2 Axially Loaded Numbers

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Tyrone Townsend
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1 xially oaded Numers 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 W 3 elongation of the spring F k W 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 10 Ga. (a) If the cale is 1 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 6 HTER xially oaded Numers Solution.- ridge section lifted y a cale 30 mm (from Tale -1) W 38 kn E 10 Ga 1 m () FTOR OF SFETY UT 06 kn (from Tale -1) max 70 kn n UT max 06 kn kn (a) STRETH OF E W (38 kn)(1 m) E (10 Ga)(30 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 Steel wire and copper wire opper wire Equal lengths and equal loads Steel: E s 30,000 ksi opper: E c 18,000 ksi (a) RTIO OF EONGTIONS (EQU DIMETERS) c E c s E s c E s s E c 18 () RTIO OF DIMETERS (EQU EONGTIONS) c s E c c E s s E c d c E s d s d c d E s d c E s 30 s E c d s E c 18 ore c c E s s 1.9

3 SETION. hanges in engths of xially oaded Memers 65 rolem.- 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 1.6 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.- age supported y a cale 1 d 300 mm d 150 mm 1.6 m 10.5 m E 10,700 kn ENGTH OF E 1 1 (d ) 1 (d ),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

66 HTER xially oaded Numers 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.

-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.

4 66 HTER xially oaded Numers 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 h height of valve (compressed length of the spring) d diameter of discharge hole pressure in tank EQUTE FORES ND SOVE FOR h: F k( h) p max d h p max d k 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 SUSTITUTE NUMERI VUES: 3 x x k kx tan (800 Nm)(150 mm) 8 N 118 mm tan 3

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.

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 1 3 F 3 displacement of point displacement of point 1 elongation of first spring shortening of second spring lso, 1 F 1 k SOVE THE EQUTIONS: 1 1 3k 3k 3k ; F k Eliminate and otain : 0 9k 3k

6 68 HTER xially oaded Numers 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? 5 5 Solution.-8 Truss with horizontal load From force triangle, F (tension) 3m 3900 mm E 00 Ga F F R = 5 5 M 0givesR FREE-ODY DIGRM OF JOINT Force triangle: 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 E max max max max 1.5 mm max (650 kn) 1.5 mm 780 kn

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 3.1 mm MXIMUM OD SED UON EONGTION max 3.0 mm E d max E max (75 Ga)(3.1 mm ) 3.8 m 186 N MXIMUM OD SED UON STRESS s allow 60 Mas max s allow (3.1 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 00 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 Numers 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 x W F 1 1 x W F X W SUSTITUTE NUMERI VUES: W 5 N UNITS: Newtons and meters F 1 (18) 1 x x k N/m k 00 N/m x F (18) x mm h 80 mm 18 N EONGTIONS OF THE SRINGS 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 00 R REMINS HORIZONT oints and are the same distance elow the reference line (see figure aove). 1 1 h or x x x W SOVE FOR x: 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.) [d (d t) ]or d (d t) 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 1.1 in. s allow 7,000 psi (d t) d ord t d t d d or t min d d min SUSTITUTE NUMERI VUES 7.5 in. t min t min 0.63 in. 7.5 in in.

10 7 HTER xially oaded Numers rolem.-1 The horizontal rigid eam D is supported y vertical ars E and F and is loaded y vertical forces 1 00 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 = 00 kn F E.1 m D. 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 = 00 kn. m = 360 kn 0.00 mm F 0.6 m E E 11,100 mm F 9,80 mm E 00 Ga E 3.0 m F. m 1 00 kn; 360 kn FREE-ODY DIGRM OF R D 1.5 m 1.5 m.1 m D 1 = 00 kn F E F F = 360 kn M 0 (00 kn)(1.5 m) F F (1.5 m) (360 kn)(3.6 m) 0 F F 6 kn M 0 (00 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 (6 kn)(. m) (00 Ga)(9,80 mm ) 1.5 m 1.5 m.1 m D E F D E F E or E F (0.00 mm) m 0.00 mm (Downward) D F ( F E ) or D 1 5 F 7 5 E 1 5 (0.600 mm) 7 (0.00 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., 5, and 10 l.) k ros and.-1 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)

12 7 HTER xially oaded Numers 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 cot u NUMERI RESUTS (Eq. 3) 8.0 in. k 16 l/in l Sustitute into Eq. (): cot cos (Eq. ) Solve Eq. () numerically: u 35.1 Sustitute into Eq. (3): 1.78 in. cos cos u cot u k rolem.-1 Solve the preceding prolem for the following data: 00 mm, k 3. kn/m, 5, and 50 N. Solution.-1 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 5 50 N Sustitute into Eq. (): cot cos (Eq. ) Solve Eq. () numerically: u 35.1 Sustitute into Eq. (3):.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 -.) 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 -: E d d (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 (3.0 k)(50 in.) (18,000 ksi)( )(1.0 in.) ( in.) ( in.) 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 800 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 500 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 Numers Solution.3- opper ar with a tensile load Steel post S (a) DOWNWRD DISEMENT ( 180 kn) c c E c c 0.65 mm (180 kn)(.0 m) (10 Ga)(800 mm ) opper ar c.0 m c 800 mm s () s E s s mm c s 0.65 mm mm mm (90 kn)(0.5 m) (00 Ga)(500 mm ) E c 10 Ga () MXIMUM OD max ( max 1.0 mm) s 0.5 m s 500 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 0.0 in. and is loaded y forces l, 1800 l, and l. The lengths of the segments of the ar are a 60 in., 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. in. 36 in. 0.0 in. (a) HNGE IN ENGTH l 1800 l l E psi N i i a E i i XI FORES (t tension) 1 E (N N N D D ) N l 1 [(300 l)(60 in.) N l ( psi)(0.0 in. ) N D l (500 l)( 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)(0.0 in. ) 1310 l rolem.3- 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 /. 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. Solution.3- ar with a slot t thickness (a) EONGTION OF R length of ar N i i a () () () E i E(t) E( 3 t) E(t) Et Et STRESS IN MIDDE REGION s ( 3 t) Sustitute into the equation for : Et 6E t 6E 3s 7s 8E 3 t or t 3s () SUSTITUTE NUMERI VUES: s 160 Ma 750 mme 10 Ga 7(160 Ma)(750 mm) mm 8 (10 Ga)

16 78 HTER xially oaded Numers rolem.3-5 Solve the preceding prolem if the axial stress in the middle region is,000 psi, the length is 30 in., and the modulus of elasticity is psi. Solution.3-5 ar with a slot STRESS IN MIDDE REGION s ( 3 t) 3t or t 3s t thickness (a) EONGTION OF R length of ar N i i a () E i E(t) Et Et () () E ( 3 t) E(t) SUSTITUTE INTO THE EQUTION FOR : Et 6E t 6E 3s 7s 8E () SUSTITUTE NUMERI VUES: s,000 psi 30 in. E psi 7(,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 00 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.0 mm? 1 = 00 kn = 70 kn = 3.75 m = 3.75 m Solution.3-6 Steel columns in a uilding 1 = 00 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 ) (00 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.0 mm 0 additional shortening of the two columns due to the load 0 0 ( ) max.0 mm mm 0.79 mm lso, E E E 1 Solve for 0 : 0 E 0 SUSTITUTE NUMERI VUES: E Nm 3.75 m 11, m 3, m 0,00 N. 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..0 ft.0 ft = 5000 l Solution.3-7 d 1 = 0.75 in l E psi ft8 in. ar in tension d = 0.50 in..0 ft.0 ft (a) EONGTION OF NONRISMTI R N i i a 1 E i i E a i (5000 l)(8 in.) psi 1 (0.75 in) 1 (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 1 1 (d 1 d ) 8 [(0.75 in.) (0.50 in.) ] in. () E p in. (5000 l)()(8 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.

2 Axially Loaded Members

2 Axially Loaded Members 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