Week 13 - Review 2013 (Part 2)
Chapter 7 Plasticity and Failure Strain energy density (strain energy per unit volume) ( SED) 1 U xx xx yy yy zz zz 2( xy xy yz yz zx zx ) 2 1 U ( 12 22 32 ) 2 ( 1 2 2 3 3 1 ) 2 E 2 1 1 K I 1 ( 1 2 )2 ( 2 3 )2 ( 3 1 )2 2 6 E
SED due to volume change caused by hydrostati c stress
Shape change due to shear stress, called distortion energy density (DED)
Tresca criterion Tresca observed that material flow seems to be along the direction of maximum shear stress. 1 Tresca stress: Tresca 1 3 2 Tresca criterion: 1 3 Y
Failure criterion (Ductile materials): 1 3
Y
Von Mises criterion: plastic yielding occurs when the distortion energy density equal or exceeds that of the same material under uniaxial tension.
Von Mises stress: vm
1
( 1 2 ) 2 ( 2 3 ) 2 ( 3 1 ) 2 2
Von Mises criterion ( 1 2 )2 ( 2 3 ) 2 ( 3 1 ) 2 2Y 2 2
2
2
Von Mises Failure criterion (Ductile materials): ( 1 2 ) ( 2 3 ) ( 3 1 ) Brittle materials There is no yield for brittle material in general.
1
The maximum normal stress theory
f
f is the failure normal stress
f is
1
The maximum normal strain theory
f
the failure normal strain.
Since from Hooke ’s law: 1
1 E
1 ( 2 3 )
1
1 ( 2 3 )
f
2Y 2
Week 13 - Review 2013 (Part 2)
Chapter 8 Finite Element Method 1D elements: Spring element, Bar element, Beam element and their combination Spring element Bar element Beam element
dof=2:
dof =2:
k k ui f i k k u f j j
1 ui f i L 1 1 u j f j
EA 1
dof =4
2D element: CST (T3), LST (T6), Q4, Q8 element Displacement Linear: u b1 b2 x b3 y v b4 b5 x b6 y (6 d.o.f.)
Shape function 1 N i ( i i x i y ) 2 A (i=1,2,3) 1 x1 y1 1 A 1 x2 y2 2 1 x3 y3
Strain Constant strain u xx b2 x v yy b6 y
N 1 (2 1)
Fully-linear xx b2 2b4 x b5 y
T3, 3-node constant strain triangle (CST) Quadratic: u b1 b2 x b3 y 2
2
b4 x b5 xy b6 y v b7 b8 x b9 y
T6 quadratic triangular element linear strain triangle ( LST )
N 3 (1 )
2(1 ) 1
b10 x 2 b11 xy b12 y 2 N 4 4 N 5 4 (1 ) (12 d.o.f.) N 4(1 )
yy b9 b11 x 2b12 y 2 xy b3 b5
b3 2b10 x 2b6 2b11 y
6
Bilinear u b1 b2 x b3 y b4 xy v b5 b6 x b7 y b8 xy
Linear Quadrilateral Element (Q4)
N 2 (2 1)
u v 2 xy y x b3 b5
(8 d.o.f.)
N 1 N 2 N 3 N 4
2
1 4 1 4 1 4 1 4
(1 )(1 ) ,
Half-linear xx
(1 )(1 ) , yy
(1 )(1 ) , 2 xy
(1 )(1 )
u x v y
b2 b4 y b7 b8 x
v x
u y
b6 b8 y b3 b4 x
Week 13 - Review 2013 (Part 2) Quadratic u b1 b2 x b3 y 7
3
4 y
b4 x 2
b5 xy b6 y
2
b7 x 2 y b8 xy 2 8
6 1 x
5
2
Quadratic Quadrilateral Element (Q8)
v
b9 b10 x b11 y
b12 x 2
b13 xy b14 y
b15 x
2
2
y b16 xy 2
N 1
N 2
N 3
N 4
4 1 4 1 4 1 4
N 5
(16 d.o.f.)
1
N 6 N 7 N 8
1 2 1 2 1 2 1 2
(1 )(1 )(1 ) (1 )(1 )(1 ) (1 )(1 )(1 ) (1 )(1 )(1 )
(1 2 )(1 )
Half-quadratic xx b2 x 2b4 x b5 y
2b7 xy b8 y 2 yy b11 y b13 y 2b14 y 2 b15 x 2b16 xy
xy b3 b5 x 2b6 y
b7 x 2 2b8 xy
(1 )(1 2 )
b10 x 2b12 x b13 y
(1 2 )(1 )
2b15 xy b16 y
2
(1 )(1 2 )
3D element: Tet (4-node and 10-node) and Brick (8-node and 20-node ) 4-node tet 10-node tet 8 node brick 20 node brick
3D elements dof
12
30
24
60
Example 17 (Quiz 2012). In the following bar – spring structure as shown, use finite element method to (1) Derive global equilibrium equation; (2) Determine displacement at the free end point O, u1, mid point u2 and reaction force in the right hand side wall (point C). (3) Sketch the displacement distribution in the x-coordinate. Assume that EA=1, L=1, k s=6. Soln: Step 1: Elemental stiffness matrices k k 1 EA 1 1 1 1 K 1 1 L 1 1 1 1 , k k 1 1
k s k s 6 6 6 6 k k s s
A
K 2
k K 3 s k s
,
P
EA=1
1m
Step 2: Expanded elemental stiffness matrices 0 1 1 0 0 0
k s
4P
O
k s 6 6 k s 6 6
B
C x
k s 1m
0 0 0 K 1 1 1 0, K 2 0 6 6, K 3 0 6 6 0 0 0 0 6 6 0 6 6 0 u1 F 1 1 1 Step 3 (sub-question (1)): Global FE equation; K u F : 1 13 12 u2 F 2 0 12 12 u3 F 3 Step 4: Apply boundary conditions, u3 0 0 u1 F 1 1 1 1 13 12 u F 1 1 u1 F 1 P 2 2 1 13 u F 4 P 2 2 0 12 12 0 F 3
3
Week 13 - Review 2013 (Part 2)
1 1 u1 P u1 u2 P 1 13 u 4 P 2 u1 13u2 4 P u1 3 P / 4 Displacement vector: u P / 4 2 u3 0
u1 3 P / 4 u2 P / 4
Step 5 Plot displacement functions:
u( x) nk 1 N k ( )uk N i ( )ui N j ( )u j (1 )ui u j u
, in which x / L Element 1 (AB): local Cartesian coordinate
3P/4
x
x (1) is the same as global Cartesian, i.e.
x (1)
-P/4
x
u AB ( x) (1 x (1) / L)u A ( x (1) / L)u B 1 x(3 P / 4) x( P / 4) 3 P / 4 Px Element 2 (BC): local Cartesian x ( 2 ) differs from global Cartesian, i.e. x ( 2 ) x L x 1
u BC ( x) (1 x ( 2) / L)u B ( x ( 2) / L)uC 1 ( x 1)( P / 4) x 1(0) Px / 4 P / 2 Step 6: Reaction force at the wall (use the dropped first equation) 3 P / 4 RC F 3 0
12 12 P / 4 3P 0
EA=1
A P
B
k s
C
Stopper
O
(4) If remove the load at node B and consider a stopper k s 1m 1m on the right hand-side as shown. Determine the displacements and reaction from the stopper and C. 0 u1 P 1 1 1 13 12 R 1 1 u1 P u1 P u1 P 1 13 R u 13 R 2 1 2 2 R2 12 P 0 12 12 0 F 3
P Displacement vector: RC F 3 0 12 12 12 0
u1 P u2 u3 0
Example 18 (Quiz 2012): Two 3-node triangular elements are used to model a plane stress problem. The stress distributions are shown below. Which are the possible correct plots and why?
yy 1
x
1
x
2
yy
yy
4
1
2
2
4
4 3
(a)
y
3
y
(b)
3
y (c)
a) should be the answer. Note that these two 3-node triangular elements are the constant strain triangular (CST) elements (need mention this, otherwise lose 3 marks) , which have constant strain and stress.
4
x
Week 13 - Review 2013 (Part 2) Example 19 (Quiz 2013) A uniform beam L=1 L=1 ( EI = 1) is fully clamped at the left end (point A, node 1), and supported at both the mid A EI=1 EI=1 B span (point B, node 2) and the right end 1 2 (point C, node 3) by rollers. A bending M moment (M = 14 Nm) is applied at the midspan as shown. a. Express the global equilibrium equation using a finite element formulation. b. Determine the slopes at the mid-span and at the right end. c. Find all reaction forces and moments. d. In order to make 3=1, what moment loading M 3 is needed? Soln Step 1: Element stiffness matrices 6 L 12 6 L 12 6 12 6 12
6 4 K 1 K 2 3 6 L 12 6 L 12 6 L 12 2 6 L 4 L2 6 2 6 L 2 L EI 6 L
4 L2
6 L
2 L2
6 12
6
C 3
6 4 2
Step 2: Install the Global FE equilibrium equation 6 12 6 0 0 v1 12
F 1 6 M 4 6 2 0 0 1 1 12 6 24 0 12 6 v2 F 2 K u 6 2 0 8 6 2 2 M 2 0 0 12 6 12 6 v3 F 3 0 6 2 6 4 3 M 3 0
Step 3: Apply Boundary condition and external loadings 6 12 6 0 0 0 F 1 12 6 12 6 0 0 0 F 1 12 6 0 M 6 6 2 4 0 0 0 6 2 4 0 0 M 1 1 12 6 24 0 12 6 0 F 2 12 6 24 0 12 6 0 F 2 6 2 0 8 6 2 6 2 0 8 6 2 14 2 14 2 0 0 0 12 6 12 6 0 F 3 0 12 6 12 6 0 F 3 6 4 3 0 0 6 2 6 4 3 0 0 0 6 2 0 8 2 2 14
0 2 4 3
2 2 Step 4: Solve for the unknowns: 3 1 Step 5: Reaction forces and moments 6 12 6 0 0 0 F 1 F 1 6 0 0 12 6 12 6 0 M M 2 2 4 6 2 4 0 0 0 0 1 1 12 6 24 0 12 6 0 F 2 ; F 2 0 6 2 0 6 2 6 M 8 2 8 2 6 2 0 8 6 2 14 2 2 3 1 14 6 0 6 6 0 12 6 12 6 0 F 3 F 3 6 6 6 4 3 0 M 3 2 4 4 0 6 2 0 2 0 Step 6: Add boundary condition of 3=1 into the global FE equilibrium equation
5
x
Week 13 - Review 2013 (Part 2)
L=1
A
L=1
1
M 2
C
EI=1
B
EI=1
2
M 3
3
x
6 12 6 0 0 0 F 1 12 M 6 4 0 0 0 6 2 1 12 6 24 0 12 6 0 F 2 6 2 0 8 6 2 2 14 0 0 12 6 12 6 0 F 3 0 6 2 6 4 1 M 3 0
6 12 6 0 0 0 F 1 12 6 4 6 2 0 0 0 M 1 12 6 24 0 12 6 0 F 2 6 2 0 8 6 2 2 14 0 0 12 6 12 6 0 F 3 0 6 2 6 4 1 M 3 0 8 2 2 14 2 12 / 8 1.5 . From the last equation: M 3 2 2 4 2 1.5 4 7
Example 20 (Quiz 2011): If one Gaussian point was used for the 2D Q8 element, determine the displacements at this Gaussian point if the nodal displacements y C(4,6) u 1 0 1 1 0 1 2 1 are given as v 2 1 0 1 1 0 1 0 D(0,3) Soln: The displacement functions as per Q8’s shape functions N i and nodal displacements ui and vi can be expressed as: 4
u N i ui , i 1
x
4
v N i vi
A(0,0)
i 1
B(6,0)
8
u( , ) u(0,0) N i ui N 1u1 N 2u2 N 3u3 N 4u4 N 5u5 N 6u6 N 7u7 N 8u8 i 1
0.25 (1 0)(1 0)(1 0 0) 1 0.25 (1 0)(1 0)(1 0 0) 0 0.25 (1 0)(1 0)(1 0 0) (1) 0.25 (1 0)(1 0)(1 0 0) (1) 0.5 (1 02 )(1 0) (0) 0.5 (1 0)(1 02 ) (1) 0.5 (1 02 )(1 0) (2) 0.5 (1 0)(1 02 ) (1) u(0,0) 2.25 8
v( , ) v (0,0) N i vi N 1v1 N 2 v2 N 3v3 N 4 v4 N 5v5 N 6v6 N 7 v7 N 8v8 i 1
0.25 (1 0)(1 0)(1 0 0) 2 0.25 (1 0)(1 0)(1 0 0) 1 0.25 (1 0)(1 0)(1 0 0) (0) 0.25 (1 0)(1 0)(1 0 0) (1) 0.5 (1 02 )(1 0) (1) 0.5 (1 0)(1 02 ) (0) 0.5 (1 02 )(1 0) (1) 0.5 (1 0)(1 02 ) (0) v(0,0) 0.5 Example 21 (Quiz 2012): For the Q4 element as shown in the Cartesian coordinate system, the nodal displacement of the element that was generated from the FEA is u 0,0, 0,0, 2,2, 1,1T . Use shape function to calculate the displacement (u, v) at the mid point E on side BE. Soln:
u
4
0, 0, 2, 1T ,
u N i ui , i 1
v
y
C(4,6)
D(0,3) E(4,3) x
0, 0, 2, 1T A(0,0)
4
v N i vi
B(4,0)
i 1
N 1 (1 )(1 ) / 4 , N 2 (1 )(1 ) / 4 , N 3 (1 )(1 ) / 4 , N 4 (1 )(1 ) / 4
6
Week 13 - Review 2013 (Part 2) At the mid point E: 1, 0 4
1
i 1
4
u N i ui
(1 )(1 )u1
1 4
(1 )(1 )u2
1 4
(1 )(1 )u3
1
1
1
1
4
4
4
4
1 4
(1 )(1 )u4
(1 1)(1 0)(0) (1 1)(1 0)(0) (1 1)(1 0)(2) (1 1)(1 0)(1) 1 4
1
i 1
4
v N i vi
(1 )(1 )v1
1
1 4
(1 )(1 )v2
1
1 4
(1 )(1 )v3
1
1
4
4
1 4
(1 )(1 )v4
(1 1)(1 0)(0) (1 1)(1 0)(0) (1 1)(1 0)(2) (1 1)(1 0)(1) 1
4 4 Displacement: (1.0, 1.0)
y
Example 22 (Quiz 2011) if four Gaussian points are used in a Q4 C(4,4) D(0,4) element, determine the coordinates of Gaussian point G1 in G4 G3 Cartesian coordinate system of iso-parametric element Q4 as shown. Compute Jacobian matrix [J] and its determinant |J|. Soln G1 G2 Step 1: Node numbering: Node (1,2,3,4) = Node (A,B,C,D) x A(0,0) B(4,0) Step 2: Coordinate in terms of Shape functions 1 1 1 1 N 1 (1 )(1 ) , N 2 (1 )(1 ) , N 3 (1 )(1 ) , N 4 (1 )(1 ) 4 4 4 4 Cartersian coordinate ( x, y) in terms of natural coordinate (, ) 4 1 1 1 1 x N i xi (1 )(1 ) x1 (1 )(1 ) x2 (1 )(1 ) x3 (1 )(1 ) x4 4 4 4 4 i 1
* * * *
1
1
1
1
4 1
4
4
4
(1 )(1 ) x A (1 )(1 ) x B (1 )(1 ) xC (1 )(1 ) x D 1
1
1
4
4
4
(1 )(1 )0 (1 )(1 )4 (1 )(1 )4 (1 )(1 )0
4 x( , ) 2 2 4
1
i 1
4
y N i yi
(1 )(1 ) y1
1 4
(1 )(1 ) y2
1 4
(1 )(1 ) y3
1
1
1
1
4 1
4
4
4
1 4
(1 )(1 ) y4
(1 )(1 ) y A (1 )(1 ) y B (1 )(1 ) yC (1 )(1 ) y D 1
1
1
(1 )(1 )0 (1 )(1 )0 (1 )(1 )4 (1 )(1 )4
4 4 4 4 y( , ) 2 2 Thus G1 coordinate: x( 0.5774, 0.5774) 2 2 2 2 (0.5774) 0.8452
y( 0.5774, 0.5774) 2 2 2 2 (0.5774) 0.8452 Step 3: Jacobian matrix x y 2 2 2 2 2 0 J x y 2 2 2 2 0 2 Jacobian determinant: det J J 2 2 4
7
Week 13 - Review 2013 (Part 2) Example 23 (Exam 2009) Given: P = 50 kN, k = 200 kN/m, L = 3 m, E = 210 GPa, I = 2 10-4 m4. Find: Deflections, rotations and reaction forces. Soln: This is a combined problem of beam and spring elements Step 1: The system has 4 nodes as well as 2 beam elements and 1 spring element. k ui f i which is related to nodes k k u f j j
Step 2: Spring element has stiffness matrix is k
#3 and #4 with displacement v3 and v4 6 L 12 6 L v2 f 2Y 12 6 L 12 6 L v1 f 1Y 12 Beam 1: EI 6 L 4 L2 6 L 2 L2 1 m1 Beam 2: EI 6 L 4 L2 6 L 2 L2 2 m2 3 6 L v3 f 3Y L 12 6 L 12 L3 12 6 L 12 6 L v2 f 2Y 2 2 2 2 6 L 2 L 6 L 4 L 2 m2 6 L 2 L 6 L 4 L 3 m3
Step 3: The global FE equations can be assembled as (where k ' L3k /( EI ) ) Beam element 1
Beam element 2 spring element
Step 4: Apply the boundary conditions v1 1 v2 v4 0, M 2 M 3 0, F 3Y P
8 L2 6 L 2 L2 2 0 The reduced FE equation becomes: EI 6 L 12 k ' 6 L v3 P 3 L 2 6 L 4 L2 3 0 2 L 2 3 0.002492rad 2 PL Step 5 Solve for the reduced FE equation: v3 7 L 0.01744m EI (12 7k ' ) 3 9 0.007475rad
Step 6: Reaction force and moment can be found from the eliminated equations: F 1Y 69.78kN M 69.78kN m Free-body 1 F 2Y 116.2kN F 4Y 3.488kN
diagram can be drawn as
8
