1. Solve Problem 1.3-3(c) 2. Solve Problem 2.2-2(b)

Transcription

1 . Sole Problem.-(c). Sole Problem.-(b). A two dimensional trss shown in the figre is made of alminm with Yong s modls E = 8 GPa and failre stress Y = 5 MPa. Determine the minimm cross-sectional area of each member so that the trss is safe with safety factor.5.

2 6 kn Pin joint 9 cm cm Roller spport Soltion: et s assme the cross-sectional areas be.5 cm. We will modify them later, if necessary. Element nmbers, node nmbers and the coordinate system sed here are referred to the following figre. N y E E N E N Connectiity table Elem N # N # E (N/cm ) A(cm ) (cm) θ l Element stiffness matri after applying transformation to the global coordinates: cos θ cos θsin θ cos θ cos θsin θ EA cos θsin θ sin θ cos θsin θ sin θ [ k ] = cos θ cos θsin θ cos θ cos θsin θ cos θsin θ sin θ cos θsin θ sin θ Element : [ k ] =

3 Element : Element : 5 5 [ k ] = [ k ] = The three element stiffness matrices are assembled to the global stiffness matri: [ K ] = Node is fied ( = = ), and Node is fied in the ertical direction ( = ). The known bondary conditions are denoted in the following matri eqation: R R y 5 5 = R y Deleting those rows and colmns corresponding to zero displacements, we hae = 8. The aboe eqation can be soled for nknown displacements, as { } = {..} cm Using the displacements and soling for the forces in each member sing Eq. (.5):

4 () e AE AE j i yj y i P = Δ = ( j i ) ( j i ) + () () () P = 5 kn; P = kn; P = 5 kn Now, the element force will remain constant een if the cross-sectional areas are changed becase the system is statically determinate. Ths, the minimm cross-sectional areas can be obtained from gien safety factor and failre stress. = = = N Y ma MPa S cm F Element : Element : Element : A A () P () 5 = = =.5 cm ma A () P () = = = cm ma () P () 5 = = =.5 cm ma Note that Element is a zero-force member. Ths, theoretically it can be remoed. Howeer, the trss will be nstable if member is remoed. In many designs, the minimm cross-sectional area is reqired.. It is desired to se FEM to sole the two plane trss problems shown in the figre below. Assme AE = 6 N, = m. Before soling the global eqations, [ K]{ Q} = { F }, find the determinant of [ K ]. Does [ K ] hae an inerse? Eplain yor answer. y, N, N

5 Soltion: After deleting rows and colmns corresponding to the fied DOFs, the aboe two trsses hae the following global matri eqations: Trss : Trss : = = After finding the determinant of the two stiffness matrices, the first set of bondary conditions are not alid for a static problem becase the determinant of [K] is zero, and therefore the matri cannot be inerted. The roller bondary condition of Trss allows rotation of the entire trss as a rigid-body. Ths no niqe soltion for displacements can be obtained. 5. In the D bar shown below, the temperatre of Element is o C aboe the reference temperatre, while Element is in the reference temperatre. An eternal force of, N is applied in the - direction (horizontal direction) at Node. Assme E = Pa, A = - m, and α = -5 / o C., N m m (a) Write down the stiffness matrices and thermal force ectors for each element. (b) Write down the global matri eqations. (c) Sole the global eqations to determine the displacement at Node. (d) Determine the forces in each element. State whether it is tension or compression. (e) Show that force eqilibrim is satisfied at Node Soltion: (a) Since the trss joints (nodes) are allowed to moe only in the horizontal direction, we can consider the two members as niaial bar elements, which hae two DOFs. The element stiffness matrices and thermal load ectors become ( ) k =, ( ) k =

6 () { T } = AEαΔ T = () f, { f T } = AEαΔ T = (b) The global strctral stiffness matri is Then, the FE eqations are K s = = R F = = + = R,,, (c) By deleting the first and last rows and colmns, we hae a scalar eqation for, as []{ } = {, } The soltion of the aboe eqation is =.5 mm. () () (d) Changes in element length: Δ =.5 mm, Δ =.5 mm. Using P = EA( Δ / αδ T), the element forces can be fond, as ( ) Δ P = AE( αδ T) = 5 = 5 N(tension) ( ) Δ P = AE( αδ T) = (.5.) = 5, N(compression) (e) At Node : F =5 5 + =