EMA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 6 Shearing Stress in Beams & Thin-Walled Members

EMA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 6 Shearing Stress in Beams & Thin-Walled Members

Beams Bending & Shearing EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 6 Shearing Stress for Beams

ntroduction n a long beam with transverse loading, x due to bending moment is often the primary design factor m ~ ~ m Mc n a short beam with transverse loading, due to shearing force often becomes the primary design factor as maximum bending moment is usually small V m A D V P/2 0 -P/2 M PL/4 0 x or 0 0.5L P 0.5L F Shear Diagram 0.5L L Moment Diagram 0.5L L m ~ m Mc V m A E x x x

Shearing Stress in Short Beams with Transverse Loading (1) Equations of equilibrium for shearing components: y component: z component: τ xy da = V τ xz da = 0 Note: for pure bending, there is NO transverse load NO transverse shearing force and NO shearing stress

Shearing Stress in Short Beams with Transverse Loading (2) w/ transverse loading Shearing stresses present For a beam subjected to transverse loading P, for a small element as labelled with da at the yz cross-section: Presence of y direction shearing stress in the yz normal cross-section xy Presence of x direction horizontal shearing stress in the xz plane yx Different from pure bending (as below): NO transverse load & NO shearing stresses

Shear on a Horizontal Plane in a Beam (1) For a prismatic beam with loads, at arbitrary distance x from A, take an element CDD C parallel to N.S.: x Length of element y 1 Distance of element from neutral surface α Element cross-section area H Horizontal shearing force on the element Equilibrium along horizontal x axis gives: H da 0 C D Side view Right/Crosssection view Side view

Shear on a Horizontal Plane in a Beam (2) H Axial normal stress σ vs. M: Define Q yda For small Δx, ΔM is related to local shear V H da 0 VQ C Horizontal shear force ΔH x D σ = My H M M D D M M C C yda First moment w.r.t. the neutral axis for the element with cross-section α M dm dx x Vx Horizontal shear increases with vertical shear V, element size Q, or length Δx increases but decreases as moment of inertia increases

Shear on a Horizontal Plane in a Beam (3) H VQ x Define q as horizontal shear per unit length, also called shear flow q H x VQ Notes about symbols: V Local shearing force Q First moment w.r.t. the neutral axis, Q yda Q reaching maximum absolute value when y 1 = 0 Moment of inertia for the entire cross-section

Class Example 6.1 (1) A prismatic wide-flange beam is made of three rectangular beams connected by a series nails along its longitude axis, as illustrated. f the spacing between every two nails is 25 mm and that the vertical shear in the beam is V = 500 N, determine the shear force each nail is subjected to due to the vertical shear force V. Knowing cross-section moment of inertia = 16.20 10-6 m 4 The nails prevent the horizontal shear to separate horizontal (top & bottom) flanges from the vertical (center) web. The horizontal shearing force ΔH over Δx is: H VQ x Side view Top view Δx Δx V

Class Example 6.1 (2) Consider the top flange (dashed portion), Q yda y Q 0.06m(0.02m0.1m) 6 3 12010 m Horizontal shearing force along the interface of interest (between top flange and web) H VQ x Top flange nterface of interest x = 25 mm Horizontal shearing force born by each nail: 6 3 500N 12010 m H 0.025m 92. 6N 6 4 16.2010 m

Shearing Stress in a Beam (1) Horizontal shearing H VQ x force for element Δx Define t as the width for the element (for a horizontal cross-section), the average shearing stress along the horizontal plane τ ave is τ ave = H A = VQ x Therefore, ave VQ t 1 t x Average shearing stress increases with V and Q, but decreases as or t increases z y x n addition, xy = yx

Shearing Stress in a Beam (2) ave VQ t xy = yx Notes: Generally non-uniform xy or yx distribution: yx = xy = 0 at both top & bottom surfaces, while they may or may not be largest at the neutral surface For fixed y, often smallest at the beam center For a narrow beam with b h/4, xy or yx almost uniform across the width of the beam with variation < 0.8%

Shearing Stresses in a Narrow Rectangular Beam (1) Average shearing stress - τ ave depends on y! VQ ave t For a rectangular beam, consider element A at arbitrary distance y from neutral axis Distance from neutral axis to y = 1 element A centroid C is 2 Therefore, first moment Q for element A Q = Ay Entire cross-section moment of inertia Average (horizontal) shearing stress along interface of interest at distance y from neutral axis (c + y) = b c y 1 2 c + y = 1 2 b(c2 y 2 ) Element A = z = bh3 12 = 2 3 bc3 τ ave (y) = VQ b nterface of interest = 3 4 c2 y 2 bc 3 V

Shearing Stresses in a Narrow Rectangular Beam (2) Average shearing stress at distance y within a rectangular beam from the N.S. τ ave (y) = VQ b = 3 4 c2 y 2 bc 3 Total cross-section area A = bh = 2bc Parabolic distribution for ave (y) vs. y y = c τ ave (y) = 0 y = 0 (neutral axis), reaching maximum ave ( y) max V τ ave (y) = 3V y2 (1 2A c 2) 3V 2A = 3V 2 2bc c2 y 2 c 2 A

Shearing Stresses in Other Prismatic Beams Example: American Standard beam (S-beam) or Wideflange beam (W-beam) (1) VQ( y) Flange ave ( y) t( y) Web Web Flange: t is large Q(y)= yda changes from zero (e.g., at AB surface) to a finite number (e.g., at ~EF cross-section) xy (y) small & varies: e.g., ~0 btw. D to E, finite btw. E to F

Shearing Stresses in Other Prismatic Beams Example: American Standard beam (S-beam) or Wideflange beam (W-beam) (2) VQ( y) Flange ave ( y) t( y) Web Web: t is small Q (y)= yda does not change too much from E-F to C xy large and does not vary much in cross-section Web carries almost all the shearing load max V A web

Class Example 6.2 (1) For a prismatic beam DE with rectangular cross-section of 5 cm 20 cm subjected to P = 200 kn at center, as illustrated. Please calculate (a) the maximum shearing stress, and (b) the length for the beam L for which maximum normal stress x will be smaller than shearing stress The absolute shearing force is constant of V = 0.5P = 100 kn. For a narrow rectangular beam, maximum shearing stress max 3V 2A 3 (10010 N) 2(0.050.20) m D V 0.5P 0-0.5P 3 7 1.510 2 P 0.5L 0.5L h = 20 cm b = 5 cm Shear Diagram 0.5L L Pa 15MPa E x

Class Example 6.2 (2) For maximum axial normal stress due to bending moment to be smaller than maximum vertical shearing stress max M max Therefore, c 0.25PL 0.5h 1 3 bh 12 max 3V 2A 30.5P 2bh Simplify and we have L h 1 2 D M 0.25PL 0 P 0.5L 0.5L F h = 20 cm b = 5 cm 1 1 L h 0.2m 0. 1m 2 2 Moment Diagram 0.5L L E x

Longitudinal Shear on a Beam Element of Arbitrary Cross- section Shape For a prismatic beam with transverse loads, at distance x from A, take an element CDD C with an arbitrary cross-section shape & interface parallel to the neutral axis, follow the same derivation as before, the longitudinal/horizontal shear flow is still q H x VQ

Class Example 6.3 (1) A square box beam is constructed from four beams with nails as shown. Knowing that the spacing between nails is 1.75 in. and the beam is subjected to a vertical shear of V = 600 lb, determine the shearing force in each nail. Knowing crosssection moment of inertia = 27.42 in 4 The nails prevent the longitudinal shear Side view to separate the two horizontal beams in the center from the two vertical beams The longitudinal shearing force for an element of length Δx is: H VQ x Top view Δx Δx V

Class Example 6.3 (2) Consider the top horizontal beam (darker, shaded region) between two vertical beams, Q Q yda ya' ( 4.5/ 2 0.75/ 2) (0.75in3in) 4.22in 3 Total shearing force on the top horizontal beam over Δx VQ (Two) interfaces of interest H x in which x = 1.75 in Because the total shearing force over Δx along (two - both left & right) interfaces of interest is resisted by two nails - one on each side, the shearing force applied on (or resisted by) each nail is 3 1 1 600lb 4.22in P H 1.75in 80. 8lb 4 2 2 27.42in Top beam

Shearing Stresses in Thin-Walled Members The equations are valid for thin-walled members if loads applied in plane of symmetry H = VQ x q = H x = VQ τ ave(y) = VQ t

EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 6 Shearing Stress for Beams Homework 6.0 Read section 6.1, 6.3, and 6.4 and give an honor statement confirm reading

EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 6 Shearing Stress for Beams Homework 6.1 Three 4 cm thick and 10 cm wide boards are connected together by two parallel rows of uniformly distributed nails separated by longitude distance δ to form a beam that is subject to constant vertical shear of 1000 N. Knowing the maximum allowed shearing force for each nail is 200 N, determine the largest longitude separation δ between nails? P = 1000 N 4 cm 4 cm 4 cm Side view 10 cm δ δ δ Top view

EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 6 Shearing Stress for Beams Homework 6.2 For uniform square box beam with outer edge of 5 in and wall thickness of 0.5 in and loading shown, consider cross-section C-C and calculate (a) the largest shearing stress in that section and (b) the shearing stress at point X D 10 in C C 2 kips F E 5 in X 0.5 in 0.5 in 20 in 20 in 5 in

EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 6 Shearing Stress for Beams Homework 6.3 A beam DE of length L and rectangular cross-section with width b and height h (noting that h>4b) carries a uniformly distributed load w. Please determine the condition in terms of the relationship between span L and beam cross-section height h when maximum shearing stress m and maximum normal stress m equal each other, i.e., m = m w D E h L b

EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 6 Shearing Stress for Beams Homework 6.4 A beam with made by gluing together 4 pieces of wood planks with cross-section dimension as shown. Knowing the allowable average shearing stress in the glued joints is 200 kpa, please calculate the largest allowable vertical shearing force in the beam. 2.5 cm 5 cm 2.5 cm 10 cm 5 cm 10 cm