For maximum deflection equating the slope at the section to zero, we get EI dy dx =10x 9.16 10( x 1)
(iii) Slope at the end A, θ A : 10x -9.16-10(x-1) = 0 10x -9.16-10 (x -x+1) = 0 x = 39.16/0 =1.958 m 10(1. 958)3 EIy= 3 y max = -35/EI y max = 8.75 mm (downward) Putting x = 0 in the slope equation, EI dy dx = 9.16 θ A = dy/dx = -9.16/EI θ A = -0.0079 radians θ A = -0.417º Result: (i) Deflection at C = 8.74 mm (ii) Maximum deflection = 8.75 mm (iii) Slope at the end A, θ A = -0.417º 9.16(1.958 ) 10(1.958 1)3 3 9. A continuous beam is shown in fig. Draw the BMD indicating salient points. (Nov/Dec 004) Solution: Given: Length L 1 = 4m Length L = 8m Length L 3 = 6m Udl on BC w = 10 kn/m Point load W 1 = 40 kn Point load W = 40 kn (i) For beam BC, For beam CD, B.M due to vertical loads: Consider beam AB, B.M = W 1 ab = 40 3 1 =30 knm L 1 4 B.M = wl 8 =10(8) =80 knm 8 B.M = W L 3 = 40 6 4 4 (ii) B.M due to support moments: 60 knm
Let M A, M B, M C, M D be the support moments at A, B, C, D. Since the end A and D are simply supported M A = M D = 0 For Span AB and BC: By using Clapeyron s Equation of Three moments. M A L 1 +M B (L 1 +L )+M C L = 6a 1 x 1 L 1 6a x L 0+M B (4+8)+M C (8)= 6a 1 x 1 6a x 4 8 M B (1) +8 M C = -1.5a 1 x 1 0.75 a x 4 M B +8 M C = -1.5a 1 x 1 0.75 a x ----------- (i) a 1 x 1 = Moment of area of B.M.D due to point load = ½*4*30*/3*3 = 10 a x = Moment of area of B.M.D due to udl = /3 (Base x Altitude) x Base/ = /3 (8*80)*8/ = 1706.67 Using these values in equation (i) 4 M B +8 M C = -1.5(10) 0.75 (1706.67) 4 M B +8 M C = -1460.005 ---------------- (ii) For Span BC and CD: M B L +M C ( L +L 3 )+M D L 3 = 6a x L 6a 3 x 3 L 3 M B (8 )+M C (8+6 )+0= 6a x 6a 3 x 3 8 6 8 M B + 8 M C = - 0.75 a x - a 3 x 3 -------------- (iii) a x = Moment of area of B.M.D due to udl = /3 (Base x Altitude) x Base/ = /3 (8*80)*8/ = 1706.67 a 3 x 3 = Moment of area of B.M.D due to point load = ½ * b*h*l/3 = ½ * 6*60*6/3 = 360 Using these values in equation (iii) 8 M B + 8 M C = - 0.75 (1706.67) 360 8 M B + 8 M C = - 1640.005 ------------------ (iv) From (ii) & (iv) M C = 45.56 knm M B = 45.657 knm Result: M A = M D = 0 M C = 45.56 knm M B = 45.657 knm
10. For the fixed beam shown in fig. draw BMD and SFD. (Nov / Dec 004) Solution: (i) B.M.D due to vertical loads taking each span as simply supported: Consider beam AB as simply supported. The B.M at the centre of AB = wl 1 8 = (3) =.5kNm 8