MEE 320 Problems Chapter 4, Spring 2018

Lebanese American University · Strength of Material · 10 pages · Dr. Carine Habchi

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M 5 15 kN · m Dimensions in mm B 20 40 20 20 20 80 A PROBLEM 4.1 Knowing that the couple shown acts in a vertical plane, determine the stress at (a) point A, (b) point B. SOLUTION For rectangle: 3 1 12 I bh  Outside rectangle: 3 1 1 (80)(120) I  12 6 4 6 4 1 11.52 10 mm 11.52 10 m I      Cutout: 3 2 1 (40)(80) 12 I  6 4 6 4 2 1.70667 10 mm 1.70667 10 m I      Section: 6 4 1 2 9.81333 10 m I I I      (a) 40 mm 0.040 m yA   3 6 6 (15 10 )(0.040) 61.6 10 Pa 9.81333 10 A A My I            61.6 MPa A    (b) 60 mm 0.060 m yB     3 6 6 (15 10 )( 0.060) 91.7 10 Pa 9.81333 10 B B My I            91.7 MPa B   10 mm 10 mm 50 mm 10 mm 150 mm 150 mm A D B C 10 kN 10 kN 250 mm 50 mm PROBLEM 4.11 Two vertical forces are applied to a beam of the cross section shown. Determine the maximum tensile and compressive stresses in portion BC of the beam. SOLUTION A, mm2 y0, mm Ay0, mm3  600 30 18 103   600 30 18 103   300 5 1.5 103   1500 37.5 103  3 0 37.5 10 25mm 1500 Y    Neutral axis lies 25 mm above the base. 3 2 3 4 4 1 2 1 3 2 3 4 3 3 4 9 4 1 2 3 1 (10)(60) (600)(5) 195 10 mm 195 mm 12 1 (30)(10) (300)(20) 122.5 10 mm 12 512.5 10 mm 512.5 10 m I I I I I I I I                   top bot 3 3 3 35 mm 0.035 m 25 mm 0.025 m 150 mm 0.150 m 10 10 N (10 10 )(0.150) 1.5 10 N m                 y y a P M Pa 3 top 6 top 9 (1.5 10 )(0.035) 102.4 10 Pa 512.5 10 My I            top 102.4 MPa (compression)     3 6 bot bot 9 (1.5 10 )( 0.025) 73.2 10 Pa 512.5 10 M y I            bot 73.2 MPa (tension)    54 mm 40 mm 80 mm M PROBLEM 4.19 Knowing that for the extruded beam shown the allowable stress is 120 MPa in tension and 150 MPa in compression, determine the largest couple M that can be applied. SOLUTION A, mm2 y0, mm Ay0, mm3 d, mm  2160 27 58,320 3  1080 36 38,880 3 Σ 3240 97,200 97,200 30 mm The neutral axis lies 30 mm above the bottom.  3240  Y top bot 3 2 3 2 3 4 1 1 1 1 1 2 2 3 2 3 4 2 2 2 2 2 3 4 9 4 1 2 54 30 24 mm 0.024 m 30 mm 0.030 m 1 1 (40)(54) (40)(54)(3) 544.32 10 mm 12 12 1 1 1 (40)(54) (40)(54)(6) 213.84 10 mm 36 36 2 758.16 10 mm 758.16 10 m y y I b h A d I b h A d I I I                            | | | |     M y I M I y Top: (tension side) 6 9 3 (120 10 )(758.16 10 ) 3.7908 10 N m 0.024 M        Bottom: (compression) 6 9 3 (150 10 )(758.16 10 ) 3.7908 10 N m 0.030 M        Choose the smaller as Mall. 3 all 3.7908 10 N m M    all 3.79 kN m M    20 mm 12 mm 60 N · m z y PROBLEM 4.24 A 60-N  m couple is applied to the steel bar shown. (a) Assuming that the couple is applied about the z axis as shown, determine the maximum stress and the radius of curvature of the bar. (b) Solve part a, assuming that the couple is applied about the y axis. Use E  200 GPa. SOLUTION (a) Bending about z-axis. 3 3 3 4 9 4 1 1 (12)(20) 8 10 mm 8 10 m 12 12 20 10 mm 0.010 m 2           I bh c 6 9 (60)(0.010) 75.0 10 Pa 8 10 Mc I          75.0 MPa  3 1 9 9 1 60 37.5 10 m (200 10 )(8 10 ) M EI             26.7 m  (b) Bending about y-axis. 3 3 3 4 9 4 6 9 1 1 (20)(12) 2.88 10 mm 2.88 10 m 12 12 12 6 mm 0.006 m 2 (60)(0.006) 125.0 10 Pa 2.88 10 I bh c Mc I                    125.0 MPa  3 1 9 9 1 60 104.17 10 m (200 10 )(2.88 10 ) M EI             9.60 m  z x y C A M PROBLEM 4.31 A W200  31.3 rolled-steel beam is subjected to a couple M of moment 45 kN  m. Knowing that E  200 GPa and v  0.29, determine (a) the radius of curvature ,  (b) the radius of curvature  of a transverse cross section. SOLUTION For W 200 31.3 rolled steel section, 6 4 6 4 31.3 10 mm 31.3 10 m I      (a) 3 3 1 9 6 1 45 10 7.1885 10 m (200 10 )(31.3 10 )            M EI  139.1 m  (b) 3 3 1 1 1 (0.29)(7.1885 10 ) 2.0847 10 m            v   480 m   30 mm 6 mm 6 mm 30 mm Aluminum Brass PROBLEM 4.33 A bar having the cross section shown has been formed by securely bonding brass and aluminum stock. Using the data given below, determine the largest permissible bending moment when the composite bar is bent about a horizontal axis. Aluminum Brass Modulus of elasticity 70 GPa 105 GPa Allowable stress 100 MPa 160 MPa SOLUTION Use aluminum as the reference material. 1.0 in aluminum / 105/70 1.5 in brass     b a n n E E For the transformed section, 1 3 2 1 1 1 1 1 1 3 3 3 4 12 1.5 (30)(6) (1.5)(30)(6)(18) 88.29 10 mm 12       n I b h n A d 3 3 3 4 2 2 2 2 3 4 3 1 3 4 1 2 3 1.0 (30)(30) 67.5 10 mm 12 12 88.29 10 mm 244.08 10 mm             n I b h I I I I I I 9 4 244.08 10  m     nMy I M I ny   Aluminum: 6 1.0, 15 mm 0.015 m, 100 10 Pa      n y  6 9 3 (100 10 )(244.08 10 ) 1.627 10 N m (1.0)(0.015) M        Brass: 6 1.5, 21 mm 0.021 m, 160 10 Pa      n y  6 9 3 (160 10 )(244.08 10 ) 1.240 10 N m (1.5)(0.021) M        Choose the smaller value 3 1.240 10 N m   M 1.240 kN m   PROBLEM 4.40 A copper strip (Ec = 105 GPa) and an aluminum strip (Ea = 75 GPa) are bonded together to form the composite beam shown. Knowing that the beam is bent about a horizontal axis by a couple of moment M = 35 N  m, determine the maximum stress in (a) the aluminum strip, (b) the copper strip. SOLUTION Use aluminum as the reference material. 1.0 in aluminum / 105/75 1.4 in copper c a n n E E     Transformed section: A, mm2 nA, mm2 Ay0, mm nAy0, mm3  216 216 7.5 1620  72 100.8 1.5 151.8 Σ 316.8 1771.2 0 1771.2 5.5909 mm  316.8  Y The neutral axis lies 5.5909 mm above the bottom. 3 2 3 2 4 1 1 1 1 1 1 1 3 2 3 2 4 2 2 2 2 2 2 2 4 9 4 1 2 1.0 (24)(9) (1.0)(24)(9)(1.9091) 2245.2 mm 12 12 1.4 (24)(3) (1.4)(24)(3)(4.0909) 1762.5 mm 12 12 4839 mm 4.008 10 m                 n I b h n A d n I b h n A d I I I (a) Aluminum: 6 9 1.0, 12 5.5909 6.4091mm 0.0064091 (1.0)(35)(0.0064091) 56.0 10 Pa 4.008 10 n y nMy I                 56.0 MPa        56.0 MPa   (b) Copper: 6 9 1.4, 5.5909 mm 0.0055909 m (1.4)(35)( 0.0055909) 68.4 10 Pa 4.008 10 n y nMy I                68.4 MPa    68.4 MPa   PROBLEM 4.52 A concrete beam is reinforced by three steel rods placed as shown. The modulus of elasticity is 20 GPa for the concrete and 200 GPa for the steel. Using an allowable stress of 9 MPa for the concrete and 140 MPa for the steel, determine the largest allowable positive bending moment in the beam. SOLUTION 2 2 2 2 200 GPa 10 20 GPa 3 3 (22) 1140 mm 4 4 11400 mm s c s s E n E A d nA π π = = = = = = = Locate neutral axis 2 200 (11400)(350 ) 0 2 100 11400 3,990,00 0 x x x x x − − = + − = Solve for x: 2 11400 11400 (4)(100)(3,990,000) 150.72 mm (2)(100) x − + + = = 350 199.28 mm − x = 3 2 3 2 6 4 1 1 200 (350 ) (200)(150.72) (11400)(199.28) 3 3 681 10 mm s I x nA x = + − = + = × nMy I M I ny σ σ = ∴ = Concrete: 1.0, 150.72 in, 9 MPa n y σ = = = 6 6 (9 10 )(681 10 ) 40664 N m 407 kN m (1.0)(0.15072) M − × × = = ⋅ = ⋅ Steel: 10, 199.28, 140 MPa n y σ = = = 6 (140 10 )(681 10 6 ) 478.42 N m 47.8 kN m (10)(0.19928) M − × × = = ⋅ = ⋅ Choose the smaller value. 40.7 kN m M = ⋅  50 mm 100 mm � � � � Et � Ec 1 2 Ec M PROBLEM 4.59 The rectangular beam shown is made of a plastic for which the value of the modulus of elasticity in tension is one-half of its value in compression. For a bending moment M  600 N m,  determine the maximum (a) tensile stress, (b) compressive stress. SOLUTION 1 n  2 on the tension side of neutral axis n 1 on the compression side Locate neutral axis. 1 2 2 2 ( ) 0 2 2 1 1 ( ) 0 2 4        x h x n bx n b h x bx b h x 2 2 3 3 6 4 1 1 3 3 6 4 2 2 6 4 6 1 2 1 1 ( ) ( ) 2 2 1 0.41421 41.421 mm 2 1 58.579 mm 1 (1) 1 (50)(41.421) 1.1844 10 mm 3 3 1 1 1 ( ) (50)(58.579) 1.6751 10 mm 3 2 3 2.8595 10 mm 2.8595 10 m                                          x h x x h x x h h h x I n bx I n b h x I I I 4 (a) Tensile stress: 1 , 58.579 mm 0.058579 m 2 n y      6 6 (0.5)(600)( 0.058579) 6.15 10 Pa 2.8595 10 nMy I           6.15 MPa t    (b) Compressive stress: 1, 41.421 mm 0.041421 m n y    6 6 (1.0)(600)(0.041421) 8.69 10 Pa 2.8595 10 nMy I           8.69 MPa c     (a) (b) 100 mm 150 mm 18 mm 100 mm 150 mm 18 mm M M PROBLEM 4.65 A couple of moment M = 2 kN · m is to be applied to the end of a steel bar. Determine the maximum stress in the bar (a) if the bar is designed with grooves having semicircular portions of radius r = 10 mm, as shown in Fig. a, (b) if the bar is redesigned by removing the material to the left and right of the dashed lines as shown in Fig. b. SOLUTION For both configurations, 150 mm 100 mm 10 mm 150 1.50 100 10 0.10 100        D d r D d r d For configuration (a), Fig. 4.28 gives Ka  2.21. For configuration (b), Fig. 4.27 gives K b  1.79. 3 3 6 4 6 4 1 1 (18)(100) 1.5 10 mm 1.5 10 m 12 12 1 50 mm 0.05 m 2           I bh c d (a) 3 6 6 (2.21)(2 10 )(0.05) 147.0 10 Pa 147.0 MPa 1.5 10        KMc I    147.0 MPa  (b) 3 6 6 (1.79)(2 10 )(0.05) 119.0 10 Pa 119.0 MPa 1.5 10        KMc I    119.0 MPa 