Homework 6.1 Three thick and 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 Side view δ δ δ Top view
Homework 6.1 P = 1000 N Side view Top view δ δ δ Chosen section/element of interest (to calculate Q) Interface of interest, along which (horizontal) shearing force matters (i.e., has the tendency to bend/cut through the nails) Cross-section view EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 6 Shearing Stress for Beams
Homework 6.1 For first moment for the section/element of interest (i.e., within red dashed rectangle) Q Q = ya = = 160cm 3 Moment of inertia for the entire cross-section I I = 1440cm 4 Cross-section view y Total (horizontal) shearing force along the interface of interest for the chosen section/element must not be larger than the total shearing force that could be allowed for the two nails within the chosen section/element H = VQ I x 2F allowable δ = x 3.6cm
Homework 6.2 For uniform square box beam with outer edge of and wall thickness of 0. 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 20 in 20 in Moment of inertia for entire cross-section I = = 30.75in 4 F E X 0. 0.
Homework 6.2 Average shearing stress τ ave = VQ It For part (a), average shearing stress reaches maximum when first Q reaches maximum while t reaches minimum, which is the center or neutral axis. The chosen section/element of interest is the top half (shaded in red), and its first moment can be calculated as Max average shearing stress Q = Q o Q i = y o A o y i A i = = 7.625in 3 Total width t = 0. + 0. = 1.0 in τ ave = VQ It EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 6 Shearing Stress for Beams Q o = 0.5P Q It Q i 0. = 248psi 0.
Homework 6.2 For part (b) 0. the chosen section/element of interest is the top bar (shaded in red), and its first moment Q can be calculated as Q = ya = = 5.625in 3 0. Total width t = 0. + 0. =?? in Average shearing stress at X τ ave = VQ It = = 183psi 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
Homework 6.3 From chapter 5 slides, the shear and bending moment diagrams are obtained Max axial normal stress σ max = M maxc I For narrow rectangular beam max shearing stress τ max = 3V max 2A When σ max = τ max Conclusion (short beam) = f(w, L, b, h) = g(w, L, b, h) L h = 1 V 0.5wL EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 6 Shearing Stress for Beams D 0-0.5wL M 0.125wL 2 R D = 0.5wL 0 Shear Diagram 0.5L w Moment Diagram 0.5L R E = 0.5wL E x x
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. The section/element of interest is colored in red and the interface of interest are the left and right side of the section The allowable shearing force per unit length for the glue interface τ ave_allowable 5cm First moment for the section/element Q = ya = = 375cm 3 Moment of inertia for the entire cross-section I = = 9167cm 4 2.5 cm 5 cm 5 cm 2.5 cm
Homework 6.4 Horizontal shear flow (or shear force per unit length) satisfy 2.5 cm 5 cm 2.5 cm q = VQ I Therefore, 2τ ave_allowable 5cm 5 cm V = 4889N