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SHEAR CENTER OF CLOSED SECTION
AIM:Shear center of a closed section determination.
THEORY:
For any unsymmetrical section there exists a point at which any vertical
force does not produce a twist of that section. This point is known as shear
center.
The location of this shear center is important in the design of beams of
closed sections when they should bend without twisting. The shear center is
important in the case of a closed section like an aircraft wing, where the lift
produces a torque about the shear center. Similarly the wing strut of a semi
cantilever wing is a closed tube of aerofoil section. A thin walled ‘D’ section
with its web vertical has a horizontal axis of symmetry and the shear center lies
on it. The aim of the experiment is to determine its location on this axis if the
applied shear to the tip section is vertical (i.e., along the direction of one of the
principal axes of the section) and passes through the shear center tip, all other
sections of the beam do not twist.
APPARATUS REQUIRED:
A thin uniform cantilever beam of ‘D’ section as shown in the figure. At
the free end extension pieces are attached on either side of the web to
facilitate vertical loading.
Two dial gauges are mounted firmly on this section, a known distance
apart, over the top flange. This enables the determination of the twist, if
any, experienced by the section.
A steel support structure to mount the ‘D’ section as cantilever.
Two loading hooks each weighing about 0.1 Kg.
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PROCEDURE:
1. Mount two dial gauges on the flange at a known distance apart at the free
end of the beam (see fig). Set the dial gauge readings to zero.
2. Place a total of say 1.2 kilograms load at A (loading hook and 5 load
pieces will make up this value). Note the dial gauge readings (nominally,
hooks also weigh a 100 grams each). Note down dial gauge reading.
3. Now remove one load piece from the hook at A and place it at hook B.
The total vertical load on this section remains 1.2 kilogram. Record the
dial gauge readings.
4. Transfer carefully all the load pieces to B from A one by one. Note each
time the dial gauge readings. This procedure ensures that while the
magnitude of the resultant vertical force remains the same its line of
action shifts by a known amount along AB every time when a load piece
is shifted. Calculate the distance ‘e’ (see fig) of the line of action from
the web thus:
(AB) x (Wa-Wb)
eexp --------------------- where WV = (Wa + Wb)
2Wv
For every load case calculate the algebraic difference between the dial
gauge readings suffered by the section.
Though a nominal value of two kilograms for the total load is suggested it
can be less. In that event the number of readings taken will reduce
proportionately.
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TABLE
Dimensions of the beam and the section :
Length of the beam (L) : 265mm
Height of the web (h) : 80mm
Thickness of the sheet (t) : 0.8mm
Distance between the two hook stations (AB) : 310mm
Weights – 5 Nos. : 0.2kgs
S. No.Wa
in Kg
Wb
in Kg
Dial gauge readings
(d1-d2)Algebraic difference
e = AB(Wa-Wb) / 2 Wv
d1 mm d2 mm
Plot e Vs (d1-d2) curve and determine where this meets the e axis and
locate the shear center.
RESULT:
The shear center obtained experimentally is compared with the
theoretical value.
PRECAUTIONS:
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1. For the section supplied there are limits on the maximum value of loads
to obtain acceptable experimental results. Beyond these the section could
undergo excessive permanent deformation and damage the beam forever.
Do not therefore exceed the suggested values for the loads.
2. The dial gauges must be mounted firmly. Every time before taking the
readings. Tap the set up (not the gauges) gently several times until the
reading pointers on the gauges settle down and do not shift any further.
This shift happens due to both backlash and slippages at the points of
contact between the dial gauges and the sheet surfaces and can induce
errors if not taken care of. Repeat the experiments with identical settings
several times to ensure consistency in the readings.
SHEAR CENTER FOR CLOSED SECTION: (Experimental result)(For Guidance only)
Wv =1.2 kg, AB = (distance of point A to point B) = 31 cm
S. No.
Wa
in Kg
Wb
in Kg
Dial gauge readings (d1-d2)
e =AB(Wa-Wb) / (2Wv)d1(anticlock)750 d2(anticlock)350
Algebraic difference
1 1.1 0.1 700 316 384 12.92 0.9 0.3 702 314 388 7.75
3 0.7 0.5 705 311 394 2.5
4 0.5 0.7 707 308 399 -2.5
5 0.3 0.9 709 305 404 -7.75
6 0.1 1.1 711 302 409 -12.9
SAMPLE CALCULATION (ONLY FOR GUIDANCE)
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e =AB(Wa-Wb) / (2Wv)
e =31(0.9-0.3) / (2×1.2)
e = 7.75A curve is plotted between (d1-d2) Vs e.From the graph e = 19mm.
Result:The shear center obtained experimentally, well with the theoretical value. e (expt) =19mm (from graph)e (theo) = 20mm (e(theo) is 20 mm from calculation)
eth > eexp and nearer to the value.
This shift happens due to both backlash and slippages at the points of contact
between the dial gauges and the sheet surfaces and can induce errors if not taken
care of. Repeat the experiments with identical settings several times to ensure
consistency in the readings.
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