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Deflection of Statically Determinate Frames | Virtual Work Method

Structures deform when subjected to various conditions such as external loads, changes in temperature, differential settlement, or fabrication errors. Structural deformation is often considered in terms of rotation (slope) or translation (deflection).

The slope is the angle (in radians) that the elastic curve makes with the original neutral axis of the member, while deflection is the movement the structural member makes from the original neutral axis. In the design of structures, it is necessary to control the deflection for aesthetics and appropriate functionality (like cracking of finishes).

Statically determinate frames will deflect under the action of externally applied load, but will not be sensitive to indirect actions such as temperature difference and differential settlement. The deflection of statically determinate frames can easily be obtained using the virtual work method. For practical analysis of normal frames, the deflection due to shear can be ignored.

In beams and frames, bending moment is the major cause of deflection, while in trusses, axial forces are the major cause of deflection. Deformation due to shear is often very small and is neglected in most simple analyses. In the solved example that is downloadable in this post, the frame shown above is completely analysed, and the horizontal deflection at point B is obtained using the virtual work method. The analysis was carried out assuming linear elastic material response. Both the integration method (Mohr’s integral) and graphical method (Vereschagin’s diagram combination method) were used in the analysis.

Bending moment, shear force, and axial force contribute to the total deformation of a frame and the formula is given by;


δi = 1/EI∫Mm ds + k/GA∫Qq ds + 1/EA∫Nn ds ——- (a)

Where;
1/EI∫Mm d= Deflection due to bending moment
k/GA∫Qq ds = Deflection due to shear force
1/EA∫Nn d = Deflection due to axial force

Of all these, the highest contributor to deflection of statically determinate frames is the bending moment. Oftentimes, deformation due to shear force and axial forces are neglected, while in some analyses, they can be quite significant. The analysis below gives an example of the computation of deformation in a frame neglecting the shear and axial forces.

Hence displacement using Mohr’s integral is given by;


δi = 1/EI∫Mm ds ————– (b)


Where;
M = Moment due to externally applied load
m = Moment due to unit load applied at the point where the displacement is sought
EI = Flexural rigidity

The steps involved in the analysis of deflection of statically determinate frames can be summarised as follows:

  1. Analyse the frame completely and draw the final bending moment diagram due to externally applied load.
  2. Remove the externally applied load, and replace it with a virtual unit load, in the direction and location that the deflection is sought. For slopes, you place a unit moment at the point you wish to obtain the rotation.
  3. Analyse the frame completely under the action of the virtual load and draw the virtual bending moment diagram.
  4. Combine the bending moment equations or diagrams of the external and virtual loads and compute the value of the deflection.

Worked Example

For the frame that is loaded as shown below;
(i) Find the support reactions
(ii) Draw the internal stresses diagram
(iii) Obtain the horizontal displacement at point B using both graphical and integration methods. (EI = Constant);

Simply supported frame

Solution
The degree of static indeterminacy is given by;
RD = (3m + r) – 3n – S
m = 3 (three members)
r = 3 (three reactions)
n = 4 (four nodes)
S = 0 (no internal hinge)

RD = 3(3) + 3 – 3(4) = 0
This shows that the structure is statically determinate and stable.

(i) Support Reactions
Let ΣMD = 0; clockwise negative
(Ay × 6) + (5 × 3) – (10 × 62/2) = 0
Ay = 27.5 kN

Let ΣMA = 0; clockwise negative
(Dy × 6) − (5 × 3) – (10 × 62/2) = 0
Dy = 32.5 kN

Let ΣFX = 0;
-Dx + 5 = 0, therefore,
Bx = 5 kN

(ii) Internal Stresses

Section 1-1 (A-1B) (0 ≤ y ≤ 3)

section 1 1

Bending Moment
My = 0 —- (1)

Shear
Qy = 0

Axial
Ny + 27.5 = 0
Ny = −27.5 kN

NA – N1B = −27.5 kN

Section 2-2 (1UP – BB) (3.0 ≤ y ≤ 6.0)

2 2

Bending Moment
My = −5(y−3)
∴ My = −5y + 15 —- (2)

At y = 3m,
M1UP = −5(3) + 15 = 0

At y = 6m
MBB = -5(6) + 15 = −15 kNm

Shear
Qy + 5 = 0
Qy = −5 kN
Q1UP – QBB = −5 kN

Axial
Ny + 27.5 = 0
Ny = −27.5 kN
NA – N1B = −27.5 kN

Section 3-3 (BR – CL) (0 ≤ x ≤ 6.0)

3 3

Mx = (27.5x) – (10 × x2/2) − (5 × 3)
Mx = −5x2 + 27.5x – 15 —– (3)

At x = 0,
MBR = −15 kNm

At x = 6m
MCL = −5(62) + 27.5(6) – 15 = −30 kNm

∂Mx/∂x = Qx = −10? + 27.5

At ∂Mx/∂x = 0, bending moment is maximum
Hence, −10x + 27.5 = 0
x = 27.5/10 = 2.75m
Hence the maximum sagging moment occurs at 2.75m

Mmax = −5(2.752) + 27.5(2.75) – 15 = 22.8125 kNm

Shear
Qx = – 10x + 27.5

At x = 0
QBR = 27.5 kN

At x = 6m
QCL = 27.5 – 10(6) = −32.5 kN

Axial
Nx + 5 = 0
Nx = −5 kN
NBR – NCL = −5 kN

Section 4-4 (D – CB) (0 ≤ y ≤ 6)

4 4

Bending Moment
My = −5y —– (4)

At y = 0,
MD = 0

At y = 6m
MCB = −5(6) = −30 kNm

Shear
Qy − 5 = 0
Qy = 5 kN
QD – QCB = +5 kN

Axial
Ny + 32.5 = 0
Ny = −32.5 kN
ND – NCB = −32.5 kN

The internal stresses diagram of the frame is shown the figure below. The sign conventions are so important because of how they influence the results we obtain for the deflections. The internal stresses equations labelled (1), (2), (3), and (4) will be used for the integration method.

Final bending moment diagram

(iii) Horizontal displacement at point B
We will now remove the externally applied load, and place a unit horizontal load at point B of the structure. We will analyse the structure as usual and obtain the bending moment equations, and plot the bending moment diagram. In this case, we will neglect the axial and shear forces.

Deflection of statically determinate frame

Support Reactions
Let ΣMD = 0; clockwise negative
(Ay × 6) + (1 × 6) = 0
Ay = −1.0

Let ΣMA = 0; clockwise negative
(Dy × 6) − (1 × 6) = 0
Dy = 1.0

Let ΣFX = 0;
−Dx + 1 = 0, therefore, Bx = 1.0

Internal Stresses due to Virtual unit Load (Bending moment only)

Section A-B (0 ≤ y ≤ 6)
My = 0 (no bending moment on the section) —- (1 & 2)

Section B-C (0 ≤ x ≤ 6)
Mx = −x ————- (c)
At x = 0, MBR = 0
At x = 6m, MCL = −6

Section D-C (0 ≤ y ≤ 6)
My = −y —————- (d)
At y = 0, MD = 0
At y = 6m, MCB = −6

The bending moment diagram due to the virtual unit load at point B is shown below;

bnm virtual load

Deflection Using Mohr’s Integral

Therefore using the integration method, we can obtain the deflection at point B integrate section by section. Note that section A – B will be zero all through because in the bending moment diagram due to the unit load, there is no bending moment at the section and hence, everything there goes to zero;

δi = 1/EI ∫Mm ds

Therefore, horizontal displacement at point B is obtained by combining sections B – C and C – D

Mohrs integral

Deflection Using Vereschagin’s Rule (Graphical Method)


To achieve this we have to combine the diagrams for sections B-C and C-D for the external load and for the virtual load as shown below.

diagram combination

Realise that the shape in the original moment diagram for section B-C can be split as shown below;

splitting of diagrams

Some diagram combination equations are shown below;

diagram combination equations

Hence;

ghy

δB = (1/6 × 15 × 6 × 6) − (1/3 × 45 × 6 × 6) + (1/3 × 30 × 6 × 6) + (1/3 × 30 × 6 × 6)
δB = 270/EI metres

Download the final solved example in PDF here and feel free to ask questions.

Internal Forces in Frames Due to Temperature Difference

If a statically indeterminate structure is subjected to change in temperature, internal stresses are induced. This can be found in structures such as cooling towers and chimneys, where the internal and external temperatures are different. This implies that the temperature at the two extreme fibres of the material section are not the same and since the structure is restrained, internal stresses are developed due to thermal strain. We can use force method to obtain the magnitude of such stresses, in order to make appropriate designs.


Instead of the normal canonical equation for externally applied load, we now replace the free terms with deformation due to temperature change as shown below;

cannonical

The deformation of a structure at a point due to temperature difference is given by;

Delta%2B1t

Where;
α = coefficient of thermal expansion of the material
h = depth of the member
∆t = change in temperature
tav = average temperature for the member
∫M ̅ds = Area of the bending moment diagram for the member
∫N ̅ds = Area of the axial force diagram for the member

In summary, the steps to follow are:
(1) Determine the degree of redundancy in the structure, and write the appropriate canonical equation.
(2) Draw your basic or primary system as usual
(3) Obtain unit bending moment, shear, and axial force diagrams
(4) Calculate the unit displacement using Mohr’s integral or Vereschagin’s method
(5) Calculate the free terms of the canonical equation (displacement due to temperature difference)
(6) Solve the canonical equation
(7) Plot your internal stresses diagram.

SOLVED EXAMPLE


A portal frame is fixed at the column base A, and hinged at the far end of the beam C as shown above. The column of the frame has a cross-section of 30cm x 30cm while the beam has a depth of 60cm and width of 30cm. The temperature inside the frame (t1) is 50°C while the temperature outside (t2) is 21°C. Draw the bending moment, shearing force, and axial force diagram for the temperature difference action on the frame (E = 2.17 × 107 KN/m2, α = 11 × 10-6 /°C).


SOLUTION
Geometric properties and temperature data;
Moment of inertia of column IC = (bh3)/12 = (0.3 × 0.33)/12 = 6.75 × 10-4 m4
Moment of inertia of beam IB = (bh3)/12 = (0.3 × 0.63)/12 = 5.4 × 10-4 m4
We desire to work in terms of IC such that IC/IB = 0.125
Hence flexural rigidity of the column, EIC = (2.17 × 107) × (6.75 × 10-4) = 14647.5 KNm2

For the frame above;
∆t = t2 – t1 = 50 – 21 = 29°C; tav = (50 + 21)/2 = 35.5°C
For member AB, ∆t/h = 29/0.3 = 96.667; For member BC, ∆t/h = 29/0.6 = 48.333

Basic system
The next step in the analysis is to reduce the structure to a basic system, which is a system that must be statically determinate and stable. The frame is statically indeterminate to the second order, which means that are we are going to remove two redundant supports.

Degree of static indeterminacy is given by;
RD = (3m + r) – 3n – s ————————— (2)
So that RD = (3 × 2) + 5 – (3 × 3) – 0 = 2

By choice, I am deciding to remove the two reactive forces at support C of the frame to obtain the basic system as shown below;

Basic%2BSYSTEM

Analysis of case 1; X1 = 1.0, X2 = 0
The simple static analysis of the structure loaded with case 1 loading is as shown below in terms of the internal stresses diagram.

Unit%2BState%2BBending%2BMoment%2BDiagram
unit%2Bstate%2Bshear%2Bforce%2Bdigram

 Analysis of case 2; X1 = 0, X2 = 1.0
The simple static analysis of the structure loaded with case 1 loading is as shown below in terms of the internal stresses diagram.

State%2B2%2BBENDING%2BMOMENT
State%2B2%2Bshear%2Bforce

The appropriate canonical equation for this structure is therefore;

δ11X1 + δ12X2 + Δ1t = 0
δ21X1 + δ22X2 + Δ2t = 0

Computation of the influence coefficients
Influence coefficients are based on Mohr’s integral such that δi = 1/EI ∫Mmds. When we wish to handle this by using the graphical method (making use bending moment diagrams), we directly employ Verecshagin’s rule which simply states that when we are combining two diagrams of which one must be of a linear form (due to the unit load) and the other of any other form, the equivalent of Mohr’s integral is given by the area of the principal combiner multiplied by the ordinate which its centroid makes with the linear diagram. The rule can also work vice versa. This process has been adopted in this work.

δ11 (Deformation at point 1 due to unit load at point 1)
This is obtained by the bending moment of case 1 combining with itself. This is shown below.

d11

δ11 = (5 × 5 × 4) + (1/3 × 5 × 5 × 5 × 0.125) = 105.208

δ21 = δ12 (Deformation at point 2 due to unit load at point 1 which is equal to deformation at point 1 due to unit load at point 2 based on Maxwell’s theorem and Betti’s law)
This is obtained by the bending moment diagram of case 1 combining with bending moment diagram of case 2. This is shown below.

d12

δ21 = (1/2 × 5 × 4 × 4) = 40

δ22 (Deformation at point 2 due to unit load at point 2)
This is obtained by the bending moment diagram of case 2 combining with itself. This is shown below.

d22

δ22 = (1/3 × 4 × 4 × 4) = 21.333

Influence coefficients due to temperature difference

Case 1 (Take a good look at the bending moment and axial force diagrams)

Δ1t/EIC = α∆t/h ∫M ̅ds + αtav ∫N ̅ds
Considering the first term of the equation for members AB and BC = [(11 × 10-6 × 96.667 × 5 × 4) + (11 × 10-6 × 48.333 × (1/2) × 5 × 4)] = 0.02791
Considering the second term of the equation for member AB = (11 × 10-6 × 35.5 × 1 × 4) = 0.001562
Hence, Δ1t = α∆t/h ∫M ̅ds + αtav ∫N ̅ds = 0.02791 + 0.001562 = 0.029473EIC
Δ1t = 0.029473 × 14647.5 = 431.691

Case 2 (Take a good look at the bending moment and axial force diagrams)

Δ2t/EIC = α∆t/h ∫M ̅ds + αtav ∫N ̅ds
Considering the first term of the equation for members AB = (11 × 10-6 × 96.667 × (1/2) × 4 × 4) = 0.0085067
Considering the second term of the equation for member BC = (11 × 10-6 × 35.5 × -1 × 5) = -0.0019525
Hence, Δ2t = α∆t/h ∫M ̅ds + αtav ∫N ̅ds = 0.0085067 – 0.0019525 = 0.0065542EIC
Δ2t = 0.0065542 × 14647.5 = 96.0026

The appropriate canonical equation now becomes;

105.208X1 + 40X2 = -431.691
40X1 + 21.333X2 = -96.0026


On solving the equations simultaneously; X1 = -8.332 KN; X2 = 11.122 KN

The final value of the internal stresses is given by the equations below;
Mdef = M1X1 + M2X2
Qdef = Q1X1 + Q2X2
Ndef = N1X1 + N2X2

Final Moment (Mdef)
MA = (-8.332 × 5) + (11.122 ×4) = 2.828 KNm
MB = (-8.332 × 5) = -41.660 KNm
MC = hinged support = 0

final%2BBMD

Final Shear Force (Qdef)
QA – QBB = (11.122 × -1) = -11.122 KN
QBR – QCL = (-8.332 × 1) = -8.332 KN

FINAL%2BSHEAR%2BFORCE%2BDIAGRAM

Final Axial Force (Ndef)
NA – NBB = (-8.332 × 1) = -8.332 KN
NBR – NCL = (11.122 × -1) = -11.122 KN

FINAL%2BAXIAL%2BFORCE%2BDIAGRAM

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