**1.0 Introduction**

Settlement of foundations occur due to load from buildings. Geotechnical engineers are usually interested in seeing that the settlements are uniform, while at the same ensuring that they are limited to a certain depth. In some cases, it is usually anticipated that differential settlement might occur. The geotechnical engineer has a responsibility of determining the magnitude of the settlement, while the structural engineer investigates the effect of the settlement on the structural members. If the structure is statically determinate, internal stresses are not induced due to differential settlement, but if it is statically indeterminate, then internal stresses due to the settlement are induced. This is usually treated as a special load case in design of structures, and this is what this post explores.

**2.0 Case Study**

Let us consider a 6m cantilever that is supported on a roller at the free end. The roller support undergoes a vertical settlement of 25mm (downwards). We are to determine the additional bending moment at the fixed support due to the settlement. We will approach this problem using force and stiffness methods.

**Solution**

*(a) By force method*A propped cantilever (roller support) is indeterminate to the 1st order. The adopted basic system for the structure is a simple 6m cantilever. We will now replace the redundant vertical reaction with a unit load, and the resulting bending moment diagram is as shown below.

The appropriate cannonical equation is given by;

Î´

_{11}X

_{1}+ Î´

_{1∆}= 0

Where;

Î´

_{11 }= Deformation at point 1 (free end) due to unit load at point 1

Î´

_{1∆}= Deformation at point 1, due to settlement of Support = - EI(∆.S)

By Vereschagin's rule;

Î´

_{11 }= (1/3) × 6 × 6 × 6 = 72

On the other hand;

Î´

_{1∆}= - EI(∆.S) = -22500 (1 × -0.025) = 562.5

Substituting into the canonical equation and solving;

72X

_{1}+ 562.5 = 0

On solving;

X

_{1}= - 7.8125 kN (downward reaction)

With this reaction now known, the moment at support A due to the sinking of support;

M

_{A}= - 7.8125 × 6 = -46.875 kNm (hogging moment)

**(b) By stiffness method**

With support A fully fixed and support B simply supported, the general equation for bending moment at support A due to external load and settlement of support is given by;

M

_{AB}= [(3EIÎ¸

_{A}/L) - (3EI∆/L

^{2})] + F

_{AB}

Where F

_{AB}is the fixed end moment due to externally applied load = 0

Also Î¸

_{A}= 0 since support A is fully fixed.

∆ = 25mm = -0.025m (negative due to downward settlement).

M

_{AB}= [((3 × 22500)/6) × (-0.025/6)] = -46.875 kNm

Now, we obtained the answer above expressly because of very known facts. Since the bending moment at B zero, we may not pass through the stress of obtaining the rotation at support B. However for the sake of knowledge, let us apply a more general method which a lot of people might be more familiar with. Here, we will obtain the rotation (slope) at support B, and use it to obtain the bending moment at support A. We are initially assuming that all supports are fixed in the development of the slope deflection equations.

M

_{AB}= EI/L[(4Î¸

_{A }+ 2Î¸

_{B }- 6∆/L)] + F

_{AB}

M

_{BA}= EI/L[(2Î¸

_{A }+ 4Î¸

_{B }- 6∆/L)] + F

_{BA}

For equilibrium and compatibility;

M

_{BA}= 0

We also know that Î¸

_{A }= 0 (fixed support), and there is no external load being considered;

22500/6[4Î¸

_{B }- (6 × -0.025)/6] = 0

15000Î¸

_{B }+ 93.75 = 0

Therefore, Î¸

_{B }= -0.00625 radians

Substituting the value of Î¸

_{B }into the slope deflection equation for M

_{AB}

M

_{AB}= EI/L[(2Î¸

_{B }]

M

_{AB}= 3750[(2 × -0.00625)] = 46.875 kNm

Thank you for reading to the end.

Great work obinna ranks. More grace to u.

ReplyDeleteThank you so much

DeleteThank you so much Ovie

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