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.

1

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.

2

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:

3

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)

4

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.

5

6