2 Axially Loaded Members

Transcription

1 xially oaded Memers hanges in engths of xially oaded Memers rolem.-1 The T-shaped arm shown in the figure lies in a vertical plane and pivots aout a horizontal pin at. The arm has constant cross-sectional area and total weight W. vertical spring of stiffness k supports the arm at point. Otain a formula for the elongation of the spring due to the weight of the arm. k Solution.-1 T-shaped arm FREE-ODY DIGRM OF RM F tensile force in the spring F W 3 W 3 W 3 M 0 F() W 3 W 3 3 W () 0 3 F 4W 3 elongation of the spring F k 4W 3k rolem.- steel cale with nominal diameter 5 mm (see Tale -1) is used in a construction yard to lift a ridge section weighing 38 kn, as shown in the figure. The cale has an effective modulus of elasticity E 140 Ga. (a) If the cale is 14 m long, how much will it stretch when the load is picked up? () If the cale is rated for a maximum load of 70 kn, what is the factor of safety with respect to failure of the cale? 63

2 64 HTER xially oaded Memers Solution.- ridge section lifted y a cale 304 mm (from Tale -1) W 38 kn E 140 Ga 14 m () FTOR OF SFETY UT 406 kn (from Tale -1) max 70 kn n UT max 406 kn kn (a) STRETH OF E W (38 kn)(14 m) E (140 Ga)(304 mm ) 1.5 mm rolem.-3 steel wire and a copper wire have equal lengths and support equal loads (see figure). The moduli of elasticity for the steel and copper are E s 30,000 ksi and E c 18,000 ksi, respectively. (a) If the wires have the same diameters, what is the ratio of the elongation of the copper wire to the elongation of the steel wire? () If the wires stretch the same amount, what is the ratio of the diameter of the copper wire to the diameter of the steel wire? Steel wire opper wire Solution.-3 Steel wire and copper wire opper wire Equal lengths and equal loads Steel: E s 30,000 ksi opper: E c 18,000 ksi () RTIO OF DIMETERS (EQU EONGTIONS) c s E c c E s s E c 4 d c E s 4 d s ore c c E s s Steel wire (a) RTIO OF EONGTIONS (EQU DIMETERS) c E c s E s d c d s E s E c d c E s 30 d s E c c E s s E c 18

3 SETION. hanges in engths of xially oaded Memers 65 rolem.-4 y what distance h does the cage shown in the figure move downward when the weight W is placed inside it? onsider only the effects of the stretching of the cale, which has axial rigidity E 10,700 kn. The pulley at has diameter d 300 mm and the pulley at has diameter d 150 mm. lso, the distance m, the distance 10.5 m, and the weight W kn. (Note: When calculating the length of the cale, include the parts of the cale that go around the pulleys at and.) 1 age W Solution.-4 age supported y a cale 1 d 300 mm d 150 mm m 10.5 m E 10,700 kn ENGTH OF E (d ) 1 (d ) 4,600 mm 1,000 mm 36 mm 36 mm 6,07 mm W kn EONGTION OF E T (11 kn)(6,07 mm) 6.8 mm E (10,700 kn) TENSIE FORE IN E T W 11 kn W OWERING OF THE GE h distance the cage moves downward h mm rolem.-5 safety valve on the top of a tank containing steam under pressure p has a discharge hole of diameter d (see figure). The valve is designed to release the steam when the pressure reaches the value p max. If the natural length of the spring is and its stiffness is k, what should e the dimension h of the valve? (Express your result as a formula for h.) h d p

4 66 HTER xially oaded Memers Solution.-5 Safety valve p max pressure when valve opens natural length of spring ( > h) h k stiffness of spring FORE IN OMRESSED SRING F k( h) (From Eq. -1a) d RESSURE FORE ON SRING p max d 4 h height of valve (compressed length of the spring) d diameter of discharge hole p pressure in tank EQUTE FORES ND SOVE FOR h: F k( h) p max d 4 h p max d 4k rolem.-6 The device shown in the figure consists of a pointer supported y a spring of stiffness k 800 N/m. The spring is positioned at distance 150 mm from the pinned end of the pointer. The device is adjusted so that when there is no load, the pointer reads zero on the angular scale. If the load 8 N, at what distance x should the load e placed so that the pointer will read 3 on the scale? x k 0 Solution.-6 ointer supported y a spring FREE-ODY DIGRM OF OINTER x F = k 8N k 800 N/m 150 mm displacement of spring F force in spring k M 0 x (k) 0or x k et angle of rotation of pointer tan x k k x tan SUSTITUTE NUMERI VUES: 3 (800 Nm)(150 mm) x tan 3 8 N 118 mm

5 SETION. hanges in engths of xially oaded Memers 67 rolem.-7 Two rigid ars, and D, rest on a smooth horizontal surface (see figure). ar is pivoted end and ar D is pivoted at end D. The ars are connected to each other y two linearly elastic springs of stiffness k. efore the load is applied, the lengths of the springs are such that the ars are parallel and the springs are without stress. Derive a formula for the displacement at point when the load is acting. (ssume that the ars rotate through very small angles under the action of the load.) D Solution.-7 Two ars connected y springs DISEMENT DIGRMS D D k stiffness of springs displacement at point due to load FREE-ODY DIGRMS F 1 F F 1 F D F 1 tensile force in first spring F compressive force in second spring EQUIIRIUM M 0 F 1 F 0 F 1 F M D 0 F 1 F 0 F F 1 Solving, F F 3 displacement of point displacement of point 1 elongation of first spring shortening of second spring lso, 1 F 1 4 k SOVE THE EQUTIONS: k 3k Eliminate and otain : 0 9k 3k ; F k 3k

6 68 HTER xially oaded Memers rolem.-8 The three-ar truss shown in the figure has a span 3 m and is constructed of steel pipes having cross-sectional area 3900 mm and modulus of elasticity E 00 Ga. load acts horizontally to the right at joint. (a) If 650 kn, what is the horizontal displacement of joint? () What is the maximum permissile load max if the displacement of joint is limited to 1.5 mm? Solution.-8 Truss with horizontal load From force triangle, F (tension) 3m 3900 mm E 00 Ga F F M 0givesR FREE-ODY DIGRM OF JOINT Force triangle: R = R F F (a) HORIZONT DISEMENT 650 kn F E (650 kn)(3 m) (00 Ga)(3900 mm ) 1.5 mm () MXIMUM OD max max 1.5 mm max max max max 1.5 mm max (650 kn) 1.5 mm 780 kn E

7 SETION. hanges in engths of xially oaded Memers 69 rolem.-9 n aluminum wire having a diameter d mm and length 3.8 m is sujected to a tensile load (see figure). The aluminum has modulus of elasticity E 75 Ga. If the maximum permissile elongation of the wire is 3.0 mm and the allowale stress in tension is 60 Ma, what is the allowale load max? d Solution.-9 luminum wire in tension d mm 3.8 m E 75 Ga d 3.14 mm 4 MXIMUM OD SED UON EONGTION max 3.0 mm E d max E max (75 Ga)(3.14 mm ) 3.8 m 186 N MXIMUM OD SED UON STRESS s allow 60 Mas max s allow (3.14 mm )(60 Ma) 189 N OWE OD (3.0 mm) Elongation governs. max 186 N rolem.-10 uniform ar of weight W 5 N is supported y two springs, as shown in the figure. The spring on the left has stiffness k N/m and natural length 1 50 mm. The corresponding quantities for the spring on the right are k 400 N/m and 00 mm. The distance etween the springs is 350 mm, and the spring on the right is suspended from a support that is distance h 80 mm elow the point of support for the spring on the left. t what distance x from the left-hand spring should a load 18 N e placed in order to ring the ar to a horizontal position? k 1 1 k x W h

8 70 HTER xially oaded Memers Solution.-10 ar supported y two springs Reference line h 1 k 1 k M 0 F X W (Eq. 1) 0 F c T vert 0 F 1 F W 0 (Eq. ) SOVE EQS. (1) ND (): 1 F 1 1 x W F x W W 5 N x W SUSTITUTE NUMERI VUES: UNITS: Newtons and meters F 1 (18) 1 x x k N/m k 400 N/m x F (18) x mm EONGTIONS OF THE SRINGS h 80 mm 18 N 1 F 1 F x k NTUR ENGTHS OF SRINGS 1 50 mm 00 mm OJETIVE Find distance x for ar to e horizontal. FREE-ODY DIGRM OF R F 1 F F F x k 400 R REMINS HORIZONT oints and are the same distance elow the reference line (see figure aove). 1 1 h or x x SOVE FOR x: x W x x m x 135 mm

9 SETION. hanges in engths of xially oaded Memers 71 rolem.-11 hollow, circular, steel column (E 30,000 ksi) is sujected to a compressive load, as shown in the figure. The column has length 8.0 ft and outside diameter d 7.5 in. The load 85 k. If the allowale compressive stress is 7000 psi and the allowale shortening of the column is 0.0 in., what is the minimum required wall thickness t min? t d Solution.-11 olumn in compression t d REQUIRED RE SED UON OWE SHORTENING E in. SHORTENING GOVERNS min in. E allow MINIMUM THIKNESS t min (85 k)(96 in.) (30,000 ksi)(0.0 in.) 4 [d (d t) ]or 85 k E 30,000 ksi 8.0 ft d 7.5 in. allow 7,000 psi allow 0.0 in. REQUIRED RE SED UON OWE STRESS s 85 k 1.14 in. s allow 7,000 psi 4 d (d t) d 4 ord t d 4 t d d or t min d d min SUSTITUTE NUMERI VUES 7.5 in. t min t min 0.63 in. (d t) 7.5 in in.

10 7 HTER xially oaded Memers rolem.-1 The horizontal rigid eam D is supported y vertical ars E and F and is loaded y vertical forces kn and 360 kn acting at points and D, respectively (see figure). ars E and F are made of steel (E 00 Ga) and have cross-sectional areas E 11,100 mm and F 9,80 mm. The distances etween various points on the ars are shown in the figure. Determine the vertical displacements and D of points and D, respectively. 1.5 m 1.5 m 1 = 400 kn F E.1 m D.4 m = 360 kn 0.6 m Solution.-1 Rigid eam supported y vertical ars 1.5 m 1.5 m.1 m D SHORTENING OF R E E F E E E E (96 kn)(3.0 m) (00 Ga)(11,100 mm ) 1 = 400 kn.4 m = 360 kn mm F 0.6 m E E 11,100 mm F 9,80 mm E 00 Ga E 3.0 m F.4 m kn; 360 kn FREE-ODY DIGRM OF R D 1.5 m 1.5 m.1 m D 1 = 400 kn F E F F = 360 kn M 0 (400 kn)(1.5 m) F F (1.5 m) (360 kn)(3.6 m) 0 F F 464 kn M 0 (400 kn)(3.0 m) F E (1.5 m) (360 kn)(.1 m) 0 F E 96 kn SHORTENING OF R F F F F F E F mm DISEMENT DIGRM (464 kn)(.4 m) (00 Ga)(9,80 mm ) 1.5 m 1.5 m.1 m D E F D E F E or E F (0.400 mm) m 0.00 mm (Downward) D F ( F E ) or D 1 5 F 7 5 E 1 5 (0.600 mm) 7 (0.400 mm) mm (Downward)

11 SETION. hanges in engths of xially oaded Memers 73 rolem.-13 framework consists of two rigid ars and, each having length (see the first part of the figure). The ars have pin connections at,, and and are joined y a spring of stiffness k. The spring is attached at the midpoints of the ars. The framework has a pin support at and a roller support at, and the ars are at an angle to the hoizontal. When a vertical load is applied at joint (see the second part of the figure) the roller support moves to the right, the spring is stretched, and the angle of the ars decreases from to the angle. Determine the angle and the increase in the distance etween points and. (Use the following data; 8.0 in., k 16 l/in., 45, and 10 l.) k Solution.-13 Framework with rigid ars and a spring k 1 WITH OD span from to cos S length of spring cos u FREE-ODY DIGRM OF WITH NO OD 1 span from to cos S 1 length of spring 1 cos h F h h F h height from to sin cos u F force in spring due to load M 0 F h 0 or cos F sin (Eq. 1) (ontinued)

12 74 HTER xially oaded Memers DETERMINE THE NGE S elongation of spring S S 1 (cos cos ) For the spring: F k(s) F k(cos cos ) Sustitute F into Eq. (1): cos k(cos cos )(sin ) or cot u cos u cos 0 (Eq. ) k This equation must e solved numerically for the angle. DETERMINE THE DISTNE 1 cos cos (cos cos ) From Eq. (): Therefore, cos u cos u cot u k NUMERI RESUTS (Eq. 3) 8.0 in. k 16 l/in l Sustitute into Eq. (): cot cos (Eq. 4) Solve Eq. (4) numerically: u 35.1 Sustitute into Eq. (3): 1.78 in. cot u cos cos u cot u k rolem.-14 Solve the preceding prolem for the following data: 00 mm, k 3. kn/m, 45, and 50 N. Solution.-14 Framework with rigid ars and a spring See the solution to the preceding prolem. Eq. (): Eq. (3): cot u cos u cos 0 k k cot u NUMERI RESUTS 00 mm k 3. kn/m N Sustitute into Eq. (): cot cos (Eq. 4) Solve Eq. (4) numerically: u 35.1 Sustitute into Eq. (3): 44.5 mm

13 SETION.3 hanges in engths under Nonuniform onditions 75 hanges in engths under Nonuniform onditions rolem.3-1 alculate the elongation of a copper ar of solid circular cross section with tapered ends when it is stretched y axial loads of magnitude 3.0 k (see figure). The length of the end segments is 0 in. and the length of the prismatic middle segment is 50 in. lso, the diameters at cross sections,,, and D are 0.5, 1.0, 1.0, and 0.5 in., respectively, and the modulus of elasticity is 18,000 ksi. (Hint: Use the result of Example -4.) 3.0 k 0 in. 50 in. 0 in. D 3.0 k Solution.3-1 ar with tapered ends 3.0 k D 0 in. 50 in. 0 in. 3.0 k d d D 0.5 in. 3.0 k d d 1.0 in. E 18,000 ksi END SEGMENT ( 0 in.) From Example -4: 4 E d d 4(3.0 k)(0 in.) in. (18,000 ksi)(0.5 in.)(1.0 in.) MIDDE SEGMENT ( 50 in.) E in. EONGTION OF R a N E 1 ( in.) ( in.) in. (3.0 k)(50 in.) (18,000 ksi)( 4 )(1.0 in.) rolem.3- long, rectangular copper ar under a tensile load hangs from a pin that is supported y two steel posts (see figure). The copper ar has a length of.0 m, a cross-sectional area of 4800 mm, and a modulus of elasticity E c 10 Ga. Each steel post has a height of 0.5 m, a cross-sectional area of 4500 mm, and a modulus of elasticity E s 00 Ga. (a) Determine the downward displacement of the lower end of the copper ar due to a load 180 kn. () What is the maximum permissile load max if the displacement is limited to 1.0 mm? opper ar Steel post

14 76 HTER xially oaded Memers Solution.3- opper ar with a tensile load Steel post s c (a) DOWNWRD DISEMENT ( 180 kn) c c E c c 0.65 mm (180 kn)(.0 m) (10 Ga)(4800 mm ) opper ar c.0 m c 4800 mm s () s E s s mm c s 0.65 mm mm mm (90 kn)(0.5 m) (00 Ga)(4500 mm ) E c 10 Ga () MXIMUM OD max ( max 1.0 mm) s 0.5 m s 4500 mm E s 00 Ga max max max max 1.0 mm max (180 kn) 67 kn mm rolem.3-3 steel ar D (see figure) has a cross-sectional area of 0.40 in. and is loaded y forces l, 1800 l, and l. The lengths of the segments of the ar are a 60 in., 4 in., and c 36 in. (a) ssuming that the modulus of elasticity E psi, calculate the change in length of the ar. Does the ar elongate or shorten? () y what amount should the load 3 e increased so that the ar does not change in length when the three loads are applied? 1 3 D a c Solution.3-3 Steel ar loaded y three forces 1 3 D 60 in. 4 in. 36 in in. (a) HNGE IN ENGTH l 1800 l l E psi N i i a E i i XI FORES 1 E (N N N D D ) N l 1 [(300 l)(60 in.) N l ( psi)(0.40 in. ) N D l (500 l)(4 in.) (1300 l)(36 in.)] in. (elongation)

15 SETION.3 hanges in engths under Nonuniform onditions 77 () INRESE IN 3 FOR NO HNGE IN ENGTH 10 in. increase in force 3 The force must produce a shortening equal to in. in order to have no change in length in. E (10 in.) ( psi)(0.40 in. ) 1310 l rolem.3-4 rectangular ar of length has a slot in the middle half of its length (see figure). The ar has width, thickness t, and modulus of elasticity E. The slot has width /4. 4 t (a) Otain a formula for the elongation of the ar due to the axial loads. () alculate the elongation of the ar if the material is high-strength steel, the axial stress in the middle region is 160 Ma, the length is 750 mm, and the modulus of elasticity is 10 Ga. 4 4 Solution.3-4 ar with a slot t thickness (a) EONGTION OF R length of ar N i i a (4) () (4) E i E(t) E( 3 4t) E(t) Et Et STRESS IN MIDDE REGION s ( 3 4t) Sustitute into the equation for : Et 6E t 6E 3s 4 7s 8E 4 3 t or t 3s 4 () SUSTITUTE NUMERI VUES: s 160 Ma 750 mme 10 Ga 7(160 Ma)(750 mm) mm 8 (10 Ga)

16 78 HTER xially oaded Memers rolem.3-5 Solve the preceding prolem if the axial stress in the middle region is 4,000 psi, the length is 30 in., and the modulus of elasticity is psi. Solution.3-5 ar with a slot 4 STRESS IN MIDDE REGION s ( t) 3t or t 3s 4 t thickness 4 4 (a) EONGTION OF R length of ar N i i a (4) E i E(t) Et Et () (4) E ( 3 4 t) E(t) SUSTITUTE INTO THE EQUTION FOR : Et 6E t 6E 3s 4 7s 8E () SUSTITUTE NUMERI VUES: s 4,000 psi 30 in. E psi 7(4,000 psi)(30 in.) in. 8( psi) rolem.3-6 two-story uilding has steel columns in the first floor and in the second floor, as shown in the figure. The roof load 1 equals 400 kn and the second-floor load equals 70 kn. Each column has length 3.75 m. The cross-sectional areas of the first- and secondfloor columns are 11,000 mm and 3,900 mm, respectively. (a) ssuming that E 06 Ga, determine the total shortening of the two columns due to the comined action of the loads 1 and. () How much additional load 0 can e placed at the top of the column (point ) if the total shortening is not to exceed 4.0 mm? 1 = 400 kn = 70 kn = 3.75 m = 3.75 m Solution.3-6 Steel columns in a uilding 1 = 400 kn = 70 kn length of each column 3.75 m E 06 Ga 11,000 mm 3,900 mm (a) SHORTENING OF THE TWO OUMNS a N i i E i i N E N E (110 kn)(3.75 m) (06 Ga)(11,000 mm ) (400 kn)(3.75 m) (06 Ga)(3,900 mm ) mm mm mm 3.7 mm

17 SETION.3 hanges in engths under Nonuniform onditions 79 () DDITION OD 0 T OINT ( ) max 4.0 mm 0 additional shortening of the two columns due to the load 0 0 ( ) max 4.0 mm mm mm lso, E E E 1 Solve for 0 : 0 E 0 SUSTITUTE NUMERI VUES: E Nm 3.75 m 11, m 3, m 0 44,00 N 44. kn m rolem.3-7 steel ar 8.0 ft long has a circular cross section of diameter d in. over one-half of its length and diameter d 0.5 in. over the other half (see figure). The modulus of elasticity E psi. (a) How much will the ar elongate under a tensile load 5000 l? () If the same volume of material is made into a ar of constant diameter d and length 8.0 ft, what will e the elongation under the same load? d 1 = 0.75 in. d = 0.50 in. 4.0 ft 4.0 ft = 5000 l Solution.3-7 d 1 = 0.75 in l E psi 4ft48 in. ar in tension d = 0.50 in. 4.0 ft 4.0 ft (a) EONGTION OF NONRISMTI R a N i i E i i E a 1 i (5000 l)(48 in.) psi 1 4 (0.75 in) 1 4 (0.50 in.) R in. = 5000 l () EONGTION OF RISMTI R OF SME VOUME Original ar: V o 1 ( 1 ) rismatic ar: V p p () Equate volumes and solve for p : V o V p ( 1 ) p () p (d 1 d ) 8 [(0.75 in.) (0.50 in.) ] in. () E p in. (5000 l)()(48 in.) ( psi)( in. ) NOTE: prismatic ar of the same volume will always have a smaller change in length than will a nonprismatic ar, provided the constant axial load, modulus E, and total length are the same.

18 80 HTER xially oaded Memers rolem.3-8 ar of length consists of two parts of equal lengths ut different diameters (see figure). Segment has diameter d mm and segment has diameter d 60 mm. oth segments have length / 0.6 m. longitudinal hole of diameter d is drilled through segment for one-half of its length (distance /4 0.3 m). The ar is made of plastic having modulus of elasticity E 4.0 Ga. ompressive loads 110 kn act at the ends of the ar. If the shortening of the ar is limited to 8.0 mm, what is the maximum allowale diameter d max of the hole? 4 d 1 4 d Solution.3-8 d 4 d diameter of hole ar with a hole d 1 4 SHORTENING OF THE R a N i i E i i E a i i 4 4 E 4 (d 1 d ) 4 d 1 4 d S E 1 d 1 d 1 d 1 d NUMERI VUES (DT): d maximum allowale shortening of the ar 8.0 mm (Eq. 1) 110 kn 1. m E 4.0 Ga d mm d max maximum allowale diameter of the hole d 60 mm SUSTITUTE NUMERI VUES INTO EQ. (1) FOR ND SOVE FOR d d max : UNITS: Newtons and meters (110,000)(1.) ( ) d 1 (0.1) d 1 (0.1) (0.06) R d d m d m d max 3.9 mm rolem.3-9 wood pile, driven into the earth, supports a load entirely y friction along its sides (see figure). The friction force f per unit length of pile is assumed to e uniformly distriuted over the surface of the pile. The pile has length, cross-sectional area, and modulus of elasticity E. (a) Derive a formula for the shortening of the pile in terms of,, E, and. () Draw a diagram showing how the compressive stress c varies throughout the length of the pile. f

19 SETION.3 hanges in engths under Nonuniform onditions 81 Solution.3-9 Wood pile with friction dy y f Friction force per unit length of pile f = FROM FREE-ODY DIGRM OF IE: F vert 0c T f 0f (a) SHORTENING OF IE: t distance y from the ase: 0 c = y ompressive stress in pile (Eq. 1) N(y) axial force N(y) fy (Eq. ) N(y) dy d E d E f ydy f E 0 E () OMRESSIVE STRESS c IN IE s c N(y) fy fy dy E y t the ase (y 0): c 0 t the top(y ): s c See the diagram aove. 0 E rolem.3-10 prismatic ar of length, cross-sectional area, modulus of elasticity E, and weight W hangs vertically under its own weight (see figure). (a) Derive a formula for the downward displacement of point, located at distance h from the lower end of the ar. () What is the elongation of the entire ar? (c) What is the ratio of the elongation of the upper half of the ar to the elongation of the lower half of the ar? h Solution.3-10 rismatic ar hanging vertically dy y h W Weight of ar (a) DOWNWRD DISEMENT onsider an element at distance y from the lower end. () EONGTION OF R (h 0) W E (c) RTIO OF EONGTIONS Elongation of upper half of ar h : N(y) Wy Wydy W E ( h d ) E h N(y)dy Wydy d E E W E ( h ) h upper 3W 8E Elongation of lower half of ar: lower upper W E upper lower W W 8E 8E

20 8 HTER xially oaded Memers rolem.3-11 flat ar of rectangular cross section, length, and constant thickness t is sujected to tension y forces (see figure). The width of the ar varies linearly from 1 at the smaller end to at the larger end. ssume that the angle of taper is small. t (a) Derive the following formula for the elongation of the ar: 1 ln Et( 1 ) 1 () alculate the elongation, assuming 5 ft, t 1.0 in., 5 k, in., 6.0 in., and E psi. Solution.3-11 Tapered ar (rectangular cross section) x dx t thickness (constant) x (Eq. 1) From Eq. (1): Solve Eq. (3) for 0 : (Eq. 3) (Eq. 4) (x) t 1 t x 0 Sustitute Eqs. (3) and (4) into Eq. (): (a) EONGTION OF THE R Et( 1 ) ln 1 (Eq. 5) d dx E(x) 0 dx E 1 tx 0 d 0 0 E 1 t 0 dx 0 x () SUSTITUTE NUMERI VUES: 5ft60 in. t 10 in. 5 k in. 6.0 in. E psi 0 0 E 1 t ln x 0 E 1 t ln (Eq. ) From Eq. (5): in.

21 SETION.3 hanges in engths under Nonuniform onditions 83 rolem.3-1 post supporting equipment in a laoratory is tapered uniformly throughout its height H (see figure). The cross sections of the post are square, with dimensions at the top and at the ase. Derive a formula for the shortening of the post due to the compressive load acting at the top. (ssume that the angle of taper is small and disregard the weight of the post itself.) H Solution.3-1 Tapered post H y 1.5 Square cross sections width at 1.5 width at y width at distance y y dy SHORTENING OF EEMENT dy d dy E y SHORTENING OF ENTIRE OST d H H dy E (H 0.5y) dx From ppendix : (a x) 1 (a x) H H E 1 (0.5)(H 0.5y) R 0 H E 1 (0.5)(1.5H) 1 0.5H R H 3E 0 dy E (H 0.5y) H (1.5 ) y H (H 0.5y) H y cross-sectional area at distance y ( y ) H(H 0.5y)

22 84 HTER xially oaded Memers rolem.3-13 long, slender ar in the shape of a right circular cone with length and ase diameter d hangs vertically under the action of its own weight (see figure). The weight of the cone is W and the modulus of elasticity of the material is E. Derive a formula for the increase in the length of the ar due to its own weight. (ssume that the angle of taper of the cone is small.) d Solution.3-13 onical ar hanging vertically d EEMENT OF R dy N y N y dy y Wweight of cone TERMINOOGY N y axial force acting on element dy y cross-sectional area at element dy cross-sectional area at ase of cone d 4 V volume of cone EONGTION OF EEMENT dy d N y dy Wy dy E y E EONGTION OF ONI R 4W d E y dy d 4W d E y dy W d E V y volume of cone elow element dy 1 3 y y W y weight of cone elow element dy V y V (W) yyw N y W y

23 SETION.3 hanges in engths under Nonuniform onditions 85 rolem.3-14 ar revolves in a horizontal plane aout a vertical axis at the midpoint (see figure). The ar, which has length and cross-sectional area, revolves at constant angular speed. Each half of the ar ( and ) has weight W 1 and supports a weight W at its end. Derive the following formula for the elongation of one-half of the ar (that is, the elongation of either or ): (W 3 ge 1 3W ) in which E is the modulus of elasticity of the material of the ar and g is the acceleration of gravity. W W 1 W 1 W Solution.3-14 Rotating ar x W 1 F(x) dx angular speed cross-sectional area E modulus of elasticity g acceleration of gravity F(x) axial force in ar at distance x from point onsider an element of length dx at distance x from point. To find the force F(x) acting on this element, we must find the inertia force of the part of the ar from distance x to distance, plus the inertia force of the weight W. Since the inertia force varies with distance from point, we now must consider an element of length d at distance, where varies from x to. Mass of element dj dj W 1 g cceleration of element d W entrifugal force produced y element (mass)(acceleration) W 1 g jdj entrifugal force produced y weight W W g ( ) XI FORE F(x) j W F(x) 1 EONGTION OF R F(x) dx E 0 W 1 W ge ( x )dx dx ge 0 jx W 1 ge dx x dx R W 1 3gE W ge 3gE (W 1 3W ) g jdj W g W 1 g ( x ) W g W ge 0 dx

24 86 HTER xially oaded Memers rolem.3-15 The main cales of a suspension ridge [see part (a) of the figure] follow a curve that is nearly paraolic ecause the primary load on the cales is the weight of the ridge deck, which is uniform in intensity along the horizontal. Therefore, let us represent the central region O of one of the main cales [see part () of the figure] as a paraolic cale supported at points and and carrying a uniform load of intensity q along the horizontal. The span of the cale is, the sag is h, the axial rigidity is E, and the origin of coordinates is at midspan. (a) (a) Derive the following formula for the elongation of cale O shown in part () of the figure: q3 (1 1 6h ) 8 he 3 () alculate the elongation of the central span of one of the main cales of the Golden Gate ridge, for which the dimensions and properties are 400 ft, h 470 ft, q 1,700 l/ft, and E 8,800,000 psi. The cale consists of 7,57 parallel wires of diameter in. q y O () h x Hint: Determine the tensile force T at any point in the cale from a free-ody diagram of part of the cale; then determine the elongation of an element of the cale of length ds; finally, integrate along the curve of the cale to otain an equation for the elongation. Solution.3-15 ale of a suspension ridge O y D q h x Equation of paraolic curve: dy dx y 4hx 8hx FREE-ODY DIGRM OF HF OF E M 0 H O y D V h H x Hh q 4 0 H q 8h F horizontal 0 q H H q 8h F vertical 0 (Eq. 1) V q (Eq. )

25 SETION.3 hanges in engths under Nonuniform onditions 87 FREE-ODY DIGRM OF SEGMENT D OF E y 0 T H T x D T V F horiz 0 T H H F vert 0V T V q x 0 T V V q x q q (Eq. 3) qx (Eq. 4) TENSIE FORE T IN E x T T H T V q 8h (qx) q V q 8h 1 64h x 4 H q 8h h 4hx 4hx x T qx T H D T V (Eq. 5) (a) EONGTION OF E O d T ds E Sustitute for T from Eq. (5) and for ds from Eq. (6): E 1 q 8h 1 64h x dx For oth halves of cale: E 0 q 8h 1 64h x 4 q3 16h 1 8hE 3 4 () GODEN GTE RIDGE E 400 ft h 470 ft q 1,700 l/ft E 8,800,000 psi 7,57 wires of diameter d in. (7,57) 4 (0.196 in.) in. Sustitute into Eq. (7): in ft dx (Eq. 7) EONGTION d OF N EEMENT OF ENGTH ds T ds dy d Tds E ds (dx) (dy) dx T dx 1 8hx dx 1 64h x 4 dx 1 dy dx (Eq. 6)

26 88 HTER xially oaded Memers Statically Indeterminate Structures rolem.4-1 The assemly shown in the figure consists of a rass core (diameter d in.) surrounded y a steel shell (inner diameter d 0.8 in., outer diameter d in.). load compresses the core and shell, which have length 4.0 in. The moduli of elasticity of the rass and steel are E psi and E s psi, respectively. (a) What load will compress the assemly y in.? () If the allowale stress in the steel is ksi and the allowale stress in the rass is 16 ksi, what is the allowale compressive load allow? (Suggestion: Use the equations derived in Example -5.) Steel shell rass core d 1 d d 3 Solution.4-1 ylindrical assemly in compression Steel shell rass core d 1 d d 3 d in. E psi d 0.8 in. E s psi d in. s 4 (d 3 d ) in. 4.0 in. 4 d in. (a) DERESE IN ENGTH ( in.) Use Eq. (-13) of Example -5. or E s s E Sustitute numerical values: E s s E ( psi)( in. ) ( psi)( in. ) l in. ( l) 4.0 in l ( ) OWE OD s ksi 16 ksi Use Eqs. (-1a and ) of Example -5. For steel: E s s s E s s E s (E s s E ) s s E s ksi s ( l) l psi For rass: E s E s s E s (E s s E ) s E 16 ksi s ( l) l psi Steel governs. allow 1300 l (E s s E )

27 SETION.4 Statically Indeterminate Structures 89 rolem.4- cylindrical assemly consisting of a rass core and an aluminum collar is compressed y a load (see figure). The length of the aluminum collar and rass core is 350 mm, the diameter of the core is 5 mm, and the outside diameter of the collar is 40 mm. lso, the moduli of elasticity of the aluminum and rass are 7 Ga and 100 Ga, respectively. (a) If the length of the assemly decreases y 0.1% when the load is applied, what is the magnitude of the load? () What is the maximum permissile load max if the allowale stresses in the aluminum and rass are 80 Ma and 10 Ma, respectively? (Suggestion: Use the equations derived in Example -5.) 350 mm luminum collar rass core 5 mm 40 mm Solution.4- ylindrical assemly in compression or E a a E (E a a E ) 350 mm d d a aluminum rass 350 mm d a 40 mm d 5 mm a 4 (d a d ) mm E a 7 GaE 100 Ga 4 d mm (a) DERESE IN ENGTH ( 0.1% of mm) Use Eq. (-13) of Example -5. Sustitute numerical values: E a a E (7 Ga)(765.8 mm ) (100 Ga)(490.9 mm ) MN MN MN mm (104.3 MN) 350 mm 104. kn () OWE OD a 80 Ma 10 Ma Use Eqs. (-1a and ) of Example -5. For aluminum: E a s a E a a E a (E a a E s a ) E a 80 Ma a (104.3 MN) kn 7 Ga For rass: E s (E a a E s ) E a a E E 10 Ma (104.3 MN) 15.1 kn 100 Ga luminum governs. max 116 kn

28 90 HTER xially oaded Memers rolem.4-3 Three prismatic ars, two of material and one of material, transmit a tensile load (see figure). The two outer ars (material ) are identical. The cross-sectional area of the middle ar (material ) is 50% larger than the cross-sectional area of one of the outer ars. lso, the modulus of elasticity of material is twice that of material. (a) What fraction of the load is transmitted y the middle ar? () What is the ratio of the stress in the middle ar to the stress in the outer ars? (c) What is the ratio of the strain in the middle ar to the strain in the outer ars? Solution.4-3 rismatic ars in tension FREE-ODY DIGRM OF END TE STRESSES: s E E E s E E E (7) EQUTION OF EQUIIRIUM F horiz 0 0 (1) EQUTION OF OMTIIITY () FORE-DISEMENT RETIONS total area of oth outer ars E E Sustitute into Eq. (): E E SOUTION OF THE EQUTIONS Solve simultaneously Eqs. (1) and (4): E E E E E E Sustitute into Eq. (3): E E (3) (4) (5) (6) (a) OD IN MIDDE R 1 () RTIO OF STRESSES s s E E 1 (c) RTIO OF STRINS ll ars have the same strain Ratio 1 E 1 E E E 1 E Given: E E E E

29 SETION.4 Statically Indeterminate Structures 91 rolem.4-4 ar having two different cross-sectional areas 1 and is held etween rigid supports at and (see figure). load acts at point, which is distance 1 from end and distance from end. 1 (a) Otain formulas for the reactions R and R at supports and, respectively, due to the load. () Otain a formula for the displacement of point. (c) What is the ratio of the stress 1 in region to the stress in region? 1 Solution.4-4 ar with intermediate load 1 1 FREE-ODY DIGRM R EQUTION OF EQUIIRIUM F horiz 0 R R (Eq. 1) EQUTION OF OMTIIITY elongation of shortening of (Eq. ) FORE DISEMENT RETIONS R 1 E 1 R E R (Eqs. 3&4) Solve Eq. (1) and Eq. (5) simultaneously: R 1 R () DISEMENT OF OINT R 1 E 1 (c) RTIO OF STRESSES s 1 R 1 (tension)s R (compression) s 1 s 1 1 E( 1 1 ) (Note that if 1, the stresses are numerically equal regardless of the areas 1 and.) (a) SOUTION OF EQUTIONS Sustitute Eq. (3) and Eq. (4) into Eq. (): R 1 E 1 R E (Eq. 5)

30 9 HTER xially oaded Memers rolem.4-5 Three steel cales jointly support a load of 1 k (see figure). The diameter of the middle cale is 3 4 in. and the diameter of each outer cale is 1 in. The tensions in the cales are adjusted so that each cale carries one-third of the load (i.e., 4 k). ater, the load is increased y 9 k to a total load of 1 k. (a) What percent of the total load is now carried y the middle cale? () What are the stresses M and O in the middle and outer cales, respectively? (Note: See Tale -1 in Section. for properties of cales.) Solution.4-5 Three cales in tension 1 in. 1 in. 3 4 in. FORE-DISEMENT RETIONS M M E M O o E o SUSTITUTE INTO OMTIIITY EQUTION: M O M O E M E O M O (3, 4) (5) RES OF ES (from Tale -1) Middle cale: M 0.68 in. Outer cales: O in. (for each cale) FIRST ODING 1 1 k Each cale carries SEOND ODING 9k (additional load) O M 1 3 or 4 k. O SOVE SIMUTNEOUSY EQS. (1) ND (5): k in. o (9 k) M O in..117 k M 0.68 in. M (9 k) M O in. o FORES IN ES Middle cale: Force 4k4.767 k k Outer cales: Force 4k.117 k k (for each cale) = 9 k EQUTION OF EQUIIRIUM F vert 0 O M 0 (1) EQUTION OF OMTIIITY M O () (a) ERENT OF TOT OD RRIED Y MIDDE E ercent k (100%) 41.7% 1 k () STRESSES IN ES ( /) Middle cale: s M k 3.7 ksi 0.68 in. Outer cales: s O k 51.4 ksi in.

31 SETION.4 Statically Indeterminate Structures 93 rolem.4-6 plastic rod of length 0.5 m has a diameter d 1 30 mm (see figure). plastic sleeve D of length c 0.3 m and outer diameter d 45 mm is securely onded to the rod so that no slippage can occur etween the rod and the sleeve. The rod is made of an acrylic with modulus of elasticity E Ga and the sleeve is made of a polyamide with E.5 Ga. (a) alculate the elongation of the rod when it is pulled y axial forces 1 kn. () If the sleeve is extended for the full length of the rod, what is the elongation? (c) If the sleeve is removed, what is the elongation? d 1 d c D Solution.4-6 lastic rod with sleeve d d 1 D d 1 c 1 kn d 1 30 mm 100 mm 500 mm d 45 mm c 300 mm Rod: E Ga Sleeve: E.5 Ga Rod: 1 d mm 4 Sleeve: 4 (d d 1 ) mm E 1 1 E MN () SEEVE T FU ENGTH mm D ( mm) 500 c 300 mm 1.36 mm (c) SEEVE REMOVED E mm (a) EONGTION OF ROD art : mm E 1 1 art D: D c E 1 1 E mm (From Eq. -13 of Example -5) D 1.91 mm

32 94 HTER xially oaded Memers rolem.4-7 The axially loaded ar D shown in the figure is held etween rigid supports. The ar has cross-sectional area 1 from to and 1 from to D. (a) Derive formulas for the reactions R and R D at the ends of the ar. () Determine the displacements and at points and, respectively. (c) Draw a diagram in which the ascissa is the distance from the left-hand support to any point in the ar and the ordinate is the horizontal displacement at that point D Solution.4-7 ar with fixed ends FREE-ODY DIGRM OF R R EQUTION OF EQUIIRIUM F horiz 0 R R D EQUTION OF OMTIIITY 4 (Eq. 1) D 0 (Eq. ) ositive means elongation. FORE-DISEMENT EQUTIONS R (4) E (R )(4) 1 E 1 D R D() E( 1 ) D R D (Eqs. 3, 4) (Eq. 5) (a) RETIONS Solve simultaneously Eqs. (1) and (6): R 3 R D 3 () DISEMENTS T OINTS ND R 4E 1 6E 1 (To the right) D R D 4E 1 (To the right) 1E 1 (c) DISEMENT DIGRM Displacement 6E 1 1E 1 SOUTION OF EQUTIONS Sustitute Eqs. (3), (4), and (5) into Eq. (): R (R )() R D 0 4E 1 4E 1 4E 1 (Eq. 6) D 0 4 Distance from end

33 SETION.4 Statically Indeterminate Structures 95 rolem.4-8 The fixed-end ar D consists of three prismatic segments, as shown in the figure. The end segments have crosssectional area mm and length 1 00 mm. The middle segment has cross-sectional area 160 mm and length 50 mm. oads and are equal to 5.5 kn and 17.0 kn, respectively. (a) Determine the reactions R and R D at the fixed supports. () Determine the compressive axial force F in the middle segment of the ar D Solution.4-8 ar with three segments kn 17.0 kn D 1 00 mm 50 mm mm 160 mm 1 1 m meter FREE-ODY DIGRM R R D D EQUTION OF EQUIIRIUM F horiz 0 S d R D R 0 or R R D 8.5 kn (Eq. 1) EQUTION OF OMTIIITY D elongation of entire ar D D 0 (Eq. ) FORE-DISEMENT RETIONS R 1 E 1 R E m (Eq. 3) SOUTION OF EQUTIONS Sustitute Eqs. (3), (4), and (5) into Eq. (): R E m R E m E m R D E m 0 Simplify and sustitute 5.5 kn: R , m R 1 D m kn m (Eq. 6) (R ) E R E m E m D R D 1 E 1 R D E m (Eq. 4) (Eq. 5) (a) RETIONS R ND R D Solve simultaneously Eqs. (1) and (6). From (1): R D R 8.5 kn Sustitute into (6) and solve for R : R R 10.5 kn 1 m kn m R D R 8.5 kn.0 kn () OMRESSIVE XI FORE F F R R D 15.0 kn

34 96 HTER xially oaded Memers rolem.4-9 The aluminum and steel pipes shown in the figure are fastened to rigid supports at ends and and to a rigid plate at their junction. The aluminum pipe is twice as long as the steel pipe. Two equal and symmetrically placed loads act on the plate at. (a) Otain formulas for the axial stresses a and s in the aluminum and steel pipes, respectively. () alculate the stresses for the following data: 1 k, cross-sectional area of aluminum pipe a 8.9 in., cross-sectional area of steel pipe s 1.03 in., modulus of elasticity of aluminum E a psi, and modulus of elasticity of steel E s psi. Steel pipe luminum pipe Solution.4-9 ipes with intermediate loads 1 R E s s E a a SOUTION OF EQUTIONS Sustitute Eqs. (3) and (4) into Eq. (): R R () 0 E s s E a a Solve simultaneously Eqs. (1) and (5): R 4E s s R E a a E s E a a s E a a E s s (Eq. 5) (Eqs. 6, 7) ipe 1 is steel. ipe is aluminum. EQUTION OF EQUIIRIUM F vert 0 R R (Eq. 1) EQUTION OF OMTIIITY 0 (Eq. ) ( positive value of means elongation.) FORE-DISEMENT RETIONS R E s s R () E a a R (Eqs. 3, 4)) (a) XI STRESSES luminum: s a R E a a E a a E s s (compression) Steel: s s R 4E s s E a a E s s (tension) () NUMERI RESUTS 1 k a 8.9 in. s 1.03 in. E a psi E s psi Sustitute into Eqs. (8) and (9): s a 1,610 psi (compression) s s 9,350 psi (tension) (Eq. 8) (Eq. 9)

35 SETION.4 Statically Indeterminate Structures 97 rolem.4-10 rigid ar of weight W 800 N hangs from three equally spaced vertical wires, two of steel and one of aluminum (see figure). The wires also support a load acting at the midpoint of the ar. The diameter of the steel wires is mm, and the diameter of the aluminum wire is 4 mm. What load allow can e supported if the allowale stress in the steel wires is 0 Ma and in the aluminum wire is 80 Ma? (ssume E s 10 Ga and E a 70 Ga.) S S Rigid ar of weight W Solution.4-10 Rigid ar hanging from three wires S S W = 800 N STEE WIRES d s mm s 0 Ma E s 10 Ga UMINUM WIRES d 4mm 80 Ma E 70 Ga FREE-ODY DIGRM OF RIGID R F S F F S + W SOUTION OF EQUTIONS Sustitute (3) and (4) into Eq. (): F s F E s s E Solve simultaneously Eqs. (1) and (5): E F ( W) E E s s E s s F s ( W) E E s s STRESSES IN THE WIRES s F ( W)E E E s s s s F s ( W)E s s E E s s OWE ODS (FROM EQS. (8) ND (9)) s E (E E s s ) W s s s E s (E E s s ) W (Eq. 5) (Eq. 6) (Eq. 7) (Eq. 8) (Eq. 9) (Eq. 10) (Eq. 11) EQUTION OF EQUIIRIUM F vert 0 F s F W 0 (Eq. 1) EQUTION OF OMTIIITY s (Eq. ) FORE DISEMENT RETIONS s F s E s s F E (Eqs. 3, 4) SUSTITUTE NUMERI VUES INTO EQS. (10) ND (11): s 4 ( mm) mm 4 (4 mm) mm 1713 N s 1504 N Steel governs. allow 1500 N

36 98 HTER xially oaded Memers rolem.4-11 imetallic ar (or composite ar) of square cross section with dimensions is constructed of two different metals having moduli of elasticity E 1 and E (see figure). The two parts of the ar have the same cross-sectional dimensions. The ar is compressed y forces acting through rigid end plates. The line of action of the loads has an eccentricity e of such magnitude that each part of the ar is stressed uniformly in compression. (a) Determine the axial forces 1 and in the two parts of the ar. () Determine the eccentricity e of the loads. (c) Determine the ratio 1 / of the stresses in the two parts of the ar. e E E 1 e Solution.4-11 imetallic ar in compression E 1 E 1 E E 1 1 FREE-ODY DIGRM (late at right-hand end) e (a) XI FORES Solve simultaneously Eqs. (1) and (3): 1 E 1 E 1 E E E 1 E 1 EQUTIONS OF EQUIIRIUM F 0 1 (Eq. 1) M 0 e 1 0 (Eq. ) ( EENTRIITY OF OD Sustitute 1 and into Eq. () and solve for e: e (E E 1 ) (E E 1 ) (c) RTIO OF STRESSES EQUTION OF OMTIIITY 1 E 1 E 1 or 1 E E 1 (Eq. 3) s 1 1 s s 1 s 1 E 1 E

37 SETION.4 Statically Indeterminate Structures 99 rolem.4-1 circular steel ar (E = 00 Ga) has crosssectional area 1 from to and cross-sectional area from to (see figure). The ar is supported rigidly at end and is sujected to a load equal to 40 kn at end. circular steel collar D having cross-sectional area 3 supports the ar at. The collar fits snugly at and D when there is no load. Determine the elongation of the ar due to the load. (ssume mm, 5 mm, mm, and 300 mm.) 1 3 D 3 1 Solution.4-1 ar supported y a collar FREE-ODY DIGRM OF R ND OR D 1 R EQUIIRIUM OF R F vert 0 R R D 0 (Eq. 1) OMTIIITY (distance D does not change) (ar) D (collar) 0 (Eq. ) (Elongation is positive.) FORE-DISEMENT RETIONS R 1 E D R D 3 1 E 3 Sustitute into Eq. (): R D R 1 E 1 R D 3 E R D 3 3 D R D (Eq. 3) SOVE SIMUTNEOUSY EQS. (1) ND (3): R R D HNGES IN ENGTHS (Elongation is positive) R 1 E 1 EONGTION OF R SUSTITUTE NUMERI VUES: 40 kn 1 50 mm 5 mm 3 15 mm mm 300 mm mm RESUTS: R R D 0 kn mm mm 1 3 E( ) E E 00 Ga mm

38 100 HTER xially oaded Memers rolem.4-13 horizontal rigid ar of weight W 700 l is supported y three slender circular rods that are equally spaced (see figure). The two outer rods are made of aluminum (E psi) with diameter d in. and length 1 40 in. The inner rod is magnesium (E psi) with diameter d and length. The allowale stresses in the aluminum and magnesium are 4,000 psi and 13,000 psi, respectively. If it is desired to have all three rods loaded to their maximum allowale values, what should e the diameter d and length of the middle rod? d 1 d d 1 1 W = weight of rigid ar Solution.4-13 ar supported y three rods R 1UMINUM SUSTITUTE NUMERI VUES: E psi d in in. 1 4,000 psi 4(700 l) d (13,000 psi) d in. (4,000 psi) (0.4 in.) 13,000 psi in in in. 1 1 R MGNESIUM E psi EQUTION OF OMTIIITY 1 (Eq. 3) d?? FORE-DISEMENT RETIONS W = 700 l 13,000 psi 1 F 1 1 E 1 1 s 1 1 E 1 (Eq. 4) FREE-ODY DIGRM OF RIGID R EQUTION OF EQUIIRIUM F F vert 0 1 F F 1 F 1 F W 0 (Eq. 1) W F s E E Sustitute (4) and (5) into Eq. (3): s 1 1 E 1 s E ength 1 is known; solve for : (Eq. 5) FUY STRESSED RODS 1 s 1E s E 1 (Eq. 6) F F 1 d 1 4 d 4 Sustitute into Eq. (1): s d s d 4 W SUSTITUTE NUMERI VUES: 4,000 psi (40 in.) 13,000 psi psi psi 48.0 in. Diameter d 1 is known; solve for d : d 4W s 1d s s (Eq. )

39 SETION.4 Statically Indeterminate Structures 101 rolem.4-14 rigid ar D is pinned at point and a = 50 mm = 500 mm supported y springs at and D (see figure). The springs at and D have stiffnesses k 1 10 kn/m and k 5 kn/m, respectively, and the dimensions a,, and c are 50 mm, 500 mm, and 00 mm, respectively. D load acts at point. If the angle of rotation of the ar due to the action of the load c = 00 mm is limited to 3, what is the maximum permissile load max? k = 5 kn/m k 1 = 10 kn/m Solution.4-14 Rigid ar supported y springs a D k 1 k c NUMERI DT a 50 mm 500 mm c 00 mm k 1 10 kn/m k 5 kn/m u max 3 60 rad EQUTION OF OMTIIITY a D FORE-DISEMENT RETIONS F k 1 D F D k SOUTION OF EQUTIONS Sustitute (3) and (4) into Eq. (): F ak 1 F D k SOVE SIMUTNEOUSY EQS. (1) ND (5): F ack 1 F a k 1 D ck k a k 1 k (Eq. ) (Eqs. 3, 4) (Eq. 5) FREE-ODY DIGRM ND DISEMENT DIGRM a NGE OF ROTTION D F D c u D k a k 1 k c a k 1 k F R F D MXIMUM OD c u c (a k 1 k ) D D max u max c (a k 1 k ) SUSTITUTE NUMERI VUES: 60 rad max 00 mm [(50 mm) (10 knm) (500 mm) (5 knm)] 1800 N EQUTION OF EQUIIRIUM M 0 F (a) (c) F D () 0 (Eq. 1)

40 10 HTER xially oaded Memers rolem.4-15 rigid ar of length 66 in. is hinged to a support at and supported y two vertical wires attached at points and D (see figure). oth wires have the same cross-sectional area ( 0.07 in. ) and are made of the same material (modulus E psi). The wire at has length h 18 in. and the wire at D has length twice that amount. The horizontal distances are c 0 in. and d 50 in. (a) Determine the tensile stresses and D in the wires due to the load 340 l acting at end of the ar. () Find the downward displacement at end of the ar. c d h D h Solution.4-15 ar supported y two wires FREE-ODY DIGRM T D h D h T c D d R h 18 in. h 36 in. c 0 in. d 50 in. 66 in. E psi 0.07 in. 340 l DISEMENT DIGRM D D EQUTION OF EQUIIRIUM M 0 T (c) T D (d) (Eq. 1) EQUTION OF OMTIIITY c D d (Eq. )

41 SETION.4 Statically Indeterminate Structures 103 FORE-DISEMENT RETIONS DISEMENT T END OF R T h E D T D(h) E (Eqs. 3, 4) D d ht D E d h E(c d ) (Eq. 10) SOUTION OF EQUTIONS Sustitute (3) and (4) into Eq. (): T h ce T D(h) de ort c T D d (Eq. 5) SUSTITUTE NUMERI VUES c d (0 in.) (50 in.) 3300 in. (a) s c (c d ) (0 in.)(340 l)(66 in.) (0.07 in. )(3300 in. ) TENSIE FORES IN THE WIRES Solve simultaneously Eqs. (1) and (5): T c c d T D d c d (Eqs. 6, 7) 10,000 psi d s D (c d ) 1,500 psi (50 in.)(340 l)(66 in.) (0.07 in. )(3300 in. ) TENSIE STRESSES IN THE WIRES s T c (c d ) s D T D d (c d ) (Eq. 8) (Eq. 9) () h E(c d ) (18 in.)(340 l)(66 in.) ( psi)(0.07 in. )(3300 in. ) in. rolem.4-16 trimetallic ar is uniformly compressed y an axial force 40 kn applied through a rigid end plate (see figure). The ar consists of a circular steel core surrounded y rass and copper tues. The steel core has diameter 30 mm, the rass tue has outer diameter 45 mm, and the copper tue has outer diameter 60 mm. The corresponding moduli of elasticity are E s 10 Ga, E 100 Ga, and E c 10 Ga. alculate the compressive stresses s,, and c in the steel, rass, and copper, respectively, due to the force. = 40 kn opper tue 30 mm rass tue Steel core 45 mm 60 mm

42 104 HTER xially oaded Memers Solution.4-16 Trimetallic ar in compression opper s compressive force in steel core compressive force in rass tue c compressive force in copper tue FREE-ODY DIGRM OF RIGID END TE EQUTION OF EQUIIRIUM F vert 0 s c (Eq. 1) EQUTIONS OF OMTIIITY s c s (Eqs. ) FORE-DISEMENT RETIONS s s E s s E c c E c c SOUTION OF EQUTIONS Sustitute (3), (4), and (5) into Eqs. (): s c rass Steel (Eqs. 3, 4, 5) SOVE SIMUTNEOUSY EQS. (1), (6), ND (7): E s s s E s s E E c c E E s s E E c c E c c c E s s E E c c OMRESSIVE STRESSES et E E s s E E c c s s s s E s E s E E s c c c E c E SUSTITUTE NUMERI VUES: 40 kn E s 10 Ga E 100 Ga E c 10 Ga d 1 30 mm d 45 mm d 3 60 mm s 4 d mm 4 (d d 1 ) mm c 4 (d 3 d ) mm E N s E E s s c s E c c E s s (Eqs. 6, 7) s s E s 1.8 Ma E s E 10.4 Ma E s c E c 1.5 Ma E

43 SETION.5 Thermal Effects 105 Thermal Effects rolem.5-1 The rails of a railroad track are welded together at their ends (to form continuous rails and thus eliminate the clacking sound of the wheels) when the temperature is 60 F. What compressive stress is produced in the rails when they are heated y the sun to 10 F if the coefficient of thermal expansion / F and the modulus of elasticity E psi? Solution.5-1 Expansion of railroad rails The rails are prevented from expanding ecause of their great length and lack of expansion joints. Therefore, each rail is in the same condition as a ar with fixed ends (see Example -7). The compressive stress in the rails may e calculated from Eq. (-18). T 10F 60F 60F s E( T ) ( psi)( F)(60F) s 11,700 psi rolem.5- n aluminum pipe has a length of 60 m at a temperature of 10. n adjacent steel pipe at the same temperature is 5 mm longer than the aluminum pipe. t what temperature (degrees elsius) will the aluminum pipe e 15 mm longer than the steel pipe? (ssume that the coefficients of thermal expansion of aluminum and steel are a / and s /, respectively.) Solution.5- luminum and steel pipes INITI ONDITIONS a 60 m s m a / FIN ONDITIONS T 0 10 T 0 10 s / luminum pipe is longer than the steel pipe y the amount 15 mm. T increase in temperature a a (T) a s s (T) s a a luminum pipe or, a (T) a a s (T) s s Solve for T: T ( s a ) a a s s Sustitute numerical values: a a s s m/ 15 mm 5 mm T m T T 0 T Steel pipe s s From the figure aove: a a s s

44 106 HTER xially oaded Memers rolem.5-3 rigid ar of weight W 750 l hangs from three equally spaced wires, two of steel and one of aluminum (see figure). The diameter of the wires is 1 8 in. efore they were loaded, all three wires had the same length. What temperature increase T in all three wires will result in the entire load eing carried y the steel wires? (ssume E s psi, s / F, and a / F.) S S W = 750 l Solution.5-3 ar supported y three wires increase in length of a steel wire due to load W/ S steel W 750 l d 1 8 in. Rigid ar S S aluminum s d in. 4 E s psi E s s 368,155 l s /F a /F Initial length of wires W S S increase in length of aluminum wire due to temperature increase T a (T) For no load in the aluminum wire: 1 3 s ( T) W E s a ( T ) s or W T E s s ( a s ) Sustitute numerical values: T W E s s 750 l ()(368,155 l)( F) 185F NOTE: If the temperature increase is larger than T, the aluminum wire would e in compression, which is not possile. Therefore, the steel wires continue to carry all of the load. If the temperature increase is less than T, the aluminum wire will e in tension and carry part of the load. W W 1 increase in length of a steel wire due to temperature increase T s (T)