MEE 320 Problems Chapter 1, Spring 2018

PROBLEM 1. Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Knowing that 1 d = 50 mm and 2 = 30 d mm, find the average normal stress at the midsection of (a) rod AB, (b) rod BC. SOLUTION (a) Rod AB 3 2 2 3 2 3 2 1 3 6 3 40 30 70 kN 70 10 N (50) 1.9635 10 mm 1.9635 10 m 4 4 70 10 35.7 10 Pa 1.9635 10 AB P A d P A π π σ − − = + = = × = = = × = × × = = = × × 35.7 MPa σ AB =  (b) Rod BC 3 2 2 2 6 2 2 3 6 6 30 kN 30 10 N (30) 706.86 mm 706.86 10 m 4 4 30 10 42.4 10 Pa 706.86 10 π π σ − − = = × = = = = × × = = = × × BC P A d P A 42.4 MPa σ BC =  4 0.2 m 0.25 m 0.4 m 20 kN C B A D E PROBLEM 1.7 Each of the four vertical links has an 8  36-mm uniform rectangular cross section and each of the four pins has a 16-mm diameter. Determine the maximum value of the average normal stress in the links connecting (a) points B and D, (b) points C and E. SOLUTION Use bar ABC as a free body. 3 3 3 3 0 : (0.040) (0.025 0.040)(20 10 ) 0 32.5 10 N Link is in tension. 0 : (0.040) (0.025)(20 10 ) 0 12.5 10 N Link is in compression.                  C BD BD B CE CE M F F BD M F F CE Net area of one link for tension (0.008)(0.036 0.016)   6 2 160 10  m   For two parallel links, 6 2 net 320 10  m   A (a) 3 6 6 net 32.5 10 101.563 10 320 10 BD BD F A         101.6 MPa  BD   Area for one link in compression  (0.008)(0.036) 6 2 288 10  m   For two parallel links, 6 2 576 10  m   A (b) 3 6 6 12.5 10 21.701 10 576 10 CE CE F A            21.7 MPa  CE    PROBLEM 1.1 When the force P reached 8 kN, the wooden specimen shown failed in shear along the surface indicated by the dashed line. Determine the average shearing stress along that surface at the time of failure. SOLUTION Area being sheared: 2 6 2 90 mm 15 mm 1350 mm 1350 10 − m = × = = × A Force: 3 8 10 N = × P Shearing stress: 3 6 6 8 10 5.93 10 Pa 1350 10 τ − × = − = × × P A τ = 5.93 MPa  7 16 mm 750 mm 750 mm 12 mm 50 mm B A C P � PROBLEM 1.25 Knowing that   40° and P  9 kN, determine (a) the smallest allowable diameter of the pin at B if the average shearing stress in the pin is not to exceed 120 MPa, (b) the corresponding average bearing stress in member AB at B, (c) the corresponding average bearing stress in each of the support brackets at B. SOLUTION Geometry: Triangle ABC is an isoseles triangle with angles shown here. Use joint A as a free body. Law of sines applied to force triangle: sin 20 sin110 sin50 sin110 sin 20 (9)sin110 24.727 kN sin 20 AB AC AB P F F P F             PROBLEM 1.25 (Continued) (a) Allowable pin diameter. 2 2 4 2 2 2       AB AB AB P F F F A d d where 3 24.727 10 N   FAB 3 2 6 2 6 2 (2)(24.727 10 ) 131.181 10 m (120 10 )          FAB d 3 11.4534 10  m   d 11.45 mm  (b) Bearing stress in AB at A. 3 6 2 3 6 6 (0.016)(11.4534 10 ) 183.254 10 m 24.727 10 134.933 10 Pa 183.254 10                b AB b b A td F A 134.9 MPa  (c) Bearing stress in support brackets at B. 3 6 2 1 3 6 2 6 (0.012)(11.4534 10 ) 137.441 10 m (0.5)(24.727 10 ) 89.955 10 Pa 137.441 10                AB b A td F A 90.0 MPa  75 mm 150 mm 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45� P' P PROBLEM 1.30 Two wooden members of uniform rectangular cross section are joined by the simple glued scarf splice shown. Knowing that the maximum allowable shearing stress in the glued splice is 620 kPa, determine (a) the largest load P that can be safely applied, (b) the corresponding tensile stress in the splice. SOLUTION 3 2 3 2 0 3 0 90 45 45 (150)(75) 11.25 10 mm 11.25 10 m 620 kPa 620 10 Pa sin 2 2 A P A                     (a) 3 3 2 0 (2)(11.25 10 )(620 10 ) sin2 sin90         A P 3 13.95 10 N   P  13.95 kN  (b) 2 3 2 3 0 cos (13.95 10 )(cos45 ) 11.25 10         P A 3 620 10 Pa     620 kPa  A B C D 480 mm 908 w P PROBLEM 1.38 Link BC is 6 mm thick, has a width w  25 mm, and is made of a steel with a 480-MPa ultimate strength in tension. What was the safety factor used if the structure shown was designed to support a 16-kN load P? SOLUTION Use bar ACD as a free body and note that member BC is a two-force member. 3 3 0: (480) (600) 0 600 (600)(16 10 ) 20 10 N 480 480          A BC BC M F P F P Ultimate load for member BC: U U F   A 6 3 (480 10 )(0.006)(0.025) 72 10 N     FU Factor of safety: 3 3 72 10 F.S. 20 10     U BC F F F.S.  3.60  PROBLEM 1.46 Three steel bolts are to be used to attach the steel plate shown to a wooden beam. Knowing that the plate will support a 110-kN load, that the ultimate shearing stress for the steel used is 360 MPa, and that a factor of safety of 3.35 is desired, determine the required diameter of the bolts. SOLUTION For each bolt, 110 36.667 kN 3 = = P Required: (F.S.) (3.35)(36.667) 122.83 kN U P P = = = 2 2 4 3 3 6 4 4 (4)(122.83 10 ) 20.8 10 m (360 10 ) π τ π πτ π − = = = × = = = × × U U U U U U P P P A d d P d d = 20.8 mm  180 mm 200 mm Top view Side view Front view 8 mm 20 mm 8 mm 8 mm 12 mm 12 mm B C B D D A B C A P PROBLEM 1.55 In the structure shown, an 8-mm-diameter pin is used at A, and 12-mm-diameter pins are used at B and D. Knowing that the ultimate shearing stress is 100 MPa at all connections and that the ultimate normal stress is 250 MPa in each of the two links joining B and D, determine the allowable load P if an overall factor of safety of 3.0 is desired. SOLUTION Statics: Use ABC as free body. 10 0 : 0.20 0.18 0 9      B A A M F P P F 10 0 : 0.20 0.38 0 19      A BD BD M F P P F Based on double shear in pin A, 2 2 6 2 (0.008) 50.266 10 m 4 4        A d 6 6 3 3 2 (2)(100 10 )(50.266 10 ) 3.351 10 N . . 3.0 10 3.72 10 N 9            U A A A F F S P F Based on double shear in pins at B and D, 2 2 6 2 (0.012) 113.10 10 m 4 4        A d 6 6 3 3 2 (2)(100 10 )(113.10 10 ) 7.54 10 N . . 3.0 10 3.97 10 N 19            U BD BD A F F S P F Based on compression in links BD, for one link, 6 2 (0.020)(0.008) 160 10  m    A 6 6 3 3 2 (2)(250 10 )(160 10 ) 26.7 10 N . . 3.0 10 14.04 10 N 19            U BD BD A F F S P F Allowable value of P is smallest,  3 3.72 10 N   P P  3.72 kN 