7.4 SHEAR FLOW IN BUILT-UP MEMBERS
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Transcript of 7.4 SHEAR FLOW IN BUILT-UP MEMBERS
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7. Transverse Shear7.4 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 member’s length
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7. Transverse Shear7.4 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
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7. Transverse Shear7.4 SHEAR FLOW IN BUILT-UP MEMBERS• Thus shear flow is
q = VQ/I Equation 7-6q = shear flow, measured as a force per unit length
along the beamV = internal resultant shear force, determined from
method of sections and equations of equilibriumI = moment of inertia of entire x-sectional area
computed about the neutral axisQ = ∫A’ y dA’ = y’A’, 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’
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7. Transverse Shear7.4 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
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7. Transverse Shear7.4 SHEAR FLOW IN BUILT-UP MEMBERSIMPORTANT• 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
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7. Transverse ShearEXAMPLE 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.
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7. Transverse ShearEXAMPLE 7.4 (SOLN)
Section propertiesNeutral axis (centroid) is located from bottom of the beam. Working in units of meters, we have
y = y A y= ... = 0.1968 m
Moment of inertia about neutral axis is
I = ... = 87.52(10-6) m4
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7. Transverse ShearEXAMPLE 7.4 (SOLN)
Section propertiesSince the glue at B and B’ holds the top board to the beam, we have
Likewise, glue at C and C’ holds inner board to beam, so
QB = y’B A’B = [0.305 m 0.1968 m](0.250 m)(0.01 m)
QB = 0.270(10-3) m3
QC = y’C A’C = ... = 0.01025(10-3) m3
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7. Transverse ShearEXAMPLE 7.4 (SOLN)
Shear flowFor B and B’, we have
Similarly, for C and C’,
q’B = VQB /I = [850 kN(0.270(10-3) m3]/87.52(10-6) m4
q’B = 2.62 MN/m
q’C = VQC /I = [850 kN(0.0125(10-3) m3]/87.52(10-6) m4
q’C = 0.0995 MN/m
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7. Transverse ShearEXAMPLE 7.4 (SOLN)
qB = 1.31 MN/m
qC = 0.0498 MN/m
Shear flowSince 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
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7. Transverse Shear7.5 SHEAR FLOW IN THIN-WALLED MEMBERS
• We can use shear-flow equation q = VQ/I to find the shear-flow distribution throughout a member’s x-sectional area.
• We assume that the member has thin walls, i.e., wall thickness is small compared with height or width of member
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7. Transverse Shear7.5 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
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7. Transverse Shear7.5 SHEAR FLOW IN THIN-WALLED MEMBERS
• To determine distribution of shear flow along top right flange of beam, shear flow is
Equation 7-8q = (b/2 x)
Vt d2I
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7. Transverse Shear7.5 SHEAR FLOW IN THIN-WALLED MEMBERS
• Similarly, for the web of the beam, shear flow is
Equation 7-9q = +0.5(d2/4 y2)
VtI
db2
[ ]
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7. Transverse Shear7.5 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
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7. Transverse Shear7.5 SHEAR FLOW IN THIN-WALLED MEMBERS
• Directional sense of q is such that shear appears to “flow” through the x-section, inward at beam’s top flange, “combining” and then “flowing” downward through the web, and then separating and “flowing” outward at the bottom flange
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7. Transverse Shear7.5 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
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7. Transverse ShearEXAMPLE 7.7
Thin-walled box beam shown is subjected to shear of 10 kN. Determine the variation of shear flow throughout the x-section.
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7. Transverse ShearEXAMPLE 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 = y’A’ = 0 since y’ = 0.
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7. Transverse ShearEXAMPLE 7.7 (SOLN)
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
Because QB = 0, then qB = 0
QC = y’A’ = (3.5 cm)(5 cm)(1 cm) = 17.5 cm3
qC = VQC/I = ... = 95.1 N/mm
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7. Transverse ShearEXAMPLE 7.7 (SOLN)
Shear flow at D is computed using the three dark-shaded rectangles shown. We have
QD = y’A’ = ... = 30 cm3
qC = VQD/I = ... = 163 N/mm
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7. Transverse ShearEXAMPLE 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)
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7. Transverse Shear*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
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7. Transverse Shear*7.6 SHEAR CENTER
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7. Transverse Shear*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 V=P 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 e = (Ff d)/P
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7. Transverse Shear*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 member’s x-sectional area
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7. Transverse Shear*7.6 SHEAR CENTERIMPORTANT• 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
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7. Transverse Shear*7.6 SHEAR CENTERProcedure for analysisShear-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 segment’s 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
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7. Transverse Shear*7.6 SHEAR CENTERProcedure for analysisShear-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
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7. Transverse Shear*7.6 SHEAR CENTERProcedure for analysisShear 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
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7. Transverse Shear*7.6 SHEAR CENTERProcedure for analysisShear 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
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7. Transverse ShearEXAMPLE 7.8
Determine the location of the shear center for the thin-walled channel section having the dimensions as shown.
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7. Transverse ShearEXAMPLE 7.8 (SOLN)
Shear-flow resultantsVertical 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.
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7. Transverse ShearEXAMPLE 7.8 (SOLN)
Shear-flow resultantsX-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 + 2[bt(0.5h)2] = (0.5th2)[(h/6) + b]
Thus, q at the arbitrary position x is
q = =VQI
V(b – x)h[(h/6) + b]
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7. Transverse ShearEXAMPLE 7.8 (SOLN)
Shear-flow resultantsHence,
The same result can be determined by first finding (qmax)f, then determining triangular area 0.5b(qmax)f = Ff
Ff = ∫0 q dx = … =Vb2
2h[(h/6) + b]b
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7. Transverse ShearEXAMPLE 7.8 (SOLN)
Shear centerSumming moments about pt A, we require
Ve = Ff h
b2
[(h/3) + 2b]e =
…
As stated previously, e depends only on the geometry of the x-section.
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7. Transverse ShearCHAPTER 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
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7. Transverse ShearCHAPTER 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
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7. Transverse ShearCHAPTER 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