Unit-II: Bending and Shear and Torsion of Thin Walled Beams
Unit-II: Bending and Shear and Torsion of Thin Walled Beams
Unit-II: Bending and Shear and Torsion of Thin Walled Beams
• Realizing that at any point on the neutral axis, the bending strain and stress are
zero, we can use the general bending stress equation to find its orientation.
Setting the stress to zero and solving for the slope y/x gives
SHEAR FLOW AND SHEAR CENTRE
Restrictions:
1. Shear stress at every point in the beam must be less than the elastic limit of the
material in shear.
2. Normal stress at every point in the beam must be less than the elastic limit of the
material in tension and in compression.
3. Beam's cross section must contain at least one axis of symmetry.
4. The applied transverse (or lateral) force(s) at every point on the beam must pass
through the elastic axis of the beam. Recall that elastic axis is a line connecting
cross-sectional shear centers of the beam. Since shear center always falls on the
cross-sectional axis of symmetry, to assure the previous statement is satisfied, at
every point the transverse force is applied along the cross-sectional axis of
symmetry.
5. The length of the beam must be much longer than its cross sectional dimensions.
6. The beam's cross section must be uniform along its length.
Shear Center
If the line of action of the force passes through the Shear Center of
the beam section, then the beam will only bend without any twist.
Otherwise, twist will accompany bending.
The shear center is in fact the centroid of the internal shear force
system. Depending on the beam's cross-sectional shape along its
length, the location of shear center may vary from section to
section. A line connecting all the shear centers is called the elastic
axis of the beam. When a beam is under the action of a more
general lateral load system, then to prevent the beam from
twisting, the load must be centered along the elastic axis of the
beam.
Shear Center
• The two following points facilitate the determination of the shear center
location.
1. The shear center always falls on a cross-sectional axis of symmetry.
2. If the cross section contains two axes of symmetry, then the shear center is
located at their intersection. Notice that this is the only case where shear
center and centroid coincide.
SHEAR STRESS DISTRIBUTION
RECTANGLE T-SECTION
SHEAR FLOW DISTRIBUTION
EXAMPLES
• For the beam and loading shown, determine:
(a) the location and magnitude of the maximum transverse shear force 'Vmax',
(b) the shear flow 'q' distribution due the 'Vmax',
(c) the 'x' coordinate of the shear center measured from the centroid,
(d) the maximun shear stress and its location on the cross section.
Stresses induced by the load do not exceed the elastic limits of the material.
NOTE:In this problem the applied transverse shear force passes through the centroid of the cross
section, and not its shear center.
Shear Flow Analysis for Unsymmetric
Beams
• SHEAR FOR EQUATION FOR UNSUMMETRIC SECTION IS
SHEAR FLOW DISTRIBUTION
• For the beam and loading shown, determine:
• (a) the location and magnitude of the maximum
transverse shear force,
• (b) the shear flow 'q' distribution due to 'Vmax',
• (c) the 'x' coordinate of the shear center
measured from the centroid of the cross section.
• Stresses induced by the load do not exceed the
elastic limits of the material. The transverse
shear force is applied through the shear center at
every section of the beam. Also, the length of
each member is measured to the middle of the
adjacent member.
Beams with Constant Shear Flow Webs
Assumptions:
1. Calculations of centroid, symmetry, moments of area and
moments of inertia are based totally on the areas and
distribution of beam stiffeners.
2. A web does not change the shear flow between two adjacent
stiffeners and as such would be in the state of constant shear
flow.
• Let's begin with a simplest thin-walled stiffened beam. This means a beam with two
stiffeners and a web. Such a beam can only support a transverse force that is
parallel to a straight line drawn through the centroids of two stiffeners. Examples of
such a beam are shown below. In these three beams, the value of shear flow would
be equal although the webs have different shapes.
• The reason the shear flows are equal is that the distance between two adjacent
stiffeners is shown to be 'd' in all cases, and the applied force is shown to be equal
to 'R' in all cases. The shear flow along the web can be determined by the following
relationship
Important Features of
Two-Stiffener, Single-Web Beams:
• Derivation
Consider a thin-walled member with a closed cross section subjected to
pure torsion.
Examining the equilibrium of a small cutout of
the skin reveals that
Angle of Twist
By applying strain energy equation due to shear and Castigliano's
Theorem the angle of twist for a thin-walled closed section can be
shown to be
positive, we obtain
• To find the angle of twist, we could use either of the two twist formulas given above. It is
also possible to express the angle of twist equation similar to that for a circular section
Shear Stress Distribution and Angle of Twist for Multiple-Cell Thin-
Wall Closed Sections
• In the figure above the area outside of the cross section will be designated as cell (0).
Thus to designate the exterior walls of cell (1), we use the notation 1-0. Similarly for cell
(2) we use 2-0 and for cell (3) we use 3-0. The interior walls will be designated by the
names of adjacent cells.
• the torque of this multi-cell member can be related to the shear flows in exterior walls
as follows
For elastic continuity, the angles of twist in all cells must be
equal
• For the thin-walled single-cell rectangular beam and loading shown, determine
(a) the shear center location (ex and ey),
(b) the resisting shear flow distribution at the root section due to the applied load of
1000 lb,
(c) the location and magnitude of the maximum shear stress
Unit-III
Structural Idealisation of Thin Walled Beams
Structural Idealization
• Consider the two-spar wing
section shown. The stringers
and spar carry most of the
direct stresses while the skin
carries the shear stresses.
• M @ Right Side
• The above expression does not account for booms. How can we
deal with booms that cause discontinuity in the skin and therefore
interrupt the shear flow?
Shear of open-section beams
• Sx and Sy produce direct stresses due to bending in the booms (and skin)
and shear stresses in the skin
Shear of open-section beams
• rth boom has a cross-sectional area Br
• Shear flows in skin adjacent to it are q1 and q2.
• This gives:
Shear of open-section beams
• Recall:
Shear of open-section beams
Sy n
qs
I xx
B y
r 1
r r
Open C/S Sample Problem
Calculate Ixx: (Only consider direct stress carrying areas) I.e. Booms
4.8 10 3 n n
qs
48 106
r r
B y
r 1
10 4
Br yr r 1
At the outside of boom 1, qs = 0. As boom 1 is crossed, the shear flow changes to:
4
q12 0 10 300 200 6 N/mm
q1
Open C/S Sample Problem
There will be no further changes in shear flow until the next boom (2) is crossed.
q3
At the outside of boom 4, the shear flow is zero (qs = 0) as expected
S1
q34 6 N/mm
12 N/mm
6 N/mm
S3
Closed C/S Sample Problem
• For the single cell beam, the
booms carry the direct stresses
and the walls carry only the
shear stress. A vertical shear
load of 10 kN acts through the
vertical plane between booms 3
and 6. Calculate the shear flow
distribution
Closed C/S Sample Problem
Centroid on Horizontal Axis of Symmetry. Ixy = 0
Also Sx = 0, tD = 0
Sy n
qs
I xx
B y
r 1
r r qs , 0
Ixx can be calculated from the direct stress carrying area of the booms
I xx 2 B1 30 2 B2 100 2 B3 100 2 B4 50 2
Substituting B1,…B4 gives Ixx = 13.86 x 106 mm4
10 103 n n
6 r r
qs B y qs ,0 7.22 10 Br yr qs ,0
4
13.86 10 r 1 r 1
Closed C/S Sample Problem
Introduce a cut in the wall 23 and calculate the basic shear flow around the walls
Since the tD = 0
qb , 23 0
B3 y3
qb ,34 7.22 10 4 400 100 28.9 N/mm
B4 y4
q b,45 28.9 7.22 10 4 100 50 32.5 N/mm
• Taking moments about the intersection of the line of action of shear load and
horizontal axis:
Draw out the shear flow distribution to determine the sign of the moment generated
by the shear flow on each segment
Closed C/S Sample Problem
q p s
[qb,81 480 60 2qb,12 240 170 2qb, 23 100 240
2qb, 43 120 100 qb,54 120 100] 2 97,200qs ,0 0