7.4 SHEAR FLOW IN BUILT-UP MEMBERS

1
7.4 SHEAR FLOW IN BUILT-UP MEMBERS

• Occasionally, in engineering practice, members
  are built-up from several composite parts in
  order to achieve a greater resistance to loads,
  some examples are shown.
• If loads cause members to bend, fasteners may be
  needed to keep component parts from sliding
  relative to one another.
• To design the fasteners, we need to know the
  shear force resisted by fastener along members
  length

2
7.4 SHEAR FLOW IN BUILT-UP MEMBERS

• This loading, measured as a force per unit
  length, is referred to as the shear flow q.
• Magnitude of shear flow along any longitudinal
  section of a beam can be obtained using similar
  development method for finding the shear stress
  in the beam

3
7.4 SHEAR FLOW IN BUILT-UP MEMBERS

• Thus shear flow is

Equation 7-6
q shear flow, measured as a force per unit
length along the beam V internal resultant
shear force, determined from method of sections
and equations of equilibrium I moment of
inertia of entire x-sectional area computed about
the neutral axis Q ?A y dA yA, where A
is the x-sectional area of segment connected to
beam at juncture where shear flow is to be
calculated, and y is distance from neutral axis
to centroid of A
4
7.4 SHEAR FLOW IN BUILT-UP MEMBERS

• Note that the fasteners in (a) and (b) supports
  the calculated value of q
• And in (c) each fastener supports q/2
• In (d) each fastener supports q/3

5
7.4 SHEAR FLOW IN BUILT-UP MEMBERS

• IMPORTANT
• Shear flow is a measure of force per unit length
  along a longitudinal axis of a beam.
• This value is found from the shear formula and is
  used to determine the shear force developed in
  fasteners and glue that holds the various
  segments of a beam together

6
EXAMPLE 7.4
Beam below is constructed from 4 boards glued
together. It is subjected to a shear of V 850
kN. Determine the shear flow at B and C that must
be resisted by the glue.
7
EXAMPLE 7.4 (SOLN)
Section properties Neutral axis (centroid) is
located from bottom of the beam. Working in units
of meters, we have
Moment of inertia about neutral axis is
I ... 87.52(10-6) m4
8
EXAMPLE 7.4 (SOLN)
Section properties Since the glue at B and B
holds the top board to the beam, we have
QB yB AB 0.305 m ? 0.1968 m(0.250 m)(0.01
m) QB 0.270(10-3) m3
Likewise, glue at C and C holds inner board to
beam, so
9
EXAMPLE 7.4 (SOLN)
Shear flow For B and B, we have
qB VQB /I 850 kN(0.270(10-3)
m3/87.52(10-6) m4 qB 2.62 MN/m
Similarly, for C and C,
qC VQC /I 850 kN(0.0125(10-3)
m3/87.52(10-6) m4 qC 0.0995 MN/m
10
EXAMPLE 7.4 (SOLN)
Shear flow Since two seams are used to secure
each board, the glue per meter length of beam at
each seam must be strong enough to resist
one-half of each calculated value of q. Thus
qB 1.31 MN/m qC 0.0498 MN/m
11
7.5 SHEAR FLOW IN THIN-WALLED MEMBERS

• We can use shear-flow equation q VQ/I to find
  the shear-flow distribution throughout a members
  x-sectional area.
• We assume that the member has thin walls, i.e.,
  wall thickness is small compared with height or
  width of member

12
7.5 SHEAR FLOW IN THIN-WALLED MEMBERS

• Because flange wall is thin, shear stress will
  not vary much over the thickness of section, and
  we assume it to be constant. Hence,

q ? ? t
Equation 7-7

• We will neglect the vertical transverse component
  of shear flow because it is approx. zero
  throughout thickness of element

13
7.5 SHEAR FLOW IN THIN-WALLED MEMBERS

• To determine distribution of shear flow along top
  right flange of beam, shear flow is

Equation 7-8
14
7.5 SHEAR FLOW IN THIN-WALLED MEMBERS

• Similarly, for the web of the beam, shear flow is

Equation 7-9
15
7.5 SHEAR FLOW IN THIN-WALLED MEMBERS

• Value of q changes over the x-section, since Q
  will be different for each area segment A
• q will vary linearly along segments (flanges)
  that are perpendicular to direction of V, and
  parabolically along segments (web) that are
  inclined or parallel to V
• q will always act parallel to the walls of the
  member, since section on which q is calculated is
  taken perpendicular to the walls

16
7.5 SHEAR FLOW IN THIN-WALLED MEMBERS

• Directional sense of q is such that shear appears
  to flow through the x-section, inward at beams
  top flange, combining and then flowing
  downward through the web, and then separating and
  flowing outward at the bottom flange

17
7.5 SHEAR FLOW IN THIN-WALLED MEMBERS

• IMPORTANT
• If a member is made from segments having thin
  walls, only the shear flow parallel to the walls
  of member is important
• Shear flow varies linearly along segments that
  are perpendicular to direction of shear V
• Shear flow varies parabolically along segments
  that are inclined or parallel to direction of
  shear V
• On x-section, shear flows along segments so
  that it contributes to shear V yet satisfies
  horizontal and vertical force equilibrium

18
EXAMPLE 7.7

• Thin-walled box beam shown is subjected to shear
  of 10 kN. Determine the variation of shear flow
  throughout the x-section.

19
EXAMPLE 7.7 (SOLN)

• By symmetry, neutral axis passes through center
  of x-section. Thus moment of inertia is

I 1/12(6 cm)(8 cm)3 ? 1/12(4 cm)(6 cm)3 184
cm4
Only shear flows at pts B, C and D needs to be
determined. For pt B, area A 0 since it can be
thought of located entirely at pt B.
Alternatively, A can also represent the entire
x-sectional area, in which case QB yA 0
since y 0.
20
EXAMPLE 7.7 (SOLN)
Because QB 0, then qB 0
For pt C, area A is shown dark-shaded. Here mean
dimensions are used since pt C is on centerline
of each segment. We have
QC yA (3.5 cm)(5 cm)(1 cm) 17.5 cm3
qC VQC/I ... 95.1 N/mm
21
EXAMPLE 7.7 (SOLN)
Shear flow at D is computed using the three
dark-shaded rectangles shown. We have
QD yA ... 30 cm3
qC VQD/I ... 163 N/mm
22
EXAMPLE 7.7 (SOLN)
Using these results, and symmetry of x-section,
shear-flow distribution is plotted as shown.
Distribution is linear along horizontal segments
(perpendicular to V) and parabolic along vertical
segments (parallel to V)
23
7.6 SHEAR CENTER

• Previously, we assumed that internal shear V was
  applied along a principal centroidal axis of
  inertia that also represents the axis of symmetry
  for the x-section
• Here, we investigate the effect of applying the
  shear along a principal centroidal axis that is
  not an axis of symmetry
• When a force P is applied to a channel section
  along the once vertical unsymmetrical axis that
  passes through the centroid C of the x-sectional
  area, the channel bends downwards and also twist
  clockwise

24
7.6 SHEAR CENTER
25
7.6 SHEAR CENTER

• When the shear-flow distribution is integrated
  over the flange and web areas, a resultant force
  of Ff in each flange and a force of VP in the
  web is created
• If we sum the moments of these forces about pt A,
  the couple (or torque) created by the flange
  forces causes the member to twist
• To prevent the twisting, we need to apply P at a
  pt O located a distance e from the web of the
  channel, thus

? MA Ff d Pe
26
7.6 SHEAR CENTER

• Express Ff is expressed in terms of P ( V) and
  dimensions of flanges and web to reduce e as a
  function of its x-sectional geometry
• We name the pt O as the shear center or flexural
  center
• When P is applied at the shear center, beam will
  bend without twisting
• Note that shear center will always lie on an axis
  of symmetry of a members x-sectional area

27
7.6 SHEAR CENTER

• IMPORTANT
• Shear center is the pt through which a force can
  be applied which will cause a beam to bend and
  yet not twist
• Shear center will always lie on an axis of
  symmetry of the x-section
• Location of the shear center is only a function
  of the geometry of the x-section and does not
  depend upon the applied loading

28
7.6 SHEAR CENTER

• Procedure for analysis
• Shear-flow resultants
• Magnitudes of force resultants that create a
  moment about pt A must be calculated
• For each segment, determine the shear flow q at
  an arbitrary pt on segment and then integrate q
  along the segments length
• Note that V will create a linear variation of
  shear flow in segments that are perpendicular to
  V and a parabolic variation of shear flow in
  segments that are parallel or inclined to V

29
7.6 SHEAR CENTER

• Procedure for analysis
• Shear-flow resultants
• Determine the direction of shear flow through the
  various segments of the x-section
• Sketch the force resultants on each segment of
  the x-section
• Since shear center determined by taking the
  moments of these force resultants about a pt (A),
  choose this pt at a location that eliminates the
  moments of as many as force resultants as possible

30
7.6 SHEAR CENTER

• Procedure for analysis
• Shear center
• Sum the moments of the shear-flow resultants
  about pt A and set this moment equal to moment of
  V about pt A
• Solve this equation and determine the moment-arm
  distance e, which locates the line of action of V
  from pt A
• If axis of symmetry for x-section exists, shear
  center lies at the pt where this axis intersects
  line of action of V

31
7.6 SHEAR CENTER

• Procedure for analysis
• Shear center
• If no axes of symmetry exists, rotate the
  x-section by 90o and repeat the process to obtain
  another line of action for V
• Shear center then lies at the pt of intersection
  of the two 90o lines

32
EXAMPLE 7.8

• Determine the location of the shear center for
  the thin-walled channel section having the
  dimensions as shown.

33
EXAMPLE 7.8 (SOLN)

• Shear-flow resultants
• Vertical downward shear V applied to section
  causes shear to flow through the flanges and web
  as shown. This causes force resultants Ff and V
  in the flanges and web.

34
EXAMPLE 7.8 (SOLN)

• Shear-flow resultants
• X-sectional area than divided into 3 component
  rectangles a web and 2 flanges. Assume each
  component to be thin, then moment of inertia
  about the neutral axis is

I (1/12)th3 2bt(0.5h)2 (0.5th2)(h/6) b
Thus, q at the arbitrary position x is
35
EXAMPLE 7.8 (SOLN)

• Shear-flow resultants
• Hence,

The same result can be determined by first
finding (qmax)f, then determining triangular area
0.5b(qmax)f Ff
36
EXAMPLE 7.8 (SOLN)

• Shear center
• Summing moments about pt A, we require

Ve Ff h

As stated previously, e depends only on the
geometry of the x-section.
37
CHAPTER REVIEW

• Transverse shear stress in beams is determined
  indirectly by using the flexure formula and the
  relationship between moment and shear (V
  dM/dx). This result in the shear formula?
  VQ/It.
• In particular, the value for Q is the moment of
  the area A about the neutral axis. This area is
  the portion of the x-sectional area that is held
  on to the beam above the thickness t where ? is
  to be determined

38
CHAPTER REVIEW

• If the beam has a rectangular x-section, then the
  shear-stress distribution will be parabolic,
  obtaining a maximum value at the neutral axis
• Fasteners, glues, or welds are used to connect
  the composite parts of a built-up section. The
  strength of these fasteners is determined from
  the shear flow, or force per unit length, that
  must be carried by the beam q VQ/I
• If the beam has a thin-walled x-section then the
  shear flow throughout the x-section can be
  determined by using q VQ/I

39
CHAPTER REVIEW

• The shear flow varies linearly along horizontal
  segments and parabolically along inclined or
  vertical segments
• Provided the shear stress distribution in each
  element of a thin-walled section is known, then,
  using a balance of moments, the location of the
  shear center for the x-section can be determined.
• When a load is applied to the member through this
  pt, the member will bend, and not twist