# Ch.10 - ENGINEERING MECHANICS — STATICS 2nd Ed 10—1...

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Unformatted text preview: ENGINEERING MECHANICS — STATICS, 2nd. Ed. 10—1* Determine the second moment of area for the isosceles triangle shown in Fig. P10-1 with respect to (a) The base of the triangle (the x-axis). (b) An axis through the centroid parallel to the base. SOLUTION From similar triangles: h_v w=3§(h—y) = j y2 (w dy) A b 2 h(h y_) y dy W. F. RILEY AND L. D. STURGES | ILL A in my...“ 1 l3 E Lia/2. 1I I Lula,l |—_x ENGINEERING MECHANICS - STATICS.I 2nd. Ed. 10—5* Determine the second moment of area for the shaded region shown in Fig. PIG-5 with respect to (a) The x—axis. (b) The y-axis. SOLUTION .4 (a) From the curve: 204.8 in? (b) From the results of Example Problem 10-1: 3 6 dy = 3 ydy 1 -l .1. 3 bh ‘ 3 X 3 I dI A y 780.2 in? W. F. RILEY AND L. 4 3 205 in. e 730 in? D. STURGES ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES Determine the second moment of area for the shaded region shown in Fig. P10—7 with respect to (a) The x-axis. (b) The y-axis. SOLUTION (a) From the results of Example Problem 10-1: 1 3 _ l 3' 3 bh ‘ 3 y [A dIX 4.267 in? e 4.27 in? (b) From the curve: 2 y 2 l x [x1 2 dx] 5/2 dx X I: I: 36.57 in? a 36.6 in? :: 1 ~ﬁwwww m- .ﬁm ENGINEERING MECHANICS — STATICS, 2nd. Determine the second moment of area for the shaded region shown in Fig. P10—8 with respect to (a) The x-axis. (b) The y-axis. SOLUTION (a) From the curve: (b) From the results of = 0.223(106) mm Example Problem 10-1: 3 3 25 I % D 1000 — 3 -1 -1 bh _ 3 x dy — 3(1000 y 3/2 (1000y3/2) dy Ed. )dy W. F. RILEY AND L. D. STURGES 4 2 5/2 25 6 4 [5 y ] = 0.417(10 ) mm 0 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES A homogeneous, triangular block of weight W has height h and base width b as shown in Fig. P9-7. Develop an expression for the coefficient of friction between the block and the surface for which impending motion by slipping and tipping occur simultaneously if (a) Force E acts in the direction shown on the figure. (b) The direction of force F is reversed. SOLUTION (a; From the free—body diagram for the triangular block: +T‘ZF =A -w=0 Y n +—>2F=p—A=o x f For slipping: P Af(max) = HA“ 3 . . _ _ _ 2 For t1pping. + C 2MB - Pth W[3] For simultaneous slipping and tipping: From which: (bl From the free-body diagram for the triangular block: + T 2F 5' + —+ 2F X For slipping: = Af(max) = MA“ r- a For tipping: + C'ZMB = Pth - W[§9] — For simultaneous slipping and tipping: From which: 4 m._:‘.i.,"_ Niwﬂ 4,. “(f ,,,,_,_,:V "swans, furry- mw.rmaw,ﬁ.w “r...” aw..-“ Hawaii-71's,. ._-,'m’/ am an“. , a. ENGINEERING MECHANICS — STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 10-22 Determine the radii of gyration for the shaded area Shown in Fig. P10-22 with respect to the x— and y—axes shown on the figure. , -_ - - 1.4 q "' x L———L————J Fig.P10-22 1 SOLUTION — a Sln 2L L n_ . A = I a sin —5 dx L o : a[— L cos E§]L = ELE n L o N dI - l a3 sin3 E5 dx x - 3 L L P . I = J d1 = I l a3 51D 2; dx x A X L _a_3_L ELL—OSBBL he;ng 4La‘ ' 3 n 3” 9Tr k _/_x 4La3/9 “2-12a Ans x ‘ A 2La/F - 3 ' 2 L 2 n I zj x dA=I x (asin'ﬂdx Y A rd L’ - a - XzL cos £3 + EL LE— cos EX + E X 5'” EX L ' n L n n2 r L L 1 L o 3 3 3 _ L 4L _La 2 _ a [E— - —-§] — 3 (n ' 4) n n k _ £1 L3a(n2 — 4)/TF3 1/2 _ [EL “2 _ 4 ‘ A s y ‘ A 2La /n _ 2” n I ENGINEERING MECHANICS — STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 10—34* Determine the second moments with respect to x (horizontal) and y (vertical) axes through the centroid of the shaded area shown in Fig. P10-34. H40 mm élx 20 mm Fig. P1 0-34 SOLUTION ,‘ ‘3 4 (50)2(1200) = 3.040(106) mm _ _1__ ' — 12(GONZO) 3 = %§(20)(80) = 0.353(106) mm4 1_ 12 (60)(20)3 (-50)2(1200) = 3.040(106) mm‘1 3.040(106) + 0.853(106) + 3.040(106) 4 4 6.933(106) mm 2 6.933(106) mm 20(60) + 20(30) + 20(60) = 4000 mm2 20(60)(30) + 20(80)(10) + 20(60)(30) = 88,000 mm3 88 000 __J___ — : ‘000 22.0 mm yC 60 mm (By Symmetry) %§(60)3(20) + (8)2(1200) = 0.4368(106) mm‘1 %§(20)3(80) (—12)2(1600) 0.2337(1oé) mm‘1 %§(50)3(20) (3)2(1200) = 0.4353(106) mm‘1 = 0.4368(106) + 0.2337(105) + 0.4368(106) 1.1573(106) mm4 a 1.157(106) mm“ ENGINEERING MECHANICS - STATICS, 2nd. Ed. W.'F. RILEY AND L. D. STURGES 10-37* Determine the second moments with respect to x (horizontal) and y (vertical) axes through the centroid of the two 10 x 1-in. steel plates that are welded to the flanges of an 518 x 70 I-beam as shown in Fig. PIG—37. SOLUTION From Table 10—2A: For an 818 X 70 beam 4 I = 926 in. I = 24.1 in.4 10(1)3 } 2 ” 1xC = 926 + 2[ 12 + (9.5) (10)] 2733 in? e 2730 in? 1(10)3] 4 24.1 + 2[ 12 = 190.8 in. - Ans. ENGINEERING MECHANICS - STATICS, 2nd. Ed. Determine the second moments with respect to x (horizontal) and y (vertical) axes through the centroid of the four L8 x 8 x 1 angles that are riveted together to form the cross section shown in Fig. P10-39. SOLUTION ,a From Table 10-2A: For an L8 x 8 x 1 Angle: A = 15.00 in? . 4 yc - 89.0 in. xc = 2.37 in. 4139.0) + 4(2.37)2(15.00) 693.01 in? a 693 in? 4(39.o> + 4(2.37>2(15.00) 693.01 in? a 593 in? W. F. RILEY AND L. D. STURGES ENGINEERING MECHANICS — STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 10—43 Determine the second moments with respect to x (horizontal) and y (vertical) axes through the centroid of the three 510 x 35 I-beams that are welded together to form the cross section shown in Fig.-P10-43. SOLUTION ” For an 310 X 35 section: 10.3 in? 10.00 in. r- - 4 141 in. I = 8.36 in? 0.594 in. I = 2(147) + 8.36 4 4 z 302 in. 302.35 in. __i —_-. dY1 — dy3 _ 2(10.00 + 0.094) — 0.297 in. ch =_147 + 2[8.36 +(5.297)2(10.3)] = 741.72 in? a 742 in? ...
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Ch.10 - ENGINEERING MECHANICS — STATICS 2nd Ed 10—1...

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