Question

The two-span continuos beam (ABC) shown in the sketch below carries a uniformly distributed load, w, in the left side span AB. Assume that EI is constant. Use the method of virtual work, together with the principle of relative displacements, to determine the following quantities in terms of E,I,L and w:

The two-span continuous beam (ABC) shown in the sketch below carries a uniformly distributed load, w, in the left side span AB. Assume that El of the beam is constant. Use the method of virtual work, together with the principle of relative displacements, to determine the following quantities in terms of E, I, L, and w: (a) The vertical deflection at midspan (Point D) of the left span AB, SD. (b) The rotation at the left support of the beam, A. (c) The rotation at the middle support of the beam, OB. (d) The vertical deflection at midspan (Point E) of the right span BC, SE. (e) Sketch the deflected shape of the entire continuous beam. (f) Find the numerical values of the deflections and rotations in (a) through (d) if the values of beam properties are E = 29,000 ksi, I = 3,000 in*, L = 40 ft, and w= 2 kips/ft. Express the deflections in inch units and the rotations in degrees. Note: In parts (a) through (d), express your answer in the lowest common fractional denominator. ! E A HTB A

Answer 1

Page SO First, we need to obtain the force response of this structure. Later, using principle of virtual work, the displacement and rotations can be calculated. → Force response wring principle of virtual work, say, the reaction at A is redundant. Note: this primary structure might look like unstable. No restraint in horizontal direction. Don't worry it. does not matter as it is not the complete structure Real displacements (curvatures) Virtual forces (bending moments)

Page ① DIL+ fu x, = D. =0. Compatibility condition to ensure displacement at hinge support in original structure is on DiL = (MG m, dx . DIL = 13 (L) ( ut ) ( 3 ) 2 + Ź With ) (4) (2/2). used area multiplication method to integrate areas. for GC-C) (3) (-)) x, = -2/2 (WL) (+78) = -7 16 w 1 x, = -1 WL I ve sign meens TVA = 771 WL1 reaction is upwards. Now calculate all support reactions of given statically indeterminate structure. Take moment equilibrium about B. VA (L) - WL ( 2 ). #Vc (A) =0. 75 WL (L) - WL (4) A Ve ( e ) =0 I reaction at c is Tue=-11 downwards e ut.

VA = 7 7 3 we ve = wt { Fu 20. VA + VS the = WL I WL tvo - will = WL 1 lo → of I WL 10WL Ms =- Wh (L) = - Wit Cove means C- hogging). the - WC Super impose + EMDR to obtain force responk given Statically indeterminate Staula This is the resultant BMD.

Page ① Vertical deflection at point D. Apply a unit load at D, and principle of visitial work use admissible ot requirfields. The We can consider any statically admissible force response & force fields. This force response does not require to satisfy Compatibility condition. So, for the unit load at D. Rat; Pro Reto (statically admissible !) BMD say RA=0. RB+ Rc =1; Rc CL) + 1C ) =0. Re = -1/2 RB = 3/2 so unit load of 1 kN at D statically admissible force resultants are RA=0 RB = 3 2 Re = - 1 2 BMD -ve 142 ore This is virtual forces ( bending moments) Real curvatures of the given beam can be obtained as mex) mex) is known as the fire response is already calculated

Page 6 (M ) m EL Z dx. virtual Вм. real curvatures + $0)6)(u) content + $(5)( ) ) Displacement at is 7WY I Do = 768 ET principal principle of So using the virtual work. พ Do = TEEL

Pase (b) rotation at the left support, da. Apply unit moment at A. say Rc =0. Moment equilibrium about A. VB (L) + 1 = 0. Vo = -7 2. VA TVB 20; TVA = 1 - A Å t virtual force field. D-Smox om de . LE Chade) (1) () +1(30) ) - Es [w] [78 + acq]. -- 3 WE m, 1 is 6. So - ve - 3WE 96 EL 96 EL rotation at a =- means, rotation in clockwise direction. Rotation at A = WL I clock wise . 96ET

Page ① (c) Rotation at middle support. OB. m=1 Put RA=0. Rotation at B= m da Rotation at & max. 1 1/2 (WC y 2) (33)} = wis Counter se clock wise . Rotation at B is We Counter clock wise. erano a 5 UPE Vertical deflection at midspan E Put RA=0. ; RB = 1 ; Rc = 1/2

Pase DE = S Mall me --C4) (5) c)(+). -- LC))) Het] 256 EI -ve means displacement is upwards Displacement at E = WLL upwards 256El Deflected shape (e) OB OB OA You for this AD fease substitute part the values Thank you!
The problem is solved using principle of virtual work.