2. In a laboratory test of a beam loaded by end couples, the fibers at layer AB in Figure below are found to increase 60 x 10-3 mm whereas those at CD decrease 100 x 103 mm in the 200-mm-gage length. Using E=70 GPa, determine the flexural stress in the top and bottom fibers. 200 mm 30 mm 120 mm 30 mm

Elements Of Electromagnetics
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2. In a laboratory test of a beam loaded by end couples, the fibers at layer AB in
Figure below are found to increase 60 x 10-3 mm whereas those at CD decrease
100 x 10-3 mm in the 200-mm-gage length. Using E=70 GPa, determine the
flexural stress in the top and bottom fibers.
200 mm
30 mm
| 120 mm
30 mm
Please solve according to
the exporters of a typical
solution.
4. Stress in Beams
The beam in figure (1) shows two section
EF, -0
* fo. •da = o
(ab) and (cd) that separate by distance (dx).
le
Neuteal
fece
dA-o
Figure (2) show the deflected shape
of the beam.
Since the (vtA) is the moment
of the differential area (A4) about
the neutral axis, the integral
6 = hk = ydÐ
Fig.1
Fig 3
f ydA ) is the total moment of area
Ay' -o
yd0
ef
ef pd0
hence :
Strain =
%3D
only (y) can be equal zero, Le. the neutral axis must contain the centroid of
the cross-sectional area .
EF, -0
that leads to the shear stress formula (V, = |t, *dA)
Neutral surface
yde
EF,-0
That leads to the formula of shear (t.dA=0)
EM, - 0 = S z(0, * d4) - o
pd0
2y• da - 0
O = E * ɛ
(Hook's law)
de
The integral =y dA is the product of inertia (P). which is zero
only if ) or (2) is an axis of symmetry or principal axis.
O = EL
d'd
E M, - 0.M - M,
M - fy (o. • dA)
Fig.2
Transcribed Image Text:2. In a laboratory test of a beam loaded by end couples, the fibers at layer AB in Figure below are found to increase 60 x 10-3 mm whereas those at CD decrease 100 x 10-3 mm in the 200-mm-gage length. Using E=70 GPa, determine the flexural stress in the top and bottom fibers. 200 mm 30 mm | 120 mm 30 mm Please solve according to the exporters of a typical solution. 4. Stress in Beams The beam in figure (1) shows two section EF, -0 * fo. •da = o (ab) and (cd) that separate by distance (dx). le Neuteal fece dA-o Figure (2) show the deflected shape of the beam. Since the (vtA) is the moment of the differential area (A4) about the neutral axis, the integral 6 = hk = ydÐ Fig.1 Fig 3 f ydA ) is the total moment of area Ay' -o yd0 ef ef pd0 hence : Strain = %3D only (y) can be equal zero, Le. the neutral axis must contain the centroid of the cross-sectional area . EF, -0 that leads to the shear stress formula (V, = |t, *dA) Neutral surface yde EF,-0 That leads to the formula of shear (t.dA=0) EM, - 0 = S z(0, * d4) - o pd0 2y• da - 0 O = E * ɛ (Hook's law) de The integral =y dA is the product of inertia (P). which is zero only if ) or (2) is an axis of symmetry or principal axis. O = EL d'd E M, - 0.M - M, M - fy (o. • dA) Fig.2
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