7–1 Determine the internal normal force and shear force, and the bending moment in the beam at points C and D Assume the support at B is a roller Point C is located just to the right of the 8-kip load kip 40 kip ft A C ft D ft B ft SOLUTION Support Reactions: FBD (a) a + ©MA = 0; + c ©Fy = 0; By (24) + 40 - 8(8) = Ay + 1.00 - = + ©F = : x By = 1.00 kip Ay = 7.00 kip Ax = T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) Internal Forces: Applying the equations of equilibrium to segment AC [FBD (b)], we have + ©F = : x + c ©Fy = 0; a + ©MC = 0; NC = 7.00 - - VC = MC - 7.00(8) = VC = - 1.00 kip MC = 56.0 kip # ft Ans Ans Ans Applying the equations of equilibrium to segment BD [FBD (c)], we have + ©F = 0; : x + c ©Fy = 0; a + ©MD = 0; VD + 1.00 = ND = VD = -1.00 kip Ans Ans 1.00(8) + 40 - MD = MD = 48.0 kip # ft Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 7–2 Determine the shear force and moment at points C and D 500 lb 300 lb 200 lb B A C ft ft E D ft ft ft SOLUTION Support Reactions: FBD (a) a + ©MB = 0; 500(8) - 300(8) -Ay (14) = Ay = 114.29 lb Internal Forces: Applying the equations of equilibrium to segment AC [FBD (b)], we have NC = 114.29 - 500 - VC = + c ©Fy = 0; a + ©MC = 0; Ans T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) + ©F = : x VC = -386 lb Ans MC + 500(4) - 114.29 (10) = MC = - 857 lb # ft Ans Applying the equations of equilibrium to segment ED [FBD (c)], we have + ©F = 0; : x + c ©Fy = 0; a + ©MD = 0; ND = VD - 300 = -MD - 300 (2) = Ans VD = 300 lb Ans MD = - 600 lb # ft Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 7–3 The strongback or lifting beam is used for materials handling If the suspended load has a weight of kN and a center of gravity of G, determine the placement d of the padeyes on the top of the beam so that there is no moment developed within the length AB of the beam The lifting bridle has two legs that are positioned at 45°, as shown 3m d A 3m 45° 45° B d 0.2 m 0.2 m SOLUTION E F Support Reactions: From FBD (a), a + ©ME = 0; FF162 - 2132 = FE = 1.00 kN + c ©Fy = 0; FF + 1.00 - = FF = 1.00 kN G From FBD (b), + c ©Fy = 0; FAC cos 45° - FBC cos 45° = FAC = FBC = F 2F sin 45° - 1.00 - 1.00 = T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) + ©F = 0; : x FAC = FBC = F = 1.414 kN Internal Forces: This problem requires MH = Summing moments about point H of segment EH [FBD (c)], we have a + ©MH = 0; 1.001d + x2 - 1.414 sin 45°1x2 - 1.414 cos 45°10.22 = d = 0.200 m Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 *7–4 The boom DF of the jib crane and the column DE have a uniform weight of 50 lb>ft If the hoist and load weigh 300 lb, determine the normal force, shear force, and moment in the crane at sections passing through points A, B, and C D ft F A B ft ft ft C 300 lb ft SOLUTION E + ©F = 0; : x NA = + c ©Fy = 0; VA - 450 = 0; a + ©MA = 0; -MA - 150(1.5) - 300(3) = 0; + ©F = 0; : x NB = + c ©Fy = 0; VB - 550 - 300 = 0; a + ©MB = 0; -MB - 550(5.5) - 300(11) = 0; + ©F = 0; : x VC = + c ©Fy = 0; NC - 650 - 300 - 250 = 0; a + ©MC = 0; -MC - 650(6.5) - 300(13) = 0; Ans VA = 450 lb Ans MA = -1125 lb # ft Ans T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) Ans VB = 850 lb MB = -6325 lb # ft Ans Ans Ans NC = 1200 lb MC = -8125 lb # ft Ans Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 7–5 Determine the internal normal force, shear force, and moment at points A and B in the column kN 0.4 m 0.4 m 30Њ SOLUTION A Applying the equation of equilibrium to Fig a gives 0.9 m kN + ©F = 0; : x VA - sin 30° = VA = kN Ans + c ©Fy = 0; NA - cos 30° - = NA = 13.2 kN Ans a + ©MA = 0; 8(0.4) + sin 30°(0.9) - cos 30°(0.4) - MA = MA = 3.82 kN # m Ans 1.5 m 2m B and to Fig b, VB - sin 30° = VB = kN Ans + c ©Fy = 0; NB - - - cos 30° = NB = 16.2 kN Ans T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) + ©F = 0; : x a + ©MB = 0; 3(1.5) + 8(0.4) + sin 30°(2.9) - cos 30°(0.4) - MB = MB = 14.3 kN # m kN Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 7–6 P Determine the distance a as a fraction of the beam’s length L for locating the roller support so that the moment in the beam at B is zero P C a SOLUTION a + ©MA = 0; -P a Cy = a + ©M = 0; A L/3 L 2L - a b + Cy1L - a2 + Pa = 2P B A L3 - a B L - a M = 2P 2PL a L L - a a b = L - ab = T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) a = A L3 - a B L Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 7–7 Determine the internal normal force, shear force, and moment at points C and D in the simply-supported beam Point D is located just to the left of the 2500-lb force 2500 lb 500 lb/ft A ft SOLUTION B D C ft ft ft With reference to Fig a, we have a + ©MA = 0; By(12) - 500(6)(3) - 2500(9) = By = 2625 lb a + ©MB = 0; + ©F = 0; : x 2500(3) + 500(6)(9) - A y(12) = A y = 2875 lb Ax = Using these results and referring to Fig b, we have + ©F = 0; : x NC = + c ©Fy = 0; 2875 - 500(3) - VC = VC = 1375 lb Ans a + ©MC = 0; MC + 500(3)(1.5) - 2875(3) = MC = 6375 lb # ft Ans T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) Ans Also, by referring to Fig c, we have + ©F = 0; : x + c ©Fy = 0; a + ©MD = 0; ND = VD + 2625 - 2500 = 2625(3) - MD = Ans VD = - 125 lb MD = 7875 lb # ft Ans Ans The negative sign indicates that VD acts in the opposite sense to that shown on the free-body diagram © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 *7–8 Determine the normal force, shear force, and moment at a section passing through point C Assume the support at A can be approximated by a pin and B as a roller 10 kip A ft kip 0.8 kip/ft C 12 ft B 12 ft ft SOLUTION a + ©MA = 0; - 19.2(12) - 8(30) + By (24) + 10(6) = By = 17.1 kip + ©F = 0; : x Ax = + c ©Fy = 0; Ay - 10 - 19.2 + 17.1 - = Ay = 20.1 kip NC = + c ©Fy = 0; VC - 9.6 + 17.1 - = VC = 0.5 kip a + ©MC = 0; Ans T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) + © : Fx = 0; Ans -MC - 9.6(6) + 17.1(12) - 8(18) = MC = 3.6 kip # ft Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 7–9 Determine the normal force, shear force, and moment at a section passing through point C Take P = kN B 0.1 m 0.5 m C 0.75 m 0.75 m P SOLUTION a + ©MA = 0; 0.75 m A -T(0.6) + 8(2.25) = T = 30 kN + ©F = 0; : x Ax = 30 kN + c ©Fy = 0; Ay = kN + ©F = 0; : x -NC - 30 = + c ©Fy = 0; VC + = VC = - kN a + ©MC = 0; Ans T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) NC = - 30 kN Ans - MC + 8(0.75) = MC = kN # m Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 7–10 The cable will fail when subjected to a tension of kN Determine the largest vertical load P the frame will support and calculate the internal normal force, shear force, and moment at a section passing through point C for this loading B 0.1 m 0.5 m C 0.75 m -2(0.6) + P(2.25) = P = 0.533 kN + ©F = 0; : x Ax = kN + c ©Fy = 0; Ay = 0.533 kN + ©F = 0; : x -NC - = Ans VC - 0.533 = VC = -0.533 kN a + ©MC = 0; Ans T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) NC = - kN + c ©Fy = 0; 0.75 m P SOLUTION a + ©MA = 0; 0.75 m A Ans - MC + 0.533(0.75) = MC = 0.400 kN # m Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 7–127 Determine the distance a between the supports in terms of the beam’s length L so that the moment in the symmetric beam is zero at the beam’s center w a L SOLUTION Support Reactions: From FBD (a), a +©MC = 0; w a (L + a) a b - By (a) = 2 By = w (L + a) Free body Diagram: The FBD for segment AC sectioned through point C is drawn Internal Forces: This problem requires MC = Summing moments about point C [FBD (b)], we have w w a wa a a b + (L - a) c (2a + L) d - (L + a) a b = 4 T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) a + ©MC = 0; 2a2 + 2aL - L2 = a = 0.366L Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 *7–128 The balloon is held in place using a 400-ft cord that weighs 0.8 lb/ft and makes a 60° angle with the horizontal If the tension in the cord at point A is 150 lb, determine the length of the cord, l, that is lying on the ground and the height h Hint: Establish the coordinate system at B as shown 60 h y SOLUTION Deflection Curve of The Cable: l x = ds L C + A 1>F 2H B ( s B where w0 = 0.8 lb>ft w0 ds)2 D L A Performing the integration yields FH b sinh - B (0.8s + C1) R + C2 r 0.8 FH From Eq 7–14 dy 1 = (0.8s + C1) w ds = dx FH L FH Boundary Conditions: (1) T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) x = (2) dy = at s = From Eq (2) dx (0 + C1) FH = Then, Eq (2) becomes C1 = dy 0.8s = tan u = dx FH (3) s = at x = and use the result C1 = From Eq (1) x = FH b sinh - B (0 + 0) R + C2 r 0.8 FH C2 = Rearranging Eq (1), we have s = FH 0.8 sinha xb 0.8 FH (4) Substituting Eq (4) into (3) yields dy 0.8 = sinh a xb dx FH Performing the integration y = FH 0.8 cosha x b + C3 0.8 FH y = at x = From Eq (5) = (5) FH FH cosh + C3, thus, C3 = 0.8 0.8 Then, Eq (5) becomes y = FH 0.8 B cosh a x b - R 0.8 FH (6) The tension developed at the end of the cord is T = 150 lb and u = 60° Thus © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 x *7–128 (continued) T = FH cos u 150 = FH cos 60° FH = 75.0 lb From Eq (3) dy 0.8s = tan 60° = dx 75 s = 162.38 ft Thus, l = 400 - 162.38 = 238 ft Ans Substituting s = 162.38 ft into Eq (4) 162.38 = 75 0.8 sinha xb 0.8 75 x = 123.46 ft h = T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) y = h at x = 123.46 ft From Eq (6) 0.8 75.0 (123.46) d - R = 93.75 ft B coshc 0.8 75.0 Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 7–129 The yacht is anchored with a chain that has a total length of 40 m and a mass per unit length of 18 kg/m, and the tension in the chain at A is kN Determine the length of chain ld which is lying at the bottom of the sea What is the distance d? Assume that buoyancy effects of the water on the chain are negligible Hint: Establish the origin of the coordinate system at B as shown in order to find the chain length BA A 60 d y SOLUTION ld Component of force at A is FH = T cos u = 7000 cos 60° = 3500 N s B x From Eq (1) of Example - 13 x = dy = 0, s = 0, then dx dy = (w s + C1); dx FH T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) Since 3500 a sinh - c (18)(9.81)s + C1 d + C2 b 18 (9.81) 3500 C1 = Also x = 0, s = 0, so that C2 = and the above equation becomes x = 19.82 a sinh - a or, s = 19.82 a sinha From Example - 13 s bb 19.82 x bb 19.82 dy 18 (9.81) w0 s s = s = = dx FH 3500 19.82 (1) (2) (3) Substituting Eq (2) into Eq (3) Integrating dy x x = sinh a b y = 19.82 cosha b + C3 dx 19.82 19.82 Since x = 0, y = 0, then C3 = - 19.82 Thus, y = 19.82 acosha x b - 1b 19.82 (4) Slope of the cable at point A is dy = tan 60° = 1.732 dx Using Eq (3), sAB = 19.82 (1.732) = 34.33 m Length of chain on the ground is thus ld = 40 - 34.33 = 5.67 m Ans From Eq (1), with s = 34.33 m x = 19.82 asinh - a 34.33 b b = 26.10 m 19.82 Using Eq (4), y = 19.82 acosha 26.10 b - 1b 19.82 d = y = 19.8 m Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 7–130 Draw the shear and moment diagrams for the beam ABC D A C B 45° 1.5 m SOLUTION 3m Support Reactions: The kN load can be replacde by an equivalent force and couple moment at B as shown on FBD (a) a + ©MA = 0; + c ©Fy = 0; 1.5 m kN FCD = 6.364 kN FCD sin 45°162 - 6132 - 9.00 = A y + 6.364 sin 45° - = 1.5 m A y = 1.50 kN Shear and Moment Functions: For ◊ x m, determine the equation of the catenary curve of the cable and the maximum tension developed in the cable 40 m A SOLUTION As shown in Fig a, the orgin of the x, y coordinate system is set at the lowest point of the cable Here, w(s) = 5(9.81) N>m = 49.05 N>m d2y dx Set u = = 49.05 dy + a b FH A dx dy d2y du , then , then = dx dx dx2 Integrating, T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) du 49.05 = dx FH 11 + u2 ln a u + 11 + u2 b = 49.05 x + C1 FH Applying the boundary condition u = ln a u + 11 + u2 b = u + 11 + u2 = e dy e = u = dx Since sinh x = 49.05 x FH dy = at x = results in C1 = Thus, dx 49.05 x FH 49.05 FH x - -e 49.05 x FH ex - e - x , then dy 49.05 = sinh x dx FH Integrating, y = FH 49.05 cosh a x b + C2 49.05 FH Applying the boundary equation y = at x = results in C2 = y = FH Thus, 49.05 FH 49.05 c cosh a xb - d m 49.05 FH If we write the force equation of equilibrium along the x and y axes by referring to the free-body diagram shown in Fig b, + ©F = 0; : x T cos u - FH = + c ©Fy = 0; T sin u - 5(9.81)s = © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 B *7–136 (continued) Eliminating T, dy 49.05s = tan u = dx FH (3) Equating Eqs (1) and (3) yields 49.05s 49.05 = sinh a xb FH FH s = FH 49.05 = sinh a b 49.05 FH Thus, the length of the cable is L = 45 = e FH 49.05 sinh a (20) b f 49.05 FH T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) Solving by trial and error, FH = 1153.41 N Substituting this result into Eq (2), y = 23.5 [cosh 0.0425x - 1] m Ans The maximum tension occurs at either points A or B where the cable makes the greatest angle with the horizontal Here umax = tan - a Thus, Tmax = dy 49.05 b = tan - e sinh a (20) b f = 43.74° ` dx x = 20m FH FH 1153.41 = = 1596.36 N = 1.60 kN cos umax cos 43.74° Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 7–137 The traveling crane consists of a 5-m-long beam having a uniform mass per unit length of 20 kg/m The chain hoist and its supported load exert a force of kN on the beam when x = m Draw the shear and moment diagrams for the beam The guide wheels at the ends A and B exert only vertical reactions on the beam Neglect the size of the trolley at C x 2m 5m A C SOLUTION B Support Reactions: From FBD (a), a + ©MA = 0; + c ©Fy = 0; By (5) - 8(2) - 0.981 (2.5) = A y + 3.6905 - - 0.981 = By = 3.6905 kN kN A y = 5.2905 kN Shear and Moment Functions: For … x m [FBD (b)], 5.2905 - 0.1962x - V = T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) + c ©Fy = 0; V = {5.29 - 0.196x} kN a + ©M = 0; Ans x M + 0.1962x a b - 5.2905x = M = {5.29x - 0.0981x2} kN # m Ans For m x … m [FBD (c)], + c ©Fy = 0; V + 3.6905 - 20(9.81) (5 - x) = 1000 V = {- 0.196x - 2.71} kN a + ©M = 0; 3.6905(5 - x) - Ans 20(9.81) - x b - M = (5 - x) a 1000 M = {16.0 - 2.71x - 0.0981x 2} kN # m Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 7–138 The bolt shank is subjected to a tension of 80 lb Determine the internal normal force, shear force, and moment at point C C 90 A in B SOLUTION + ©F = 0; : x NC + 80 = VC = a + ©MC = 0; MC + 80(6) = Ans Ans MC = - 480lb # in Ans T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) + c ©Fy = 0; NC = - 80 lb © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 7–139 Determine the internal normal force, shear force, and the moment as a function of 0° … u … 180° and … y … ft for the member loaded as shown ft B u C 200 lb y 150 lb SOLUTION ft For 0° … u … 180°: V + 200 cos u - 150 sin u = + Q©Fx = 0; V = 150 sin u - 200 cos u Ans A N - 200 sin u - 150 cos u = +a©Fy = 0; N = 150 cos u + 200 sin u Ans T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) -M - 150(1) (1 - cos u) + 200(1) sin u = a + ©M = 0; M = 150 cos u + 200 sin u - 150 Ans At section B, u = 180°, thus VB = 200 lb NB = - 150 lb MB = -300 lb # ft For … y … ft: + ©F = 0; : x V = 200 lb Ans + c ©Fy = 0; N = -150 lb Ans a + ©M = 0; -M - 300 - 200 y = M = - 300 - 200 y Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 ... publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information... publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information... publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information