# Solved Example on the Analysis of Internal Stresses in Frames Due to Settlement of Support In this post, an indeterminate frame whose support has been subjected to an indirect action of differential settlement has been presented and fully solved. If any statically indeterminate structure is subjected to settlement of supports, then internal forces arise in the members of the structure.

Analysis of such structures may be effectively performed by the force or stiffness method in canonical form. In the paper downloadable in this post, the frame loaded as shown below is subjected to a differential settlement of 25mm at fixed support A. The internal stresses induced were calculated using force method, stiffness method, and slope deflection method.

The portal frame shown below is fixed at the column base A, and hinged at point C as shown in Figure 1.1. The column of the frame has a square cross-section of 30cm x 30cm while the beam has a depth of 60cm and width of 30cm. The foundation of support A settles vertically by 25mm. Draw the bending moment, shearing force, and axial force diagram due to the differential settlement action on the frame (E = 2.17 × 107 KN/m2).

SOLUTION
1.0 Geometrical properties
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 × 10-3 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 KN.m2

The deformation of a structure at a point due to support settlement is given by;

δi∆ = -EIC (∆.R) ———— (1)
Where;
EIC = Flexural rigidity of the column
∆ = Settlement at the point under consideration
R = Support reaction at the point under consideration

2.0 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 (RD) 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 in Figure 1.2 below;

2.1 Analysis of case 1; X1 = 1.0, X2 = 0
The simple static analysis of the structure with case 1 loading is as shown below in terms of the internal stresses diagram. Realise that there is a unit value support reaction acting at support A (pointed downwards) and a bending moment of 5.0 units.

2.2 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. It will be so important to realise that there is no vertical support reaction at point A. A unit horizontal reaction is developed at the support to counter the applied unit load for case 2.

The appropriate canonical equation for this structure is therefore;

δ11X1 + δ12X2 + δ1∆ = 0
δ21X1 + δ22X2 + δ2∆ = 0

2.3 Computation of the influence coefficients
Influence coefficients are based on Mohr’s integral such that δi = 1/EI∫Mmds.
We can compute this by using the graphical method (making use bending moment diagrams), we directly employ Vereschagin’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 (diagram of arbitrary shape) 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.

Where;
EI = Flexural Rigidity of the section
M = Bending moment due to externally applied load
̅M = Bending moment due to unit load at the point where the deflection is sought

Elaborate Formulation of Diagram Combination Equations According to Vereschagin’s Rule

Example on Analysis of Statically Determinate Frames (Part 2)

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.

EICδ11 = (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.

EICδ21 = (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.

EICδ11 = (1/3 × 4 × 4 × 4) = 21.333

Influence coefficients due to support settlement
Case 1 (Take a good look at the support reactions)
δi∆ = -EIC(∆.R)
R = Support reaction at point A = -1.0
∆ = Support settlement in the direction of support reaction = -25 mm = -0.025m
Hence δ1 ∆ = – 14647.5 × (-1 × -0.025) = -366.1875

Case 2 (Take a good look at the support reactions again)
Horizontal support reaction at point A = R = 1.0
∆ = Support settlement/movement in the direction of support reaction = 0
Hence δ2∆ = 0

The appropriate canonical equation now becomes;

105.208X1 + 40X2 = 366.187
40X1 + 21.333X2 = 0

On solving the simultaneous equations; X1 = 12.123 KN; X2 = -22.730 KN

The final value of the internal stresses is given by the equations below;

Mdef = M1X1 + M2X2
Qdef = Q1X1 + Q2X2
Ndef = N1X1 + N2X2

Final Bending Moments (Mdef)
MA = (12.1226 × 5) + (-22.73 × 4) = -30.307 KNm
MB = (12.123 × 5) = -60.613 KNm
MC = hinged support = 0

Final Shear Force (Qdef)
QA – QBB = (-22.73 × -1) = 22.73 KN
QBR – QCL = (12.123 × -1) = -12.123 KN

Final Axial Force (Ndef)
NA – NBB = (12.123 × 1) = 12.123 KN NBR – NCL = (-22.73 × -1) = 22.730 KN