CHAPTER OBJECTIVES Develop a method for finding the shear stress in a beam having a prismatic x-section and made from homogeneous material that behaves.

CHAPTER OBJECTIVES Develop a method for finding the shear stress in a beam having a prismatic x-section and made from homogeneous material that behaves in a linear-elastic manner This method of analysis is limited to special cases of x-sectional geometry Discuss the concept of shear flow, with shear stress for beams and thin-walled members Discuss the shear center

CHAPTER OUTLINE Shear in Straight Members The Shear Formula Shear Stresses in Beams Shear Flow in Built-up Members Shear Flow in Thin-Walled Members *Shear Center

7.1 SHEAR IN STRAIGHT MEMBERS
Shear V is the result of a transverse shear-stress distribution that acts over the beam’s x-section. Due to complementary property of shear, associated longitudinal shear stresses also act along longitudinal planes of beam

As shown below, if top and bottom surfaces of each board are smooth and not bonded together, then application of load P will cause the boards to slide relative to one another. However, if boards are bonded together, longitudinal shear stresses will develop and distort x-section in a complex manner

As shown, when shear V is applied, the non-uniform shear-strain distribution over x-section will cause it to warp, i.e., not remain in plane.

Recall that the flexure formula assumes that x-sections must remain plane and perpendicular to longitudinal axis of beam after deformation This is violated when beam is subjected to both bending and shear, we assume that the warping is so small it can be neglected. This is true for a slender beam (small depth compared with its length) For transverse shear, shear-strain distribution throughout the depth of a beam cannot be easily expressed mathematically

Thus, we need to develop the formula for shear stress is indirectly using the flexure formula and relationship between moment and shear (V = dM/dx)

By first principles, flexure formula and V = dM/dx, we obtain
7.2 THE SHEAR FORMULA By first principles, flexure formula and V = dM/dx, we obtain  = VQ It Equation 7-3  = shear stress in member at the pt located a distance y’ from the neutral axis. Assumed to be constant and therefore averaged across the width t of member V = internal resultant shear force, determined from method of sections and equations of equilibrium

By first principles, flexure formula and V = dM/dx, we get:
7.2 THE SHEAR FORMULA By first principles, flexure formula and V = dM/dx, we get:  = VQ It Equation 7-3 I = moment of inertia of entire x-sectional area computed about the neutral axis t = width of the member’s x-sectional area, measured at the pt where  is to be determined Q = ∫A’ y dA’ = y’A’, where A’ is the top (or bottom) portion of member’s x-sectional area, defined from section where t is measured, and y’ is distance of centroid of A’, measured from neutral axis

The equation derived is called the shear formula
Since Eqn 7-3 is derived indirectly from the flexure formula, the material must behave in a linear-elastic manner and have a modulus of elasticity that is the same in tension and in compression Shear stress in composite members can also be obtained using the shear formula To do so, compute Q and I from the transformed section of the member as discussed in section 6.6. Thickness t in formula remains the actual width t of x-section at the pt where  is to be calculated

7.3 SHEAR STRESSES IN BEAMS
Rectangular x-section Consider beam to have rectangular x-section of width b and height h as shown. Distribution of shear stress throughout x-section can be determined by computing shear stress at arbitrary height y from neutral axis, and plotting the function. Hence, Q =  y2 b ( ) 1 2 h2 4

Rectangular x-section After deriving Q and applying the shear formula, we have  =  y2 ( ) 6V bh3 h2 4 Equation 7-4 Eqn 7-4 indicates that shear-stress distribution over x-section is parabolic.

Rectangular x-section At y = 0, we have max = 1.5 V A Equation 7-5 By comparison, max is 50% greater than the average shear stress determined from avg = V/A.

Wide-flange beam A wide-flange beam consists of two (wide) “flanges” and a “web”. Using analysis similar to a rectangular x-section, the shear stress distribution acting over x-section is shown

Wide-flange beam The shear-stress distribution also varies parabolically over beam’s depth Note there is a jump in shear stress at the flange-web junction since x-sectional thickness changes at this pt The web carries significantly more shear force than the flanges

Limitations on use of shear formula One major assumption in the development of the shear formula is that shear stress is uniformly distributed over width t at section where shear stress is to be determined By comparison with exact mathematical analysis based on theory of elasticity, the magnitude difference can reach 40% This is especially so for the flange of a wide-flange beam

Limitations on use of shear formula The shear formula will also give inaccurate results for the shear stress at the flange-web junction of a wide-flange beam, since this is a pt of sudden x-sectional change (stress concentration occurs here) Furthermore, inner regions of flanges are free boundaries, thus shear stress at these boundaries should be zero However, shear formula calculated at these pts will not be zero

Limitations on use of shear formula Fortunately, engineers are often interested in the average maximum shear stress, which occurs at the neutral axis, where b/h ratio is very small Also, shear formula does not give accurate results when applied to members having x-sections that are short or flat, or at pts where the x-section suddenly changes It should also not be applied across a section that intersects the boundary of a member at an angle other than 90o

IMPORTANT Shear forces in beams cause non-linear shear-strain distributions over the x-section, causing it to warp Due to complementary property of shear stress, the shear stress developed in a beam acts on both the x-section and on longitudinal planes The shear formula was derived by considering horizontal force equilibrium of longitudinal shear stress and bending-stress distributions acting on a portion of a differential segment of the beam

IMPORTANT The shear formula is to be used on straight prismatic members made of homogeneous material that has linear-elastic behavior. Also, internal resultant shear force must be directed along an axis of symmetry for x-sectional area For beam having rectangular x-section, shear stress varies parabolically with depth. For beam having rectangular x-section, maximum shear stress is along neutral axis

IMPORTANT Shear formula should not be used to determine shear stress on x-sections that are short or flat, or at pts of sudden x-sectional changes, or at a pt on an inclined boundary

Procedure for analysis Internal shear Section member perpendicular to its axis at the pt where shear stress is to be determined Obtain internal shear V at the section Section properties Determine location of neutral axis, and determine the moment of inertia I of entire x-sectional area about the neutral axis Pass an imaginary horizontal section through the pt where the shear stress is to be determined

Procedure for analysis Section properties Measure the width t of the area at this section Portion of area lying either above or below this section is A’. Determine Q either by integration, Q = ∫A’ y dA’, or by using Q = y’A’. Here, y’ is the distance of centroid of A’, measured from the neutral axis. (TIP: A’ is the portion of the member’s x-sectional area being “held onto the member” by the longitudinal shear stresses.)

Procedure for analysis Shear stress Using consistent set of units, substitute data into the shear formula and compute shear stress  Suggest that proper direction of transverse shear stress be established on a volume element of material located at the pt where it is computed  acts on the x-section in the same direction as V. From this, corresponding shear stresses acting on the three other planes of element can be established

EXAMPLE 7.3 Beam shown is made from two boards. Determine the maximum shear stress in the glue necessary to hold the boards together along the seams where they are joined. Supports at B and C exert only vertical reactions on the beam.

EXAMPLE 7.3 (SOLN) Internal shear Support reactions and shear diagram for beam are shown below. Maximum shear in the beam is 19.5 kN.

EXAMPLE 7.3 (SOLN) Section properties The centroid and therefore the neutral axis will be determined from the reference axis placed at bottom of the x-sectional area. Working in units of meters, we have y = = ... = m  yA  A Thus, the moment of inertia, computed about the neutral axis is, I = ... = 27.0(10-6) m4

EXAMPLE 7.3 (SOLN) Section properties The top board (flange) is being held onto the bottom board (web) by the glue, which is applied over the thickness t = 0.03m. Consequently A’ is defined as the area of the top board, we have Q = y’A’ = [(0.180 m  m  m] (0.03 m)(0.150 m) Q = (10-3) m3

EXAMPLE 7.3 (SOLN) Shear stress Using above data, and applying shear formula yields VQ It max = = ... = 4.88 MPa Shear stress acting at top of bottom board is shown here. It is the glue’s resistance to this lateral or horizontal shear stress that is necessary to hold the boards from slipping at support C.

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 member’s length

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

Thus shear flow is q = VQ/I 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’ = 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’

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

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

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.

EXAMPLE 7.4 (SOLN) Section properties Neutral axis (centroid) is located from bottom of the beam. Working in units of meters, we have y =  y A  y = ... = m Moment of inertia about neutral axis is I = ... = 87.52(10-6) m4

EXAMPLE 7.4 (SOLN) Section properties Since the glue at B and B’ holds the top board to the beam, we have QB = y’B A’B = [0.305 m  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 QC = y’C A’C = ... = (10-3) m3

EXAMPLE 7.4 (SOLN) Shear flow For B and B’, we have q’B = VQB /I = [850 kN(0.270(10-3) m3]/87.52(10-6) m4 q’B = 2.62 MN/m Similarly, for C and C’, q’C = VQC /I = [850 kN(0.0125(10-3) m3]/87.52(10-6) m4 q’C = MN/m

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 = MN/m

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 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

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

To determine distribution of shear flow along top right flange of beam, shear flow is q = (b/2  x) Vt d 2I Equation 7-8

Similarly, for the web of the beam, shear flow is q = (d2/4  y2) Vt I db 2 [ ] Equation 7-9

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

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

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

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.

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 = y’A’ = 0 since y’ = 0.

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 = y’A’ = (3.5 cm)(5 cm)(1 cm) = 17.5 cm3 qC = VQC/I = ... = 95.1 N/mm

EXAMPLE 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

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)

*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

*7.6 SHEAR CENTER

*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 e = (Ff d)/P  MA = Ff d = Pe

*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

*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

*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 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

*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

*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

*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

EXAMPLE 7.8 Determine the location of the shear center for the thin-walled channel section having the dimensions as shown.

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.

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 + 2[bt(0.5h)2] = (0.5th2)[(h/6) + b] Thus, q at the arbitrary position x is q = = VQ I V(b – x) h[(h/6) + b]

Shear-flow resultants Hence,
EXAMPLE 7.8 (SOLN) Shear-flow resultants Hence, Ff = ∫0 q dx = … = Vb2 2h[(h/6) + b] b The same result can be determined by first finding (qmax)f, then determining triangular area 0.5b(qmax)f = Ff

EXAMPLE 7.8 (SOLN) Shear center Summing 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.

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

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

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

CHAPTER OBJECTIVES Develop a method for finding the shear stress in a beam having a prismatic x-section and made from homogeneous material that behaves.

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CHAPTER OBJECTIVES Develop a method for finding the shear stress in a beam having a prismatic x-section and made from homogeneous material that behaves.

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