Finite Element Method: Element

Beam Element

¨ Beam: A long thin structure that is subject to the vertical loads. A beam shows more evident bending deformation than the torsion and/or axial deformation.

¨ Bending strain is measured by the lateral and the rotation -> Lateral deflection and rotation determine the number of DoF (Degree of Freedom).

Mechanics and Design SNU School of Mechanical and Aerospace Engineering : Beam Element

¨ Sign convention

1. Positive : Anti-clock wise rotation.

2. Positive load: y$ -direction.

3. Positive displacement: y$ -direction.

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element

¨ The governing equation dVdV -w-wdxdx$ +ˆ -dV dV= 0 =0or or w w= =- - dxdx$ ˆ dMdM VVdxdx$ ˆ+-dM dM= =0 0or or VV == dxdx$ˆ

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element

¨ Relation between beam curvature (k ) and bending moment

1 M d 2v$ M k = = or = r EI dx$ 2 EI

d 2v$ Curvature for small slope (q = dv$ / dx$ ): k = ¨ dx$ 2

r : Radius of the deflection curve, v$: Lateral displacement function along the y$ -axis direction

E : Stiffness coefficient, I : Moment of inertia along the z$-axis

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element

Solving the equation with M ,

d 2 F d 2v$I EI = - wax$f dx$ 2 HG dx$ 2 KJ

When EI is constant, and force and moment are only applied at nodes,

d 4v$ EI = 0 dx$ 4

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element

Step 1: To select beam element type

Step 2: To select displacement function

Assumption of lateral displacement

3 2 vˆ() xˆ = a1 xˆ + a 2 xˆ + a 3 xˆ + a 4

- Complete 3-order displacement function is suitable, because it has four degree of freedom (one lateral displacement and one small rotation at each node)

- The function is proper, because it satisfies the fundamental differential equation of a beam.

- The function satisfies continuity of both displacement and slope at each node.

$ $ $ ˆ Representing v$ with functions of d1y , d2 y , f1, f2 $ v$ (0) = d1y = a4

dv$a0f $ = f1 = a3 dv$

$ 3 2 v$aLf = d2 y = a1 L + a2 L + a3 L + a4 $ dvaLf $ 2 = f 2 = 3a1L + 2a2 L + a3 dx$

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element

$ $ $ ˆ Replacing a1~a4 with d1y , d2 y , f1, f2

2 1 v$ = L d$ - d$ + f$ +f$ O x$ 3 3 e 1y 2 y j 2 d 1 2 i NM L L QP 3 1 + L- d$ - d$ - 2f$ +f$ O x$ 2 +f$ x$ + d$ 2 e 1y 2 y j d 1 2 i 1 1y NM L L QP

$ Representing it in matrix form, v$ = N odt $ Rd1y U | f$ | d$ = | 1 | o t S $ V d N = N1 N2 N3 N4 | 2 y | |f$ | T 2 W

1 1 N = 2x$ 3 - 3x$ 2 L + L3 N = x$ 3 L - 2x$ 2 L2 + x$ L3 1 L3 c h 2 L3 c h 1 1 N = -2x$ 3 + 3x$ 2 L N = x$ 3 L - x$ 2 L2 3 L3 c h 4 L3 c h

where N1, N2, N3 , N4 : shape functions of the beam element

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element

Step 3: To define strain-displacement relation and stress-strain relation

Assume that the equation of strain-displacement relation is valid

du$ dv$ d 2v$ $ $ $ $ $ $ $ e x bx, yg = , u = - y => e x bx, yg = - y 2 dx$ dx$ dx$

Basic assumption: Cross-section of the beam sustains its shape after deformation by bending, and generally rotates by degree of bdv$ / dx$g .

Bending moment-lateral displacement relation and shear force-lateral displacement relation

d 2v$ d 3v$ $ $ $ mbxg = EI 2 V = EI 3 dx$ dx$

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element

Step 4: To derive an element stiffness matrix and governing equations by

¨ Element stiffness matrix and governing equations

d 3v$ 0 EI f$ = V$ = EI b g = 12d$ + 6Lf$ -12d$ + 6Lf$ 1y dx$3 L3 e 1y 1 2 y 2 j d 2v$ 0 EI m$ = - m$ = - EI b g = 6Ld$ + 4L2f$ - 6d$ + 2L2f$ 1 dx$2 L3 e 1y 1 2 y 2 j d 3v$ L EI f$ = -V$ = - EI b g = -12d$ - 6Lf$ +12d$ - 6Lf$ 2 y dx$3 L3 e 1y 1 2 y 2 j d 2v$ L EI m$ = m$ = EI b g = 6Ld$ + 2L2f$ - 6Ld$ + 4L2f$ 2 dx$2 L3 e 1y 1 2 y 2 j

- Matrix Form

$ 12 6L -12 6L $ R f1y U L ORd1y U L 12 6L -12 6L O | | 2 2 | | 2 2 m$ EI M 6L 4L -6L 2L P f$ EI M 6L 4L -6L 2L P | 1 | = M P| 1 | $ M P S V 3 S V k = 3 f$ L -12 -6L 12 -6L d$ L -12 -6L 12 -6L | 2 y | M P| 2 y | M P |m$ | M 6L 2L2 -6L 4L2 P|f$ | M 6L 2L2 -6L 4L2 P T 2 W N QT 2 W N Q

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element

Step 5: To constitute a global stiffness matrix using boundary conditions

Assemble example

Assume EI of the beam element is constant.

1000 lb load and 1000 lb - ft moment are applied at the center of the beam.

Assume load and moment were only applied at nodes.

Left end of the beam is fixed and right end is pin-connected.

The beam is divided into two elements (nodes 1, 2, and 3 as shown above figure).

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element

d1y f1 d2 y f 2 d2 y f 2 d3 y f 3 L 12 6L -12 6L O L 12 6L -12 6L O EI M 6L 4L2 -6L 2L2 P EI M 6L 4L2 -6L 2L2 P k (1) = M P k (2) = M P L3 M-12 -6L 12 -6LP L3 M-12 -6L 12 -6LP M 2 2 P M 2 2 P N 6L 2L -6L 4L Q N 6L 2L -6L 4L Q

R F1y U L 12 6L -12 6L 0 0 O Rd1y U | M | M 6L 4L2 -6L 2L2 0 0 P | f | | 1 | M P | 1 | |F2 y | EI M-12 -6L 12 +12 -6L + 6L -12 6L P |d2 y | = ´ SM V L3 M 6L 2L2 -6L + 6L 4L2 + 4L2 -6L 2L2 P Sf V | 2 | M P | 2 | |F3y | M 0 0 -12 -6L 12 -6LP |d3y | | M | M 0 0 6L 2L2 -6L 4L2 P | f | T 3 W N Q T 3 W

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element

Boundary conditions and constrains at the node 1(fixed) and node 3(pin-connected) are

f1 = 0 d1y = 0 d3y = 0

R-1000U L24 0 6L ORd2 y U EI | 1000 | = M 0 8L2 2L2 P|f | S V L3 M PS 2 V (5.2.5) | 0 | 6L 2L2 4L2 | f | T W NM QPT 3 W

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element

Example: Beam analysis using direct stiffness method

Cantilever beam supported by roller at the center

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element

Load P is applied at node 1.

Length: 2L ,Stiffness: EI

Constrains: (1) Roller at node 2, (2) Fixed at node 3.

- Global stiffness matrix

d1y f1 d2 y f 2 d3 y f 3 L12 6L -12 6L 0 0 O M 2 2 P M 4L -6L 2L 0 0 P EI M 12 +12 -6L + 6L -12 6L P K = L3 M 4L2 + 4L2 -6L 2L2 P M P M 12 -6LP M 2 P N Symmetry 4L Q

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element

RF1y U L 12 6L -12 6L 0 0 ORd1y U |M | M 6L 4L2 -6L 2L2 0 0 P| f | | 1 | M P| 1 | |F2 y | EI M-12 -6L 24 0 -12 6L P|d2 y | = SM V L3 M 6L 2L2 0 8L2 -6L 2L2 PSf V | 2 | M P| 2 | |F3y | M 0 0 -12 -6L 12 -6LP|d3y | |M | M 2 2 P|f | T 3 W N 0 0 6L 2L -6L 4L QT 3 W

Boundary conditions d2 y = 0 d3y = 0 f 3 = 0

R-PU L12 6L 6L ORd1y U EI | 0 | = M6L 4L2 2L2 P|f | S V L3 M PS 1 V | 0 | 6L 2L2 4L2 |f | T W NM QPT 2 W

7PL3 3PL2 PL2 d1y = - f1 = f 2 = 12EI 4EI 4EI

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element

¨ Substituting the obtained values to the final equation, R 7PL3 U F 12 6L -12 6L 0 0 - R 1y U L O| 12EI | |M | M 6L 4L2 -6L 2L2 0 0 P| 3PL2 | | 1 | M P| | |F2 y | EI M-12 -6L 24 0 -12 6L P| 4EI | = 2 2 2 0 SM V L3 M 6L 2L 0 8L -6L 2L PS 2 V | 2 | M P| PL | |F3y | M 0 0 -12 -6L 12 -6LP| 4EI | |M | M 0 0 6L 2L2 -6L 4L2 P| 0 | T 3 W N Q| | T 0 W 5 F = - P M = 0 F = P 1y 1 2 y 2 3 1 M = 0 F = - P M = PL 2 3y 2 3 2

F1y = - P : Force at node 1

F2 y , F3y , M3 : Reacting forces and moment at nodes

M1 , M2 : Zero

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element

¨ Calculating local nodal loads

Force at the element 1.

When f$ = k$d$

R 7PL3 U $ |- | R f1y U L 12 6L -12 6L O 12EI | 2 | | m$ | EI M 6L 4L2 -6L 2L2 P 3PL | 1 | = M P| | S f$ V L3 -12 -6L 12 -6L S 4EI V | 2 y | M P| 0 | M 2 2 P 2 |m$ | 6L 2L -6L 4L | PL | T 2 W N Q| | T 4EI W $ $ f1y = - P m$1 = 0 f 2 y = P m$ 2 = - PL

Shear moment curve

Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element

Homework: Distributed load

¨ Equivalent force

Mechanics and Design SNU School of Mechanical and Aerospace Engineering