MEE 320 Problems Chapter 6, Spring 2018

Lebanese American University · Strength of Material · 4 pages · Dr. Carine Habchi

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50 mm s s 50 mm 50 mm 100 mm PROBLEM 6.1 Three full-size 50  100-mm boards are nailed together to form a beam that is subjected to a vertical shear of 1500 N. Knowing that the allowable shearing force in each nail is 400 N, determine the largest longitudinal spacing s that can be used between each pair of nails. SOLUTION 3 3 6 4 1 1 (100)(150) 28.125 10 mm 12 12     I bh 6 4 28.125 10  m   2 (100)(50) 5000 mm   A 1 y  50 mm 3 3 6 3 1 250 10 mm 250 10  m      Q Ay 6 3 6 (1500)(250 10 ) 13.3333 10 N/m 28.125 10         VQ q I qs  2 nail F 3 nail 3 2 (2)(400) 60.0 10 m 13.3333 10       F s q s  60.0 mm  1.5 m 100 mm 200 mm 40 mm 12 mm 12 mm 150 mm 0.3 m 10 kN n a n PROBLEM 6.10 For the beam and loading shown, consider section n-n and determine (a) the largest shearing stress in that section, (b) the shearing stress at point a. SOLUTION At section n-n, V  10 kN. 1 2 3 3 2 1 1 2 2 2 2 3 3 2 6 6 6 6 4 6 4 4 1 1 4 12 12 1 1 (100)(150) 4 (50)(12) (50)(12)(69) 12 12 28.125 10 4 0.0072 10 2.8566 10 39.58 10 mm 39.58 10 m I I I b h b h A d                                          (a) 1 1 2 2 3 3 6 3 2 (100)(75)(37.5) (2)(50)(12)(69) 364.05 10 mm 364.05 10 m 100 mm = 0.100 m Q A y A y t           3 6 3 max 6 (10 10 )(364.05 10 ) 920 10 Pa (39.58 10 )(0.100) VQ It           max 920 kPa    (b) 1 1 2 2 3 3 6 3 2 (100)(40)(55) (2)(50)(12)(69) 302.8 10 mm 302.8 10 m 100 mm 0.100 m Q A y A y t            3 6 3 6 (10 10 )(302.8 10 ) 765 10 Pa (39.58 10 )(0.100) a VQ It           765 kPa a   2.4 kN 4.8 kN 7.2 kN 1 m 1 m 1 m 0.5 m 150 mm A E b B C D PROBLEM 6.18 For the beam and loading shown, determine the minimum required width b, knowing that for the grade of timber used, all 12 MPa   and all 825 kPa.   SOLUTION 0: 3 (2)(2.4) (1)(4.8) (0.5)(7.2) 0 2 kN MD A          A Draw the shear and bending moment diagrams. 3 max 3 max | | 7.2 kN 7.2 10 N | | 3.6 kN m 3.6 10 N m V M         Bending: M   S max min 3 6 6 3 3 3 | | 3.6 10 12 10 300 10 m 300 10 mm M S           For a rectangular section, 2 1 6 S bh  3 2 2 6 (6)(300 10 ) 80 mm (150) S b h     Shear: Maximum shearing stress occurs at the neutral axis of bending for a rectangular section. 2 3 2 1 8 3 1 12 3 3 3 3 1 1 1 , , 2 4 8 1 12 ( ) 3 2 ( )( ) 3 (3)(7.2 10 ) 87.3 10 m 2 (2)(150 10 )(825 10 )                     A bh y h Q Ay bh I bh t b V bh VQ V It bh bh b V b h The required value of b is the larger one. b  87.3 mm  PROBLEM 6.2 For the beam and loading shown, consider section n-n and determine the shearing stress at (a) point a, (b) the shearing stress at point b. SOLUTION 100 kN A B R = R = At section n-n, V =100 kN Locate centroid and calculate moment of inertia. Part 2 A(mm ) y(mm) 3 Ay(mm ) d(mm) 2 4 Ad (mm ) 4 I (mm )  3200 170 544000 55.4 9821312 106667  7200 90 648000 24.6 4357152 19440000 Σ 10400 1192000 14178464 19546667 2 6 1192000 114.6 mm 104.00 14178464 19546667 33.725 10 Ay Y A I Ad I = Σ = = Σ = Σ + Σ = + = × (a) 3 (20)(38)(114.6 19) 72656 mm 20 mm Qa Ay t = = − = = 6 (100000)(72656) 10.8 MPa (33.725 10 )(20) a VQ It τ = = = ×  (b) 9 (20)(76)(114.6 38) 1164.32 mm 20 mm Qb Ay t = = − = = 6 (100000)(116432) 17.3 MPa (33.725 10 )(20) a VQ It τ = = = ×  1