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IS 456:2000
BEAM DESIGN
PROBLEM 1
Singly fixed beam
L =5m B=300mm D=600mm M20 Fe415 Cl.cover =25mm
W=150 kN/m
DIAMETER OF
BAR
16 dia 25 dia 32 dia
MANUAL Ast 1868 mm2 1891
mm2
1910
mm2
dt 33 37.5 41
STAAD PRO V8I
Ast1940.5 mm
2 1894.4 mm
2 1913.50
mm2
Use dt to match the
results with STAAD
46 38 42
-In staad, Ast decreases as diameter of bar increases.
-Encounters a problem in using dt.
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PROBLEM 2
Doubly fixed beam.
L =5m B=300mm D=600mm M20 Fe415 cover =25mm
W=180 kN/m
DIAMETER OFBAR 16 dia 20 dia 32 dia
Manual Ast 2251 mm2 2257mm
2 2279mm
2
Manual Asc 219 mm 234mm 278 mm
dt 33 35 41
STAAD PRO V8I
Ast2297.86 mm
2 2286.54 mm
2 2277.3mm
2
Asc 328.78 mm2 302.9 mm
2 279.5 mm
2
Use dt to match theresults with STAAD 46 44 41
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PROBLEM 3
Simply supported doubly beam section
diameter 16 mm 20 mm 32 mm
staad Ast 2381.40 mm2 2346.20 mm
2 2360.89
mm2
staad Asc 413.63 mm2 337.47 mm
2 364.59 mm
2
use dt for
matchindg result
46 47 41
manual Ast 2332 mm 2339 mm 2362 mm
manual Asc 302 mm 317 mm 363 mm
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PROBLEM 4
4.1
Tv >TC
Sv ,By design worksheet
distance 0 mm 1250 mm 2500 mm
shear design Tv >TC Tv >TC Tv
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4.2
This type of problem occurred in simply supported beam.
By Design worksheet :
MANUAL SV
0mm 250mm 500mm 750mm 1000 mm
Tv >TC Tv >TC Tv >TC Tv >TC Tv
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PROBLEM 5 (Torsion design)
5.1
-20 mm diameter are provided in staad pro.
Eff. Cover = 25 + 10 = 35 mm
Results from manual calculations are not matched with staad pro result.
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5.2
AT SUPPORT : 0 m
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If we consider the effective cover as 42 mm then the answers are nearer
to the results of staad pro.
AT 1.250m
If we consider the effective cover as 52 mm then the answers are nearer
to the results of staad pro.
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5.3
Torsion design + Shear design
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At support
At 1.25 m
At 2.5 m
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5.4
Torsion in simply supported beam
At 2.5 m
Actually Asc does not require in design but somehow staad provides.
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At 3.75 m
Actually Asc does not require in design but somehow staad provides.
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5.5
Torsion in simply supported beam (doubly beam section)
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At 2.5 m
Required Asc = 95 mm^2 ( Manually )
Required Asc = 350.24 mm^2 ( Staad result )
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Problem 5.6
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At support : 0 m TV < TC
Shear design is verified ( matched ) with design worksheet.
AT 1.25m TV < TC
Required Sv = 300 mm ( Manually )
Provided Sv = 190 mm ( Staad result )
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At end : 5 m
Required Sv = 300 mm ( Manually )
Provided Sv = 190 mm ( Staad result )
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Problem 6 Tee beam (fixed beam)
6.1
l=5m M20 Fe415
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-This doesnt require Asc but staad provides 306 mm^2 at 0 m.
-Shear design problems are same as a beam design.
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6.2
Cantilever tee beam design
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Ast and Asc are not matched with staad pro v8i.
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6.3 cantilever
STRESS DIAGRAM :
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At support : 0 m
Flexural steel matches with staad results.
At 2.5 m
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W = -75 kN
When the applied point load is 60 kN downwards at mid span, the Ast is
provided by Staad is at bottom and when the applied point load is 75 Kndownwards at mid span, the Ast is provided at top.
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PROBLEM 7
CONTROL OF DEFLECTION
1.1 The vertical deflection limits may. Generally be assumed to besatisfied provided that the span to depth ratios are not greater than the
values obtained as below:
a) Basic values of span to effective depth ratios
for spans up to 10m:
a) cantilever-7
simply supported-20
continuous-26
basic value of span to effective depth ratio for span up to 10 m.
Allowable deflection as per IS 456
=7*M.F
By deflection chart M.F=0.82
5.74
Actual
=8.33
Actual
< maximum
8.33< 5.74 Not ok
Staad pro v8i doesnt consider this criteria and provides a design
without even any warning.
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PROBLEM 8
1.2 Slenderness Limit for Beams to Ensure Lateral Stability
A simply supported or continuous beam shall be so proportioned that the clear
distance between the lateral
restraints does not exceed 60 b whichever is less,where d is the effective depth of the beam and b the breadth of the compression
face midway between the lateral restraints. For a cantilever, the clear distance from
the free end of the cantilever to the lateral restraint shall not exceed
or 25 b
or whichever is less.
[not including this clause.]
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Problem 9
5.SIDE FACE REIINFORCEMENT
Where the depth of the web in beam exceed 750 mm sidereinforcement shall be provide along the two faces. the reinforcement
shall be not less than 0.1 % of the web area and shall be distributed
equally distributed equally on two faces at a spacing not exceed 300
mm or web thickness whichever is less.
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Udl =6 kN/m
Staad provides side face reinforcement, but it doesnt satisfy the
spacing of bars which is limited to 300mm or width of web whichever is
less.
Calculation =
Area of web=830*350=290500mm2
0.1%web area=290 mm2
Staad pro v8i provided=1-16i =402 mm2
>290 mm2
This criteria is provided in staad pro v8i.
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Problem 10
TORSION DESIGN
In Torsion Design If Depth Increse more Then 450 Side Reinforcement Are
Provided As Per Is 456.
But staad pro does not provide a side face reinforcement.
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Problem-11
Deep beam
a) A beam shall be deemed to be deep beam when the ratio of effective span to
depth to overall depth l/d is less than :
1) 2 for simply supported beam
2) 2.5 for a continuous beam
Deep beam is not designed assuming it to be a part of a continuous beam and
away from the critical section for enhanced shear, ordinary shear check is
performed.
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Problem-12
Fixed beam
Ld =
=
= 483.5 (for tension)
Ld =
= 386.83
-At end staad pro v8i provide a 0 development length.
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Fixed beam 2
Ld=
=
=483.5 (for tension)
Ld=
=386.83 (for compression)
-at end staad pro v8i provide a 0 development length.
-staad consider some time larger diameter and.sum time take a
smaller diameter.
-no meaning Ld provided at intermediate span.
-in compression staad pro v8i not increase Tbd value 25%.
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PROBLEM -13
In Staad pro v8i 2legged stirrups are provided .and also inclined stirrups
are not provided.
PROBLEM -14
Spacing Between Two Bar Are
Spmain Criteria Are Provided In Staad Pro V8i.But Spacing BetweenTwo Bar Are Taken By Staad Pro Itself.
SPMAIN design parameter are provided in staad pro v8i,still staad pro
take spacing by itself.
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At 1.5 m result are not matched.
Example 2
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Not mathched ast and asc.
Example 3
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Example 5
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