Solid And Structural Mechanics Lecture 10 and 11 PDF

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LECTURE TENBENDING MOMENT AND SHEAR FORCE IN CONTINOUS BEAMS10 IntroductionA beam supported on more than two supports is called a continuous beam (CB). CB’s have advantages over the simply (or freely) supported in thata) The maximum bending moment in case of CB is much less than in the case of SSB o...

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LECTURE TEN BENDING MOMENT AND SHEAR FORCE IN CONTINOUS BEAMS 10.1 Introduction A beam supported on more than two supports is called a continuous beam (CB). CB’s have advantages over the simply (or freely) supported in that a) The maximum bending moment in case of CB is much less than in the case of SSB of the same span and carrying same loads. b) The BM is more over the supports than at midspan in case of CB and hence the weight of the beam does not materially affect the stresses c) Average BM is less in case of CB and hence lighter materials of construction can be used to resist the BM. 10.2 The three moment theorem It is based on the application of the 1st and 2nd moment area theorems to this specific problem. Consider only two spans AB and BC of a continuous beam shown in Fig. 10.1(a). The beam may be carrying any pattern of loading Fig 10.1(b) shows the BM diagrams for each of the two spans treating both beams AB and BC to be freely supported and independent of each other. Let the beam ABC under the effect of the applied loads take up the shape the supports A and C sinking by different amounts

With respect to support B, draw a tangent to the classic curve at B and the intercepts made by this tangent on the verticals through A and C be and . Assume; Length of span Flexural rigidity for the span Flexural rigidity for the span Area of B.M. diagram for span AB (simply supported) = Area of B.M. diagram for span BC (simply supported) = Distance of the centroid of from Distance of the centroid of from Sinking of support A with respect to support Sinking of support C with respect to support Fixing moment at support A

=MA

Fixing moment at support B

= MB

Fixing moment at support C

= MC

Angle made by the tangent to elastic curve at B with the horizontal

Now

is the deflection of the point A with respect to the tangent to the elastic curve at B and according to second moment area theorem it is equal to the moment of the diagram between A and B about A.

Also

But as before

is the deflection of point C with respect to tangent to the

elastic curve at B and according to the second “moment area” theorem it is equal to the moment of the diagram about C:

Equating 10.1 and 10.2 have;

Or

Or

The above is the most general form of the theorem of three moments. NOTE. All sagging moments are taken as and all hogging moments as ; and are taken as if after sinking supports A and C lie below the support of reference B. For any two consecutive spans if the end supports lie below the central support the value of for the particular support is taken as otherwise . 10.3. Special cases of theorem of three moments: (i)

If all supports remain at the same level then both and are zero and equation 10.3 can be written as;

(ii)

If flexural rigidity for all spans is the same i.e., if and supports are at the same level then: or

The above equation gets further simplified if there are only two spans and the beam is freely supported at A and C, then

(iii)

For a fixed end of the beam the theorem of three moments can be modified by imagining a span of length and moment of inertia beyond the support and applying the theorem of three moments as usual as shown in Fig 10.2:

Figure 10.2: Clarifies the point.

10.4: Maximum Bending Moments in Simply Supported Beams (Revision) a) A concentrated Load at Midspan

a) A concentrated Load not at Midspan

b) UDL over the entire span

10.5: Geometric Properties of some shapes a)

b)

c)

d)

10.6: Worked examples 1. Draw the bending moment and shear force diagrams for the two span continuous beams shown in Fig Q1. The beam is simply supported at the supports A and C and is continuous over the support B. EI is constant.

Let be the support or fixing moments at supports A, B and C. Consider spans AB AND BC as simply supported the bending moments under both the loads are 36.KN.m.

Support moments Applying the 3-moment equation for spans AB and BC

For support reactions, take moments about B for span BC

Take moments about B for span AB

But

2. A continuous beam ABCD has three equal spans AB, BC and CD and carries a w/unit length throughout its length. The beam is freely supported on all supports which are at same level. Draw the S.F and the BM diagram.

From symmetry

In applying the theorem of three moments, we have area of B.M diagram for each of the three spans

Similarly distance of the centroid of the BM diagram for each span from either support

The three moment theorem applied to spans AB and BC gives

Fig. (c) shows the support moment diagram and by superposition, Fig (d) shows the BMD for the beam. Again by symmetry, support reactions at A and D are equal and so are the support reactions at B and C.

Take moments about C for the span CD

For equilibrium we have

but

Fig (e) shows the S.F.D By similarly of in the SFD, the SF is zero at

from A in span AB; from D is

span CD and at middle point of span BC. The +ve BM is maximum at points of zero shear. Therefore, maximum +ve B.M in span AB or CD is at

from A or D

respectively and is

In span BC the maximum +ve B.M is at mid span of BC and is (Fig (d).

Point’s contraflexure Span AB: Let the B.M be zero at a distance x from A, then

Thus the point of contraflexure in span AB is at from A and similarly in span CD it is at from D. Span BC: Let the B.M be zero at a distance x from B, then

Or

LECTURE ELEVEN CONTINUOUS BEAMS WITH FIXED END SUPPORTS OVERCHARGING AND SINKING SUPPORTS 11.1 Introduction If the beam is fixed at the left end A then an imaginary zero spans is taken to the left of A and the three moment’s theorem is as applied as usual. Similarly if the beam is fixed at the right end, then an imaginary zero spans is taken after the right and support and the three moment theorem is applied as usual.

Imaginary span of length , beyond the support NB: (i) The fixing moment at the imaginary supports of the zero span i.e. is always equal to zero. Ex1. A continuous beam ABC of uniform section with span AB as 8m and BV as 6m is fixed at A and simply supported at B and C. the beam is carrying a uniformly distributed load of 1KN/m throughout length. Find the moments along the beam and the reactions at the supports. Also draw the bending moment and shear force diagram.

Since the beam is fixed at A, therefore we assume zero spans to the length of First of all consider the beam AB as a simply supported beam. Therefore, B.M at the mid of span AB.

Similarly B.M at the mid of span BC

Now draw the M-diagram, with the heep of the above B.M as shown is Fig. (b) From the geometry of the above B.M.D we find that for the spans OA and AB

And Similarly for the span AB and BC

And

Now using three moments equations for the spans OA and AB

Again using three moments equations for the spans AB and BC

Solving for Eq. (1) and (2)

SUPPORT REACTIONS Taking moment about B

Now taking moment about A

And NB: The reaction at R may also be found by taking moment about B i.e.

11.2 CONTINUOUS BEAMS WITH OVERHANGS

In case of continuous beam overhangs on one side or on both sides then the moment for the support that the beam overhangs is determined by treating the overhanging portion as a cantilever. Rest of the analysis is done by applying the theorem of three moments as usual.

Ex.2: A continuous beam ABCDE rests on three supports A, C and D all at the same level, with equal overhangs of length x on either side. Spans BC and CD are each of length L. the beam caries of w/unit length. Find the ratio of for the three support reactions to be equal. Take EI constant. B.M diagram, treating each span as SSB and independent of other spans as shown in Fig (b)

Total load on the beam ABCD is

And if all the support reaction are equal then

Applying the three moments to the beam we have

Now taking moments about C (from the rights side)

Or

Or

11.3 SINKING OF SUPPORTS Sinking of supports modifies the BM and the SF along a continuous beam appreciably. The general equation.

Is applied but in so doing most care is exercised in adopting the sign convention.

If the intermediate support (in case of the 2 spans under consideration) sinks and its level is lower than the side supports, the are negative. If the end support (in case of the 2 spans under consideration) is at a lower level than the middle one, them is to be taken as +ve....