Week 4 Lecture - Beam Deflection & Analysis (Sept 15, 2014)

8/11/2019 Week 4 Lecture - Beam Deflection & Analysis (Sept 15, 2014)

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Aer E 3 : Aerospace Structures Laboratory

Week 4 Lecture:

Linear Analysis of a Beam

by Direct Stiffness Method

September 15, 2014

In this course, we consider uniform elastic materials

A beam is a long straight structure of elastic members

Subjected to transverse loads and bending moments Small displacement causes small change in member length, and hence

negligible axial force

All loads, reactions and deformations lie in the plane of symmetry of

its cross-section, with all the forces perpendicular to its centroidal axis

Elastic material:

transverse and bending deformations allowed

Linear Analysis by Direct Stiffness Method

Direct stiffness method is the most popular matrix method for

analyzing discrete structures like trusses, beams and frames

Concept of “unit response”:

for each member of a structure, there is this force per unit

displacement defined as stiffness coefficient, kij, which

represent the force at the location acting in the direction of member end force Q i required, along with other end forces, to

cause a unit value of displacement u j, while all other end

displacements are zero

In stiffness method, these kij’s serve as the universal “scalingfactors” for all members’ forces in a structure

Linear Analysis by Direct Stiffness Method (cont’d)

Stiffness method works by first identifying where the displacement

unknowns (called degrees-of-freedom) are located in the frame structure

Second, the structure is conceptually broken down to individual bar

members in their own local coordinates, and a local stiffness matrix is

determined for each member

The global stiffness matrix is then assembled in the global coordinates bysumming up the contributions from individual members. External joint

loads (including the equivalent loads resulted from loads not applied at

joints) are transformed to the global coordinates as well.

Solve for the unknown displacements from the global stiffness matrix

equation

From the now known displacements, reverse the process to calculate the

individual member’s internal forces and displacement as well as stresses

and strains 4

Naming and Coordinate Systems Setup

X1 2 3

X-Y: one global coordinate system for entire beam; origin is usuallyset at left end

: joint ID

: member ID

1 2 3 4

Structure Degree-of-Freedom

Every free joint has two degree-of-freedoms (DOF, allowing vertical

translation and rotation). Constrained joints have reduced DOF,depending on the constraint condition

This beam has 6 DOF’s = 2(2 DOF’s per joint) x 4(4 joints in frame) – 2 x 1(fixed end at joint 1 allows no displacement)

1 2 3 41 2 3

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Forces, Displacements and Stiffness

in Local Coordinates

Q 4x1 = k4x4 u4x1

Q : end forces

u: member end displacements

k: local stiffness

deformu3

Small displacement causes small

change in member length, and

hence negligible axial force

1 2 3 41 2 3

Local Stiffnessdue to unit displacement u1

V (=-k11)

Consider a member section x from left end,

Local Stiffnessdue to unit displacement u1 (cont’d)

2d y M

EI dx=Recall beam deflection eq. . Integrate once and twice to arrive at

Boundary conditions (1) x=0, y=1 and dy/dx=0 and (2) x=L, y=dy/dx=0lead to c1=0 and c2=1 and

21 11 1

dy xk x k c

= − + +

Local Stiffnessdue to unit displacement u2

Again use beam deflection eq. . Integrate once and twice to arrive at2

EI dx=

Boundary Conditions: At end b: x=0, dy/dx=1 and y=0.

22 12 1

dy xk x k c

= − + +

1 21 0andc c= =

Local Stiffnessdue to unit displacement u2 (cont’d)

At end e: x=L, dy/dx=0 and y=0 and we obtain

L Lk k L

= − + +

Local stiffness matrix in its entirety:

Local Stiffness Matrix

Derivations can be done similarly for other cases.

Transformation Between Global and

Local Coordinate Systems

Here we only consider beam in horizontal position, so

Member local coordinates ~ member global coordinates

We will need rotation later in frame analysis

Local Fixed-End External Forces

External forces can be applied not only at joints but also at members.

The latter, called fixed-end forces Q f , are conceptually generated as the

member were fixed at ends. Equivalent forces at the joints (resulted

from the reactions at fixed ends) are then added to the member’s end

forces

Q = ku + Q f

equivalent loads added

to member’s end forces Q f3Q f1

Q f2 Q f4

reactions at the fixed

ends become equivalent

loads at joints

external loads applied to member

member ends fixed and reactions

to the external loads generatedW

Local Fixed-End External Forces (cont’d)

Source: reference 1

Assembly of the Structure Stiffness Matrix

Using Member Code Numbers

A beam with loads at joints and at members

2 DOF’s at joint 2: 2 unknown

displacements d’s and 2external joint loads P’s

(resulting from all applied loads)

Corresponding FBD

Source: reference 1

Only 1 DOF at joint 3 (since it

has support from below): 1unknown rotation d3 and 1

external joint load P3 (resulting

from all applied loads)

Using Member Code Numbers (cont’d)

Matching member end displacements u’s with joint displacements d’s

Breaking down to FBDs of individual members or joints

Superscripts (1), (2) or

(3) denote member

Source: reference 1

Super i denotes member

(= 1, 2 or 3)

Expanding the equations for member

end forces (shown on the right) yields:

source: reference 1

see pages 8&15

Plugging these into the equation for P, from the previous slide,

gives us the desired structure stiffness relationships

P=Sd + Pf

with global beam Stiffness matrix S

Sum of equivalent joint loads

In matrix form

source: reference 1

Sum of stiffness of both

members at the free joints

Solving Beam Stiffness Assembly

Unknown displacement d can be solved from the beam stiffness

matrix equation P = Sd + Pf .

From d, we can get u by matching member end displacements.

From Q=ku+Q f to solve for Q , both in local coordinates.

Given the four components of Q we can further express the shear

forces and bending moments as functions of position along the

member. Member’s stress and strain can then be calculated from

these forces and moments

A Worked Example1000lb

1250ft-lb

60ft40ft

E=29000ksi, A=11.8in2

, I=310in4

60ft40ft

Degrees of Freedom: Unknown Displacements 3,4

A Worked Example (cont’d)member 1:

tracking with node number

corresponds to unknown d’s

L = 40’*12 in/ft = 480in, E=29000ksi, A=11.8in2, I=310in4

1 2 3 4

Warning: here only finite digits are displayed 23

A Worked Example (cont’d)member 2:

tracking with node number

L = 60’*12 in/ft = 720in, E=29000ksi, A=11.8in2, I=310in4

3 4 5 6

Warning: here only finite digits are displayed 24

A Worked Example (cont’d)

Assembly of Global Beam Stiffness Matrix

The rows and columns corresponding to the known displacementsare crossed out here to show us the structure stiffness matrix S.

1 2 3 4 5 6

No fixed-end forces: Pf = 0

Warning: here only finite

digits are displayed

Now that the unknown displacements have been found, we simply plug

these values back into F-F f =KU to find the end reaction forces in the

global system

The big K from page 25

U: no other joint

can move except3&4 (page 22)

Lab and Homework this Week:

Lab 3 – Beam Deflection and Analysis

Prelabs due in lab sessions Thursday Sept. 18 or Tuesday, Sept. 23

Homework 3 due online 11:59pm, Wednesday, Sept. 24

lab report due online 11:59pm, Friday, Sept. 26 (Thursday sections)

and Sunday, Sept. 28 (Tuesday sections) 28

Reference

Aslam Kassimali, Matrix Analysis of Structures,

second edition, Cengage Learning, 2012, Ch. 5

Week 4 Lecture - Beam Deflection & Analysis (Sept 15, 2014)

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