1-1 Determine the resultant internal normal force acting on the cross section through point A in each column In (a), segment BC weighs 180 lb/ft and segment CD 250 io/ft In (6), the column has a mass of 200 kg/m weighs (a) Hip +T ZF, =0; F-L0-3-3-18-5=0 F,=13.8kip Ans R hi | (2/ø)60) “ip iF 2/6 *ip il|Ì i sKïp LAL né 022 4@)= hạ (b) gkn +fEE =0; F-45—-4.5~5.89—6~6—§=0 F,=34.9kN Ans one b, 4k Cu“ + AVE, * S.gqka Fa From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X © 2005 R C Hibbeler Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 1-2 Determine the resultant internal torque acting on the cross sections through points C and D The support bearings at A and B allow free turning of the shaft =M, = 0; Te - 250 =0 Te = 250N-m =M, = 0; Ih =0 Ans Te Ans From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X © 2005 R C Hibbeler Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 1-3 Determine the resultant internal torque acting on the cross sections through points B and C A AL 600 the ` ụ 350 thft C tt 500 Ib-ft 1% +350-500= e© 5M, =0; >> wo Tp = 1501b-ft XS 50 (6 S09 b-$¢ Ans ot + EM,=0; Tc-500=0 Te =500 lb-ft Ans OS£ ®-Ƒt x From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X © 2005 R C Hibbeler Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *1-4 Determine the resultant internal normal and shear force in the member st (a) section a—a and (6) section b-b, each of which passes through point A The 500-Ib load is applied along the centroida! axis of the member 500 tb 500 tb (a) SEF, =0; N,-500=0 Ans Nz = 500 Ib +125F=0; | Me 500" I Ans W=0 (b) Wr =0; N,~500cos 30° =0 N, =433lb +⁄EF,=0; 30 W-500sin30?=0 W=2501b- \y 500 Ans a — x» N, Ans From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X © 2005 R C Hibbeler Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 1-5 Determine the resultant internal loadings acting on the cross section through point D of member AB 175A 30mm ' 50 mm 134.25 2/479 I15H W !3i28w My Np Segment AD - SER =0; Np+131.25=0; +12ER,=0; W+175=0; EM =0; Mp Np =-I31N Ans W=-175N + 175(0.05) = 0; Mp =-8.75 N-m Ans Ans From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X © 2005 R C Hibbeler Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 1-6 - The beam Á ¡s pin supported at A and supported by a cable 8C Determine the resultant internal loadings acting on the cross section al point D 6=tan" B =36.87° -/10 ¢ = tan '($)-3687 = 14.47 Member AB (+ >M, : = 0; Fac sin 14.47°(10) Fac — 1200(6) = = 2881.46 lb Segment BD : elk, = 0; ~Np Np AEF, = 0; — 2881.46 cos 14.47° — 1200 cos 36.87° = = ~ 3750 tb = —3.75 kip Ans Vp + 2881.46 sin 14.47° — 1200 sin 36.87°= % =0 Ans G EMp = 0; — 2881.46 sin 14.47°(6.25) ~ 1200 sin 36.87°(6.25) — Mp = Mp = Ans BX Notice that member AB is the two - force member ; therefore the shear force and moment are zero Ạ 1200 ‘6 Yy From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X © 2005 R C Hibbeler Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 1-7 Solve Prob 1-6 for the resultant internal loadings act- ing at point E `? ah |—aa—+——su—— 8=tan! B = 36.87° @ = tan (= )- 36.879 = 14.47° Member AB : C+ EM, = 0; — Fac sin 14.47°(10) - 1200(6) = Ze, £ ng N, Fgc = 2881.46 Ib Segment BE : + LF, = 0; — Ng — 2881.46 cos 14.47° ~ 1200 cos 36.87° = Ne = ~— 3750 lb = -3.75 kip X+ ER = 0; Ve + 2881.46 sin 14.47° — 1200 sin 36.87°= =0 Œ IMs = 0; Ans ‘Ans 2881.46 sin 14.47°(3) Mẹ Notice that member AB = — 1200 sin 36.87°(3) — Me = Ans is the two - force member ; therefore the shear force and moment are zero From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X © 2005 R C Hibbeler Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *1-8 The beam AB is fixed to the wall and has a uniform weight of 80 Ib/ft If the trolley supports a load of 1500 lb, determine the resultant internal loadings acting on the cross sections through points C and D 208-10 r3 zxrnn F) Segment BC : +T†1£E Ans Nc=0 CEE =0; =0, a Ye = 3.50 kip (+EM =0; 1500 tb | Ne i =0 W- 20-15 ake ee se¿.0X?P ~Me ~ 2(2.5) —1.5 (15) = Mc = —47.5kip-ft Ans Segment BD: CEF,=0; +TER =0; Np =0 Ans W-024=0 0-083)= 0-24 RIP VY = 0.240 kip (+ EMp = 0; Np » Te = 52 kip (c) TF, =0; +TXR=0; —NÑc ~ 52 cos 30° = 0; V+52sin309-26=0, Ác =-45.0kip Y=0 Ans Ans EMc=0; — 52cos 30°(0.2)+52 sin 30°(3)~ 26(3)~ Mc = Mc = 9.00 kip ft Ans From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X © 2005 R C Hibbeler Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 1-10 Determine the resultant internal loadings acting on the cross se¢tions through points D and E of the frame Member AG ( IM, : = 0; (+ EM = 0: * > EE =0, si (3) — 75(4)(5) ~ 150 cos30 (7) = 0; Fy¢ = 1003.89 Ib A, (3) — 15(4)(2) - 150cos30(4) = 0; A, = 373.201b Fae 75 C4) Ib A, ~ 3(1003.89) + 150sin30 = 0; A, = 527.33 ib For point D : EF, = 0: Np + 527.33 = Np = +TIF =0; Ans -373.20 - Vp = YW Œ EM =0: ~—5271b = —373 1b Mp + 373.201) Mp = Ans = —3731b-ft Ans For point E : * ¬kÈEF, = 0; N 150 sin30 - Ne= - Ng = 7501 +TFE, = Ans Ms f~“~—~~ yt i_—-—-~¬ Ibe ' , Bit a W — 715(3) — 150cos 30 = lý = 355 Ib & EM = 0; E isolt Ans — Mg — 75(3)(1.5) ~ 150 cos 30 (3)= 0; Mẹ = —7271b-ft Ans From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X © 2005 R C Hibbeler Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher ... D The support bearings at A and B allow free turning of the shaft =M, = 0; Te - 250 =0 Te = 250N-m =M, = 0; Ih =0 Ans Te Ans From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X... point E of the beam in Prob 1-13 + ZF, = 0; Ne+ Ng +†EEQ =0, 2943 = = —2.94kKN IM; = 0; Me ~2943-W =0 W = —2.94kN Œ Ans Ans 2242 8l M,; + 2943 (1) = Mr; = ~2.94kN-m Ans From Mechanics of Materials,... the member st (a) section a—a and (6) section b-b, each of which passes through point A The 500-Ib load is applied along the centroida! axis of the member 500 tb 500 tb (a) SEF, =0; N,-500=0 Ans