Work and energy method. Exercise 1 : Beam with a couple. Exercise 1 : Non-linear loaddisplacement. Exercise 2 : Horizontally loaded frame

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1 Work and energy method EI EI T x-axis Exercise 1 : Beam with a coupe Determine the rotation at the right support of the construction dispayed on the right, caused by the coupe T using Castigiano s nd theorem. z-axis 0,5 0,5 Exercise 1 : Non-inear oaddispacement diagram A non-inear oad-dispacement diagram is given in the figure to the right. The reation between oad and dispacement is given as: u u = c ind: a) The strain energy ( in terms of u ) b) The compementary energy ( in terms of ) u Exercise : Horizontay oaded frame Cacuate the horizonta dispacement at B using Castigiano s nd theorem. The bending stiffness of a beams is EI. Axia deformation (by the norma force) is negected. C B A Exercise 3 : Cantiever beam Cacuate the dispacement at B using the Castigiano s nd theorem. The bending stiffness of the prismatic beam is EI. A B C Exercise 4 : Cantiever beam 3 Cacuate the rotation at C caused by a force at B. The bending stiffness of the prismatic beam is EI. Use Castigiano s nd theorem ϕ C A C B x-axis ½ ½ z-axis Hans Weeman

2 Hans Weeman

3 Exercise 5 : Cantiever beam 4 In the figure a hinged beam with a distributed oad q(x) is presented with a fuy camped support at A. M=34 knm q max =7 kn/m EI beam =36000 knm A S B 6,0 m 3,0 m x The moment distribution can be described as a function of x [m]: M ( x) = 9x 4,5x 0,5x 3 [knm] Hint : Have a good ook at the coordinate system! A dispacement starting at S to the eft is positive! Questions: a) Draw the bending moment diagram for the beam. b) Determine the defection of the hinge S using the work method with a unit oad. Hans Weeman

4 Exercise 6: Strengthened beam In the figure a strengthened beam with bending stiffness EI is presented with tension bars with ength of which the force distribution has to be determined. The axia stiffness of the beam is infinite and the axia stiffness of a bars is EA. α cosα cosα sinα Questions: a) Determine the static indeterminacy of this construction. b) Describe how you wi sove the compatibiity conditions which are associated to the static redundant of the structure using Castigiano s theorem. c) Sove the static redundant using the described method and give an expression using the parameters, EI,, EA andα. Hans Weeman

5 Exercise 0 : Beam with a coupe Appy Castigiano: Exercise 1: Non-inear oad-dispacement diagram Aternative (use area s): paraboa Hans Weeman

6 Exercise : Horizontay oaded frame ind support reactions and moment distribution: The sign is not reevant The order of differentiation and integration can be changed to save time Hans Weeman

7 Exercise 3 : Cantiever beam ind the defection at B with Castigiano s second theorem. The beam is prismatic and has a bending stiffness EI. Soution: Give both forces a unique name: substitute the vaue Hans Weeman

8 Exercise 4 : Cantiever beam 3 ind the rotation at C due to a oad at B. The beam is prismatic and has a bending stiffness EI. Soution: Appy a coupe T at C in the direction of the requested rotation. Substitute T = 0 in the answer obtained. Substitute T = 0 to obtain the fina answer: The coupe T is ony a dummy so therefore this method is aso referred to as the dummy oad method Hans Weeman

9 Exercise 5 : Cantiever beam 4 a) The moment distribution is presented in bue in the graph to the right. M=34 knm q max =7 kn/m b) To find the dispacement the work method with unit oad is used: w = 0 m( x) M ( x) dx EI In this M(x) is the moment distribution due to the actua oading, and m(x) is the moment distribution due to a unit oad appied at the hinge for which the defection is asked for. This graph is presented in red. A EI beam =36000 knm S B 6,0 m 3,0 m 34 knm x The tabe from the notes to find the soution for this integra can not be used so an anaytica soution is required. However part of the distribution m(x) is zero so this reduces the compexity. Ony part AS from 0 to 6 m has to be evauated : M(x) 6 knm m(x) 1 kn 3 883,6 w = ( x) ( 9x 4,5x 0,5x )dx = = 80,1 mm EI EI Hans Weeman

10 Exercise 6 : Strengthened beam The beam is simpy staticay indeterminate. Assume the compressive force D in the vertica strut (two-force member) as the staticay unknown. A forces in the truss system under the beam can be expressed in terms of this unknown D. With these forces the compementary strain energy can be expressed in terms of the unknown D. or the beam the moment distribution can be obtained incuding the concentrated oad from the vertica strut. N α D D N N N = D sin α D N cosα A bars have their own stiffness k from which the strain energy in the strengthened part can be obtained: = 1 N with: 1 ; + D EA EA v 1 = = sin E k k k k α The compementary energy stored in bending can be found with the moment distribution in the beam due to the oad and the staticay unknown D: ( ) 3 3 D cos cosα cosα α M 1 E = dx = ( ( D) x) dx = 1EI v EI 0 0 At mid span the defection of the beam (in the direction of ) must be the same as that of the strengthening sub structure (in the direction of D) resuting in: dev 1 dev = dd d 3 3 D D sinα ( D) cos α + = EAsin α EA 6EI 3 EA cos α sin α D = 3 3 3EI + 6EI sin α + EA cos α sin α Hans Weeman

Module 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method. Version 2 CE IIT, Kharagpur

Module 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method. Version 2 CE IIT, Kharagpur odue 2 naysis of Staticay ndeterminate Structures by the atri Force ethod Version 2 E T, Kharagpur esson 12 The Three-oment Equations- Version 2 E T, Kharagpur nstructiona Objectives fter reading this