Short Notes on Structural Analysis Static Indeterminacy • If a structure cannot be analyzed for external and internal reactions using static equilibrium conditions alone then such a structure is called indeterminate structure External static indeterminacy: • It is related with the support system of the structure and it is equal to number of external reaction components in addition to number of static equilibrium equations. Internal static indeterminacy: • It refers to the geometric stability of the structure. If after knowing the external reactions it is not possible to determine all internal forces/internal reactions using static equilibrium equations alone then the structure is said to be internally indeterminate. Kinematic Indeterminacy • • It the number of unknown displacement components are greater than the number of compatibility equations, for these structures additional equations based on equilibrium must be written in order to obtain sufficient number of equations for the determination of all the unknown displacement components. The number of these additional equations necessary is known as degree of kinematic indeterminacy or degree of freedom of the structure. Three Hinged Arches (i) Three Hinged Parabolic Arch of Span L and rise 'h' carrying a UDL ovr the whole span DS 0 BMC 0 2 wl 8h H wx 2 Hy 2 where, H = Horizontal thrust MX VAx V A = Vertical reaction at A VA x wx 2 2 wl 2 Simply supported beam moment i.e., moment caused by vertical reactions. Hy = H-moment D S = Degree of static indeterminacy BM C = Bending Moment at C. (ii) Three Hinged Semicircular Arch of Radius R carrying a UDL over the whole span. wR H 2 MX wR2 [sin 2 Mmax wR2 8 BMC sin2 ] 0 Point of contraflexure = 0 (iii) Three Hinged Parabolic Arch Having Abutments at Different Levels (a) When it is subjected to UDL over whole span. HA 2( h1 h2 ) l h1 l1 h1 h2 l h2 l2 h1 BMC (b) wl 2 HB h2 0 When it is subjected to concentrated load W at crown wl H h1 h2 2 (iii) Three Hinged Semicircular Arch Carrying Concentrated Load W at Crown H V V A B W 2 Temperature Effect on Three Hinged Arches (i) h l2 4h2 T 4h h = free rise in crown height l = length of arch h = rise of arch α = coefficient of thermal expansion T= rise in temperature in 0C 1 (ii) H h Where, H = horizontal thrust Where, and h = rise of arch (iii) % Decrease in horizontal thrust h h 100 Two Hinged Arches ds My El y 2 ds El H DS = 1 Where, M = Simply support Beam moment caused by vertical force. (i) Two hinged semicircular arch of radius R carrying a concentrated load 'w' at the town. w H (ii) Two hinged semicircular arch of radius R carrying a load w at a section, the radius vector corresponding to which makes an angle α with the horizontal. H w sin2 (iii) A two hinged semicircular arch of radius R carrying a UDL w per unit length over the whole span. 4 wR 3 (iv) A two hinged semicircular arch of radius R carrying a distributed load uniformly varying from zero at the left end to w per unit run at the right end. H 2 wR 3 (v) A two hinged parabolic arch carries a UDL of w per unit run on entire span. If the span off the arch is L and its rise is h. H wl 2 8h (vi) When half of the parabolic arch is loaded by UDL, then the horizontal reaction at support is given by H H wl 2 16h (vii) When two hinged parabolic arch carries varying UDL, from zero to w the horizontal thrust is given by H wl 2 16h (viii) A two hinged parabolic arch of span l and rise h carries a concentrated load w at the crown. 25 wl H 128 h Temperature Effect on Two Hinged Arches H l T y 2 ds (i) El 4El T R2 where H = Horizontal thrust for two hinged semicircular arch due to rise in temperature by T 0C. H (ii) H 15 El0 T 8 h2 where l 0 = Moment of inertia of the arch at crown. H = Horizontal thrust for two hinged parabolic arch due to rise in temperature T 0C. Reaction Locus for a Two Hinged Arch (a) Two Hinged Semicircular Arch Reaction locus is straight line parallel to the line joining abutments and height at R 2 (b) Two Hinged Parabolic Arch y PE 1.6hL2 L Lx x 2 2 Eddy's Theorem MX y where, M X = BM at any section y = distance between given arch linear arch Trusses: Degree of Static Indeterminacy (i) DS m re 2 j where, D S = Degree of static indeterminacy m = Number of members, re = Total j = Total number of joints (ii) D S = 0 Truss is determinate external If D se = +1 & D si = –1 then D S = 0 at specified point. (iii) D S > 0 Truss is indeterminate or dedundant. reactions, Truss Member Carrying Zero forces (i) M 1 , M 2 , M 3 meet at a joint M 1 & M 2 are collinear M 3 carries zero force where M 1 , M 2 , M 3 represents member. (ii) M 1 & M 2 are non collinear and F ext = 0 M1 & M2 carries zero force. Indeterminate Truss (i) Final force in the truss member PkL AE S = P + kX and X k2L AE sign convn +ve for tension, –ve for compression where, S = Final force in the truss member K = Force in the member when unit load is applied in the redundant member L = Length of the member A = Area of the member E = Modulus of elasticity P = Force in the member when truss become determinate after removing one of the member. P = Zero for redundant member. Lack of Fit in Truss Q2L U where, U X 2AE Q = Force induce in the member due to that member which is ' pulled by force 'X'. ' too short or ' Deflection of Truss yC k L T PL AE Where, y C = Deflection of truss due to effect of loading & temp. both. If effect of temperature is neglected then PkL yC AE Coefficient of thermal expansion T = Change in temperature T = +ve it temperature is increased T = -ve it temperature is decreased P & K have same meaning as mentioned above. ' too long is Stiffness Method for Truss AB P B A AE cos BX AB L AX By [ Ay =Axial deflection of member AB. where, sin ] AB P AB = Force in member AB (Axial force) Difference between Force Method and Displacement Method Force Method 1. Unknowns are forces/reacctions. taken Displacement Method redundants 1. Unknown are taken displacement 2. To find unknown forces or redundants compatibility equations are written. 2. To find unknown displacement joint equilibrium conditions are written. 3. The number of compatibility equations needed is equal to degree of static indeterminacy. 3. The number of equilibrium conditions needed is equal to degree of kinematic indeterminacy. Type of Displacement Diagram Flexibility Stiffness (i) Axial L AE AE L (ii) Transverse Displacement (a) with far end fixed L3 12El 12El L3 3El 3El L 4El 4El L (b) with hinged far end L3 L3 (iii) Flexural Displacement (a) with far end fixed (b) with hinged far end L 3El 3El L L Gl p GlP L (iv) Torsional Displacement Castigliano's first theorem U U P& M where, U = Total strain energy Displacement in the direction of force P. Rotation in the direction of moment M. Castiglianos Second Theorem U U , P M Betti's Law Pm mn Pn nm where, P m = Load applied in the direction P n = Load applied in the direction n. mn = Deflection in the direction 'm' due to load applied in the direction 'n'. nm = Deflection in the direction 'n' due to load applied in the direction 'm'. Maxwells Reciprocal Theorem 21 m. 12 where, 12 21 deflection in the direction (2) due to applied load in the direction (1). = Deflection in the direction (1) due to applied load in the direction (2). Moment distribution method (Hardy Cross method) k • Stiffness : It is the force/moment required to be applied at a joint so as to produce unit deflection/rotation at that joint. F M or Where, K = Stiffness F = Force required to produce deflection M = Moment required to produce rotation . • Stiffness of beam (i) Stiffness of member BA when farther end A is fixed. 4El k L Where, K = Stiffness of BA at joint B. When farther end is fixed. El = Flexural rigidity L = Length of the beam M = Moment at B. (ii) Stiffness of member BA when farther end A is hinged 3El k L where, K = Stiffness of BA at joint B. When farther end is hinged • Carry over factor : Carry over moment Carry over factor = Applied moment COF may greater than, equal to or less than 1. • Standard Cases : (i) COF= 1 2 (ii) COF = 0 (iii) COF= a b Fixed Convention +ve Sagging –ve Hogging and All clockwise moment +ve and All Anti clockwise moment –ve Span length is l M AB MBA Pl 8 Pl 8 Pab2 l2 Pa2 b l2 wl 2 12 wl 2 12 wl 2 30 wl 2 20 11 2 wl 192 5 wl 2 192 5 2 wl 96 5 wl 2 96 M0 b(3a L2 l) M0 a(3b L2 MO MO 4 4 6El 6El 2 l2 0 3El l l2 l) 6El( (6l 2 1 2 ) 6El( 1 2 l2 l2 wa2 12l 2 wa2 12l 2 (4l 3a) 8l6.a 3a2 ) ) Slope deflection Method (G.A. Maney Method) In this method, joints are considered rigid. It means joints rotate as a whole and the angles between the tangents to the elastic curve meeting at that joint do not change due to rotation. The basic unknown are joint displacement and . To find and , joints equilibrium conditions and shear equations are established. The forces (moments) are found using force displacement relations. Which are called slope deflection equations. Slope Deflection Equation (i) The slope deflection equation at the end A for member AB can be written as : M 2El M AB 2 AB 3 2 A l B l (ii) The slope deflection equation at the end B for member BA can be written as : M 2El M 2 BA BA l 3 2 B A l where, L = Length of beam, El = Flexural Rigidity M AB MBA are fixed end moments at A & B respectively. M AB & M BA are final moments at A & B respectively. joint A & B respectively. Settlement of support Sign Convention M Clockwise M A n ti clockwis e Clockwis e A and B are rotation of Anti-clockwise +ve, if it produces clockwise rotation to the member & vice-versa. The number of joint equilibrium conditions will be equal to number of ' ' components & number of shear equations will be equal to number of ' ' Components.
