1 Tension, Compression, and Shear

Transcription

1 1 Tension, ompression, and Shear Normal Stress and Strain 1 roblem 1.-1 solid circular post (see figure) supports a load lb acting at the top. second load is uniformly distributed around the shelf at. The diameters of the upper and lower parts of the post are d 1.5 in. and d.5 in., respectively. (a) alculate the normal stress in the upper part of the post. (b) If it is desired that the lower part of the post have the same compressive stress as the upper part, what should be the magnitude of the load? d d Solution lb d 1.5 in. d.5 in. (a) NORML STRESS IN RT s 1 ircular post in compression 500 lb 4 (1.5 in.) (b) LOD FOR EQUL STRESSES s lb 4 (.5 in.) 040 psi Solve for : 5600 lb 040 psi 1 LTERNTE SOLUTION s 1 1 s 1 1 s s 4 d 1 1 d d d 1.8 d 4 d or 1 d lb 1 R d 1

2 HTER 1 Tension, ompression, and Shear roblem 1.- alculate the compressive stress c in the circular piston rod (see figure) when a force 40 N is applied to the brake pedal. ssume that the line of action of the force is parallel to the piston rod, which has diameter 5 mm. lso, the other dimensions shown in the figure (50 mm and 5 mm) are measured perpendicular to the line of action of the force. 5 mm iston rod 50 mm 5 mm = 40 N Solution 1.- Free-body diagram of brake pedal F 50 mm 5 mm = 40 N EQUILIRIUM OF RKE EDL M 0 F F(50 mm) (75 mm) 0 75 mm 50 mm (40 N) 75 0 N 50 F compressive force in piston rod d diameter of piston rod 5mm OMRESSIVE STRESS IN ISTON ROD (d 5 mm) s c F 0 N 11. Ma 4 (5 mm) roblem 1.-3 steel rod 110 ft long hangs inside a tall tower and holds a 00-pound weight at its lower end (see figure). If the diameter of the circular rod is 1 4 inch, calculate the maximum normal stress max in the rod, taking into account the weight of the rod itself. (Obtain the weight density of steel from Table H-1, ppendix H.) 1 in ft 00 lb

3 SETION 1. Normal Stress and Strain 3 Solution 1.-3 Long steel rod in tension d L 00 lb L 110 ft d 1 4 in. Weight density: 490 lb/ft 3 W Weight of rod (Volume) L 00 lb s max W gl gl (490 lbft 3 )(110 1 ft ft) 144 in psi 00 lb 4 (0.5 in.) 4074 psi max 374 psi 4074 psi 4448 psi Rounding, we get max 4450 psi = 00 lb roblem 1.-4 circular aluminum tube of length L 400 mm is loaded in compression by forces (see figure). The outside and inside diameters are 60 mm and 50 mm, respectively. strain gage is placed on the outside of the bar to measure normal strains in the longitudinal direction. Strain gage L = 400 mm (a) If the measured strain is , what is the shortening of the bar? (b) If the compressive stress in the bar is intended to be 40 Ma, what should be the load? Solution 1.-4 luminum tube in compression Strain gage e L 400 mm d 60 mm d 1 50 mm (a) SHORTENING OF THE R el ( )(400 mm) 0.0 mm (b) OMRESSIVE LOD 40 Ma 4 [d d 1 ] 4 [(60 mm) (50 mm) ] mm (40 Ma)(863.9 mm ) 34.6 kn

4 4 HTER 1 Tension, ompression, and Shear roblem 1.-5 The cross section of a concrete pier that is loaded uniformly in compression is shown in the figure. (a) Determine the average compressive stress c in the concrete if the load is equal to 500 k. (b) Determine the coordinates x and y of the point where the resultant load must act in order to produce uniform normal stress. 48 in. y 0 in. 16 in. 16 in. 16 in. O 0 in. 16 in. x Solution 1.-5 oncrete pier in compression y (a) VERGE OMRESSIVE STRESS c 48 in. O x y 0 in. 16 in. 16 in. 16 in. 16 in. x k s c 500 k 1.70 ksi 147 in. (b) OORDINTES OF ENTROID c From symmetry, y 1 (48 in.) 4 in. USE THE FOLLOWING RES: 1 (48 in.)(0 in.) 960 in. 4 1 (16 in.)(16 in.) 18 in. 3 (16 in.)(16 in.) 56 in ( ) in. 147 in. x x i i (see hapter 1, Eq. 1 7a) x 1 (x 1 1 x x 3 3 ) in. [(10 in.)(960 in. ) (5.333 in.)(18 in. ) (8 in.)(56 in. )] 15.8 in. roblem 1.-6 car weighing 130 kn when fully loaded is pulled slowly up a steep inclined track by a steel cable (see figure). The cable has an effective cross-sectional area of 490 mm, and the angle of the incline is 30. alculate the tensile stress t in the cable. able

5 SETION 1. Normal Stress and Strain 5 Solution 1.-6 ar on inclined track FREE-ODY DIGRM OF R W W Weight of car T Tensile force in cable R 1 R ngle of incline Effective area of cable R 1, R Wheel reactions (no friction force between wheels and rails) EQUILIRIUM IN THE INLINED DIRETION F T 0Q b T W sin 0 T W sin TENSILE STRESS IN THE LE s t T W sin SUSTITUTE NUMERIL VLUES: W 130 kn mm (130 kn)(sin 30) s t 490 mm 133 Ma roblem 1.-7 Two steel wires, and, support a lamp weighing 18 lb (see figure). Wire is at an angle 34 to the horizontal and wire is at an angle 48. oth wires have diameter 30 mils. (Wire diameters are often expressed in mils; one mil equals in.) Determine the tensile stresses and in the two wires. Solution 1.-7 FREE-ODY DIGRM OF OINT 0 y T Two steel wires supporting a lamp W = 18 lb EQUTIONS OF EQUILIRIUM x T F x 0 T cos T cos 0 F y 0 T sin T sin W d 30 mils in. d in. SUSTITUTE NUMERIL VLUES: T (0.8904) T ( ) 0 T ( ) T ( ) 18 0 SOLVE THE EQUTIONS: T lb T lb TENSILE STRESSES IN THE WIRES s T 17,00 psi s T 1,300 psi

6 6 HTER 1 Tension, ompression, and Shear roblem 1.-8 long retaining wall is braced by wood shores set at an angle of 30 and supported by concrete thrust blocks, as shown in the first part of the figure. The shores are evenly spaced, 3 m apart. For analysis purposes, the wall and shores are idealized as shown in the second part of the figure. Note that the base of the wall and both ends of the shores are assumed to be pinned. The pressure of the soil against the wall is assumed to be triangularly distributed, and the resultant force acting on a 3-meter length of the wall is F 190 kn. If each shore has a 150 mm 150 mm square cross section, what is the compressive stress c in the shores? Soil Retaining wall oncrete Shore thrust block F m 0.5 m 4.0 m Solution 1.-8 Retaining wall braced by wood shores F 1.5 m Wall m 4.0 m Shore F 190 kn area of one shore (150 mm)(150 mm),500 mm 0.05 m FREE-ODY DIGRM OF WLL ND SHORE SUMMTION OF MOMENTS OUT OINT F 1.5 m V 30 H V H 30 M 0 F(1.5 m) V (4.0 m) H (0.5 m) 0 or (190 kn)(1.5 m) (sin 30)(4.0 m) (cos 30)(0.5 m) kn compressive force in wood shore H horizontal component of V vertical component of H cos 30 V sin 30 OMRESSIVE STRESS IN THE SHORES s c kn 0.05 m 5.1 Ma

7 SETION 1. Normal Stress and Strain 7 roblem 1.-9 loading crane consisting of a steel girder supported by a cable D is subjected to a load (see figure). The cable has an effective cross-sectional area in. The dimensions of the crane are H 9 ft, L 1 1 ft, and L 4 ft. (a) If the load 9000 lb, what is the average tensile stress in the cable? (b) If the cable stretches by 0.38 in., what is the average strain? H D Girder able L 1 L Solution 1.-9 Loading crane with girder and cable H D EQUILIRIUM M 0 T V (1 ft) (9000 lb)(16 ft) 0 T V 1,000 lb T H L 1 1 ft T V H 9 ft T H T V 1 9 H 9ft L 1 1 ft L 4ft effective area of cable in lb FREE-ODY DIGRM OF GIRDER T T H L 1 L = 9000 lb T V 1 ft 4 ft = 9000 lb T tensile force in cable 9000 lb T H (1,000 lb) ,000 lb TENSILE FORE IN LE T T H T V (16,000 lb) (1,000 lb) 0,000 lb (a) VERGE TENSILE STRESS IN LE s T (b) VERGE STRIN IN LE L length of cablel H L 1 15 ft stretch of cable 0.38 in. e L 0,000 lb 4,500 psi in in. (15 ft)(1 in.ft)

8 8 HTER 1 Tension, ompression, and Shear roblem Solve the preceding problem if the load 3 kn; the cable has effective cross-sectional area 481 mm ; the dimensions of the crane are H 1.6 m, L m, and L 1.5 m; and the cable stretches by 5.1 mm. Figure is with rob Solution D Loading crane with girder and cable H H 1.6 m L 1.5 m L m effective area of cable 481 mm 3 kn L 1 L = 3 kn FREE-ODY DIGRM OF GIRDER TENSILE FORE IN LE T T V T = tensile force in cable T T H T V (90 kn) (48 kn) T H 10 kn EQUILIRIUM M 0 T V (3.0 m) (3 kn)(4.5 m) 0 T V 48 kn T H T V L 1 H T H T V T H (48 kn) kn 3.0 m 1.6 m 3.0 m 1.5 m = 3 kn (a) VERGE TENSILE STRESS IN LE s T 10 kn 1 Ma 481 mm (b) VERGE STRIN IN LE L length of cable L H L m stretch of cable 5.1 mm e 5.1 mm L 3.4 m roblem reinforced concrete slab 8.0 ft square and 9.0 in. thick is lifted by four cables attached to the corners, as shown in the figure. The cables are attached to a hook at a point 5.0 ft above the top of the slab. Each cable has an effective cross-sectional area 0.1 in. Determine the tensile stress t in the cables due to the weight of the concrete slab. (See Table H-1, ppendix H, for the weight density of reinforced concrete.) ables Reinforced concrete slab

9 SETION 1. Normal Stress and Strain 9 Solution W able Reinforced concrete slab supported by four cables H t T tensile force in a cable able : T V T H L H T V T H L (Eq. 1) L H height of hook above slab L length of side of square slab t thickness of slab weight density of reinforced concrete W weight of slab L t D length of diagonal of slab L DIMENSIONS OF LE L D W = L FREE-ODY DIGRM OF HOOK T OINT L H L length of cable H L T H T V EQUILIRIUM F vert 0 W 4T V 0 T V W 4 OMINE EQS. (1) & (): H T H L W 4 T W 4 TENSILE STRESS IN LE (Eq. ) H L W H 4 1 L H effective cross-sectional area of a cable s t T W 4 1 L H SUSTITUTE NUMERIL VLUES ND OTIN t : H 5.0 ft L 8.0 ft t 9.0 in ft 150 lb/ft in. W L t 700 lb s t W 4 1 L H,600 psi T T T T T roblem 1.-1 round bar of length L (see figure) rotates about an axis through the midpoint with constant angular speed (radians per second). The material of the bar has weight density. (a) Derive a formula for the tensile stress x in the bar as a function of the distance x from the midpoint. (b) What is the maximum tensile stress max? L x L

10 10 HTER 1 Tension, ompression, and Shear Solution 1.-1 x L Rotating ar dm d onsider an element of mass dm at distance from the midpoint. The variable ranges from x to L. dm g g dj df Inertia force (centrifugal force) of element of mass dm angular speed (rad/s) cross-sectional area weight density g g mass density We wish to find the axial force F x in the bar at Section D, distance x from the midpoint. The force F x equals the inertia force of the part of the rotating bar from D to. df (dm)(j ) g g jdj L g F x df g jdj g g (L x ) D (a) TENSILE STRESS IN R T DISTNE x s x F x g g (L x ) (b) MXIMUM TENSILE STRESS x 0s max g L x g Mechanical roperties and Stress-Strain Diagrams roblem Imagine that a long steel wire hangs vertically from a high-altitude balloon. (a) What is the greatest length (feet) it can have without yielding if the steel yields at 40 ksi? (b) If the same wire hangs from a ship at sea, what is the greatest length? (Obtain the weight densities of steel and sea water from Table H-1, ppendix H.) Solution Hanging wire of length L W total weight of steel wire S weight density of steel L 490 lb/ft 3 W weight density of sea water 63.8 lb/ft 3 cross-sectional area of wire max 40 psi (yield strength) (a) WIRE HNGING IN IR W S L s max W g SL L max s max g S (b) WIRE HNGING IN SE WTER F tensile force at top of wire F (g S g W ) Ls max F (g S g W )L L max g S g W 11,800 ft s max 40,000 psi ( )lbft 3 (144 in. ft ) 13,500 ft 40,000 psi 490 lbft (144 3 in. ft )

11 SETION 1.3 Mechanical roperties and Stress-Strain Diagrams 11 roblem 1.3- Imagine that a long wire of tungsten hangs vertically from a high-altitude balloon. (a) What is the greatest length (meters) it can have without breaking if the ultimate strength (or breaking strength) is 1500 Ma? (b) If the same wire hangs from a ship at sea, what is the greatest length? (Obtain the weight densities of tungsten and sea water from Table H-1, ppendix H.) Solution 1.3- Hanging wire of length L W total weight of tungsten wire T weight density of tungsten L 190 kn/m 3 W weight density of sea water 10.0 kn/m 3 cross-sectional area of wire max 1500 Ma (breaking strength) (a) WIRE HNGING IN IR W T L (b) WIRE HNGING IN SE WTER F tensile force at top of wire F ( T W )L s max F (g T g W )L s max L max g T g W 1500 Ma ( ) knm m s max W g TL L max s max g T 1500 Ma 190 knm m roblem Three different materials, designated,, and, are tested in tension using test specimens having diameters of in. and gage lengths of.0 in. (see figure). t failure, the distances between the gage marks are found to be.13,.48, and.78 in., respectively. lso, at the failure cross sections the diameters are found to be 0.484, 0.398, and 0.53 in., respectively. Determine the percent elongation and percent reduction in area of each specimen, and then, using your own judgment, classify each material as brittle or ductile. Gage length

12 1 HTER 1 Tension, ompression, and Shear Solution Tensile tests of three materials in ercent reduction in area (100).0 in ercent elongation L 1 L 0 (100) L L 0 L 0 L 0.0 in. ercent elongation L 1 1 (100) (Eq. 1).0 where L 1 is in inches. d 0 initial diameter 1 d 1 d 0 d in. 0 ercent reduction in area d R (100) where d 1 is in inches (100) d 1 final diameter (Eq. ) L 1 d 1 % Elongation % Reduction rittle or Material (in.) (in.) (Eq. 1) (Eq. ) Ductile? % 8.1% rittle % 37.9% Ductile % 74.9% Ductile roblem The strength-to-weight ratio of a structural material is defined as its load-carrying capacity divided by its weight. For materials in tension, we may use a characteristic tensile stress (as obtained from a stress-strain curve) as a measure of strength. For instance, either the yield stress or the ultimate stress could be used, depending upon the particular application. Thus, the strength-to-weight ratio R S/W for a material in tension is defined as R S/W in which is the characteristic stress and is the weight density. Note that the ratio has units of length. Using the ultimate stress U as the strength parameter, calculate the strength-to-weight ratio (in units of meters) for each of the following materials: aluminum alloy 6061-T6, Douglas fir (in bending), nylon, structural steel STM-57, and a titanium alloy. (Obtain the material properties from Tables H-1 and H-3 of ppendix H. When a range of values is given in a table, use the average value.) Solution Strength-to-weight ratio The ultimate stress u for each material is obtained from Table H-3, ppendix H, and the weight density is obtained from Table H-1. The strength-to-weight ratio (meters) is R SW s u(ma) g(knm 3 ) (103 ) Values of u,, and R S/W are listed in the table. u R S/W (Ma) (kn/m 3 ) (m) luminum alloy T6 Douglas fir Nylon Structural steel STM-57 Titanium alloy Titanium has a high strength-to-weight ratio, which is why it is used in space vehicles and high-performance airplanes. luminum is higher than steel, which makes it desirable for commercial aircraft. Some woods are also higher than steel, and nylon is about the same as steel.

13 SETION 1.3 Mechanical roperties and Stress-Strain Diagrams 13 roblem symmetrical framework consisting of three pinconnected bars is loaded by a force (see figure). The angle between the inclined bars and the horizontal is 48. The axial strain in the middle bar is measured as Determine the tensile stress in the outer bars if they are constructed of aluminum alloy having the stress-strain diagram shown in Fig (Express the stress in USS units.) D Solution Symmetrical framework L length of bar D L 1 distance L cot L(cot 48) L L length of bar D D luminum alloy 48 e D Use stress-strain diagram of Figure 1-13 L csc L(csc 48) L Elongation of bar D distance DE e D L e D L L L 3 distance E L 3 L 1 (L e D L) (0.9004L) L ( ) L elongation of bar D L D L D E L L3 L 3 L L Strain in bar D L L L From the stress-strain diagram of Figure 1-13: s 31 ksi

14 14 HTER 1 Tension, ompression, and Shear roblem specimen of a methacrylate plastic is tested in tension at room temperature (see figure), producing the stress-strain data listed in the accompanying table. lot the stress-strain curve and determine the proportional limit, modulus of elasticity (i.e., the slope of the initial part of the stress-strain curve), and yield stress at 0.% offset. Is the material ductile or brittle? STRESS-STRIN DT FOR ROLEM Stress (Ma) Strain Fracture Solution Tensile test of a plastic Using the stress-strain data given in the problem statement, plot the stress-strain curve: Stress (Ma) L slope 0.% offset 40 Ma =.4 Ga L proportional limit L 47 Ma Modulus of elasticity (slope).4 Ga Y yield stress at 0.% offset Y 53 Ma Material is brittle, because the strain after the proportional limit is exceeded is relatively small Strain roblem The data shown in the accompanying table were obtained from a tensile test of high-strength steel. The test specimen had a diameter of in. and a gage length of.00 in. (see figure for rob ). t fracture, the elongation between the gage marks was 0.1 in. and the minimum diameter was 0.4 in. lot the conventional stress-strain curve for the steel and determine the proportional limit, modulus of elasticity (i.e., the slope of the initial part of the stress-strain curve), yield stress at 0.1% offset, ultimate stress, percent elongation in.00 in., and percent reduction in area. TENSILE-TEST DT FOR ROLEM Load (lb) Elongation (in.) 1, , , , , , , , , , , , , , , , ,600 Fracture

15 SETION 1.3 Mechanical roperties and Stress-Strain Diagrams 15 Solution Tensile test of high-strength steel d in. L 0.00 in. 0 d in. 4 ENLRGEMENT OF RT OF THE STRESS-STRIN URVE Stress (psi) ONVENTIONL STRESS ND STRIN s 0 e L 0 70,000 L 69,000 psi (0.1% offset) L 65,000 psi Load Elongation Stress (lb) (in.) (psi) Strain e 1, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,600 Fracture 113,000 STRESS-STRIN DIGRM 150,000 Stress (psi) 100,000 60,000 50,000 RESULTS Strain 0.1% pffset 50,000 psi Slope psi roportional limit 65,000 psi Modulus of elasticity (slope) psi Yield stress at 0.1% offset 69,000 psi Ultimate stress (maximum stress) 113,000 psi ercent elongation in.00 in. L 1 L 0 L 0 (100) 0.1 in. (100) 6%.00 in. ercent reduction in area (100) 0.00 in. 4 (0.4 in.) 0.00 in. (100) 31% 50, Strain

16 16 HTER 1 Tension, ompression, and Shear Elasticity and lasticity roblem bar made of structural steel having the stressstrain diagram shown in the figure has a length of 48 in. The yield stress of the steel is 4 ksi and the slope of the initial linear part of the stress-strain curve (modulus of elasticity) is ksi. The bar is loaded axially until it elongates 0.0 in., and then the load is removed. How does the final length of the bar compare with its original length? (Hint: Use the concepts illustrated in Fig. 1-18b.) (ksi) Solution Steel bar in tension ELSTI REOVERY e E s e E Slope 4 ksi ksi E 0 R L 48 in. Yield stress Y 4 ksi Slope ksi 0.0 in. STRESS ND STRIN T OINT RESIDUL STRIN e R e R e e E ERMNENT SET e R L (0.0077)(48 in.) 0.13 in. Final length of bar is 0.13 in. greater than its original length. Y 4 ksi e 0.0 in L 48 in. roblem 1.4- bar of length.0 m is made of a structural steel having the stress-strain diagram shown in the figure. The yield stress of the steel is 50 Ma and the slope of the initial linear part of the stress-strain curve (modulus of elasticity) is 00 Ga. The bar is loaded axially until it elongates 6.5 mm, and then the load is removed. How does the final length of the bar compare with its original length? (Hint: Use the concepts illustrated in Fig. 1-18b.) (Ma)

17 SETION 1.4 Elasticity and lasticity 17 Solution 1.4- Steel bar in tension E 0 R STRESS ND STRIN T OINT Y 50 Ma e 6.5 mm L 000 mm L.0 m 000 mm Yield stress Y 50 Ma Slope 00 Ga 6.5 mm ELSTI REOVERY e E s e E Slope RESIDUL STRIN e R 50 Ma Ga e R e e E ermanent set e R L (0.0000)(000 mm) 4.0 mm Final length of bar is 4.0 mm greater than its original length. roblem n aluminum bar has length L 4 ft and diameter d 1.0 in. The stress-strain curve for the aluminum is shown in Fig of Section 1.3. The initial straight-line part of the curve has a slope (modulus of elasticity) of psi. The bar is loaded by tensile forces 4 k and then unloaded. (a) What is the permanent set of the bar? (b) If the bar is reloaded, what is the proportional limit? (Hint: Use the concepts illustrated in Figs. 1-18b and 1-19.) Solution luminum bar in tension STRESS ND STRIN T OINT s 4 k 31 ksi 4 (1.0 in.) From Fig. 1-13: e 0.04 E 0 R L 4ft 48 in. d 1.0 in. 4 k See Fig for stress-strain diagram Slope from O to is psi. ELSTI REOVERY e E s e E Slope 31 ksi psi RESIDUL STRIN e R e R e e E (Note: the accuracy in this result is very poor because e is approximate.) (a) ERMNENT SET e R L (0.037)(48 in.) 1.8 in. (b) ROORTIONL LIMIT WHEN RELODED 31 ksi

18 18 HTER 1 Tension, ompression, and Shear roblem circular bar of magnesium alloy is 800 mm long. The stress-strain diagram for the material is shown in the figure. The bar is loaded in tension to an elongation of 5.6 mm, and then the load is removed. (a) What is the permanent set of the bar? (b) If the bar is reloaded, what is the proportional limit? (Hint: Use the concepts illustrated in Figs. 1-18b and 1-19.) 00 (Ma) Solution ( L ) Magnesium bar in tension Slope (s pl) 1 e 88 Ma 44 Ga 0.00 ( L ) 1 E 0 R L 800 mm 5.6 mm ( L ) 1 initial proportional limit 88 Ma (from stress-strain diagram) ( L ) proportional limit when the bar is reloaded INITIL SLOE OF STRESS-STRIN URVE From e diagram: t point : ( L ) 1 88 Ma e 0.00 STRESS ND STRIN T OINT e 5.6 mm L 800 mm From e diagram: ( L ) 170 Ma ELSTI REOVERY e E e E s Slope (s L) Slope RESIDUL STRIN e R e R e e E (a) ERMNENT SET e R L ( )(800 mm).51 mm (b) ROORTIONL LIMIT WHEN RELODED ( L ) 170 Ma 170 Ma Ga roblem wire of length L 4 ft and diameter d 0.15 in. is stretched by tensile forces 600 lb. The wire is made of a copper alloy having a stress-strain relationship that may be described mathematically by the following equation: 18, ( ksi) (a) onstruct a stress-strain diagram for the material. (b) Determine the elongation of the wire due to the forces. (c) If the forces are removed, what is the permanent set of the bar? (d) If the forces are applied again, what is the proportional limit? in which is nondimensional and has units of kips per square inch (ksi).

19 SETION 1.5 Hooke s Law and oisson s Ratio 19 Solution Wire stretched by forces L 4ft48 in. d 0.15 in. 600 lb OER LLOY s 18,000e 0 e 0.03 (s ksi) 1 300e (a) STRESS-STRIN DIGRM (From Eq. 1) (ksi) = 54 ksi E = R R INITIL SLOE OF STRESS-STRIN URVE Take the derivative of with respect to e: (Eq. 1) ds (1 300e)(18,000) (18,000e)(300) de (1 300e) 18,000 (1 300e) t e 0, ds 18,000 ksi de Initial slope18,000 ksi LTERNTIVE FORM OF THE STRESS-STRIN RELTIONSHI Solve Eq. (1) for e in terms of : s e 0 s 54 ksi(s ksi) 18, s This equation may also be used when plotting the stress-strain diagram. (b) ELONGTION OF THE WIRE s 600 lb 48,900 psi 48.9 ksi 4 (0.15 in.) From Eq. () or from the stress-strain diagram: e el (0.0147)(48 in.) 0.71 in. STRESS ND STRIN T OINT (see diagram) 48.9 ksi e ELSTI REOVERY e E s e E Slope RESIDUL STRIN e R e R e e E (c) ermanent set e R L (0.010)(48 in.) 0.58 in. (d) roportional limit when reloaded 49 ksi 48.9 ksi ,000 ksi (Eq. ) Hooke s Law and oisson s Ratio When solving the problems for Section 1.5, assume that the material behaves linearly elastically. roblem high-strength steel bar used in a large crane has diameter d.00 in. (see figure). The steel has modulus of elasticity E psi and oisson s ratio 0.9. ecause of clearance requirements, the diameter of the bar is limited to.001 in. when it is compressed by axial forces. What is the largest compressive load max that is permitted? d

20 0 HTER 1 Tension, ompression, and Shear Solution Steel bar in compression STEEL r d.00 in. Max. d in. LTERL STRIN XIL STRIN E psi e d in d.00 in. e e n (shortening) 0.9 XIL STRESS Ee ( psi)( ) ksi (compression) ssume that the yield stress for the high-strength steel is greater than 50 ksi. Therefore, Hooke s law is valid. MXIMUM OMRESSIVE LOD max s (50.00 ksi) (.00 in.) k roblem 1.5- round bar of 10 mm diameter is made of aluminum alloy 7075-T6 (see figure). When the bar is stretched by axial forces, its diameter decreases by mm. Find the magnitude of the load. (Obtain the material properties from ppendix H.) 7075-T6 d = 10 mm Solution 1.5- luminum bar in tension d 10 mm d mm (Decrease in diameter) 7075-T6 From Table H-: E 7 Ga 0.33 From Table H-3: Yield stress Y 480 Ma XIL STRESS Ee (7 Ga)( ) Ma (Tension) ecause < Y, Hooke s law is valid. LOD (TENSILE FORE) LTERL STRIN e d mm d 10 mm s (349.1 Ma) (10 mm) kn XIL STRIN e e n (Elongation) roblem nylon bar having diameter d in. is placed inside a steel tube having inner diameter d 3.51 in. (see figure). The nylon bar is then compressed by an axial force. t what value of the force will the space between the nylon bar and the steel tube be closed? (For nylon, assume E 400 ksi and 0.4.) Steel tube Nylon bar d 1 d

21 SETION 1.5 Hooke s Law and oisson s Ratio 1 Solution Nylon bar inside steel tube d 1 d XIL STRIN e e n (Shortening) ompression d in. d 3.51 in. Nylon: E 400 ksi LTERL STRIN e d 1 d in. e in. d in. 0.4 (Increase in diameter) XIL STRESS Ee (400 ksi)( ).857 ksi (ompressive stress) ssume that the yield stress is greater than and Hooke s law is valid. FORE (OMRESSION) s (.857 ksi) (3.50 in.) k roblem prismatic bar of circular cross section is loaded by tensile forces (see figure). The bar has length L 1.5 m and diameter d 30 mm. It is made of aluminum alloy with modulus of elasticity E 75 Ga and oisson s ratio 1 3. If the bar elongates by 3.6 mm, what is the decrease in diameter d? What is the magnitude of the load? L d Solution L 1.5 m luminum bar in tension d 30 mm DERESE IN DIMETER E 75 Ga mm (elongation) XIL STRIN e L 3.6 mm m LTERL STRIN e ne ( 1 3)(0.004) (Minus means decrease in diameter) d ed (0.0008)(30 mm) 0.04 mm XIL STRESS Ee (75 Ga)(0.004) 180 Ma (This stress is less than the yield stress, so Hooke s law is valid.) LOD (TENSION) s (180 Ma) (30 mm) 4 17 kn

22 HTER 1 Tension, ompression, and Shear roblem bar of monel metal (length L 8 in., diameter d 0.5 in.) is loaded axially by a tensile force 1500 lb (see figure). Using the data in Table H-, ppendix H, determine the increase in length of the bar and the percent decrease in its cross-sectional area. Solution ar of monel metal in tension L 8 in. d 0.5 in lb From Table H-: E 5,000 ksi 0.3 XIL STRESS s 30,560 psi 1500 lb 4 (0.5 in.) ssume is less than the proportional limit, so that Hooke s law is valid. XIL STRIN e s E 30,560 psi ,000 ksi INRESE IN LENGTH e L (0.001)(8 in.) in. LTERL STRIN e ne (0.3)(0.001) DERESE IN DIMETER d e d ( )(0.5 in.) in. DERESE IN ROSS-SETIONL RE Original area: 0 d 4 Final area: 1 (d d) [d d d ( d) ] Decrease in area: 0 1 ( d)(d d) 4 ERENT DERESE IN RE ercent 0 ( )(0.4999) (0.5) (100) 0.078% (100) ( d)(d d) d (100) roblem tensile test is peformed on a brass specimen 10 mm in diameter using a gage length of 50 mm (see figure). When the tensile load reaches a value of 0 kn, the distance between the gage marks has increased by 0.1 mm. (a) What is the modulus of elasticity E of the brass? (b) If the diameter decreases by mm, what is oisson s ratio? 10 mm 50 mm

23 SETION 1.5 Hooke s Law and oisson s Ratio 3 Solution rass specimen in tension d 10 mm Gage length L 50 mm 0 kn 0.1 mm d mm XIL STRESS s 0 kn 54.6 Ma 4 (10 mm) ssume is below the proportional limit so that Hooke s law is valid. XIL STRIN e L 0.1 mm mm (a) MODULUS OF ELSTIITY E s e (b) OISSON S RTIO 54.6 Ma 104 Ga ee d ed ed n d ed mm 0.34 ( )(10 mm) roblem hollow steel cylinder is compressed by a force (see figure). The cylinder has inner diameter d in., outer diameter d 4.5 in., and modulus of elasticity E 30,000 ksi. When the force increases from zero to 40 k, the outer diameter of the cylinder increases by in. (a) Determine the increase in the inner diameter. (b) Determine the increase in the wall thickness. (c) Determine oisson s ratio for the steel. d 1 d Solution Hollow steel cylinder d in. (c) OISSON S RTIO d 4.5 in. t 0.3 in. E 30,000 ksi t d 1 d xial stress: s 4 [d d 1 ] 4 [(4.5 in.) (3.9 in.) ] 40 k (compression) in. d in. (increase) LTERL STRIN e d d in. 4.5 in. (a) INRESE IN INNER DIMETER d 1 e d 1 ( )(3.9 in.) in. s 40 k in ksi (compression) ( Y ; Hooke s law is valid) xial strain: e s E 10,105 ksi 30,000 ksi (b) INRESE IN WLL THIKNESS t e t ( )(0.3 in.) in. n e e

24 4 HTER 1 Tension, ompression, and Shear roblem steel bar of length.5 m with a square cross section 100 mm on each side is subjected to an axial tensile force of 1300 kn (see figure). ssume that E 00 Ga and v 0.3. Determine the increase in volume of the bar kn 100 mm.5 m 100 mm Solution Square bar in tension Find increase in volume. Length: L.5 m 500 mm Side: b 100 mm Force: 1300 kn E 00 Ga 0.3 XIL STRESS s b 1300 kn s 130 Ma (100 mm) Stress is less than the yield stress, so Hooke s law is valid. XIL STRIN e s E 130 Ma 00 Ga DERESE IN SIDE DIMENSION e ne b e b ( )(100 mm) mm FINL DIMENSIONS L 1 L L mm b 1 b b mm FINL VOLUME V 1 L 1 b 1 5,006,490 mm 3 INITIL VOLUME V Lb 5,000,000 mm 3 INRESE IN VOLUME V V 1 V 6490 mm 3 INRESE IN LENGTH L el ( )(500 mm) 1.65 mm

25 SETION 1.6 Shear Stress and Strain 5 Shear Stress and Strain roblem n angle bracket having thickness t 0.5 in. is attached to the flange of a column by two 5 8-inch diameter bolts (see figure). uniformly distributed load acts on the top face of the bracket with a pressure p 300 psi. The top face of the bracket has length L 6 in. and width b.5 in. Determine the average bearing pressure b between the angle bracket and the bolts and the average shear stress aver in the bolts. (Disregard friction between the bracket and the column.) p b L t Solution F ngle bracket bolted to a column b p pressure acting on top of the bracket 300 psi F resultant force acting on the bracket pbl (300 psi) (.5 in.) (6.0 in.) 4.50 k t L ERING RESSURE ETWEEN RKET ND OLTS b bearing area of one bolt dt (0.65 in.) (0.5 in.) in. s b F 4.50 k 7.0 ksi b (0.315 in. ) Two bolts d 0.65 in. t thickness of angle 0.5 in. b.5 in. L 6.0 in. VERGE SHER STRESS IN THE OLTS s Shear area of one bolt 4 d 4 (0.65 in.) in. t aver F 4.50 k 7.33 ksi s ( in. )

26 6 HTER 1 Tension, ompression, and Shear roblem 1.6- Three steel plates, each 16 mm thick, are joined by two 0-mm diameter rivets as shown in the figure. (a) If the load 50 kn, what is the largest bearing stress acting on the rivets? (b) If the ultimate shear stress for the rivets is 180 Ma, what force ult is required to cause the rivets to fail in shear? (Disregard friction between the plates.) / / Solution 1.6- / / Three plates joined by two rivets t s b b dt 78.1 Ma 50 kn (0 mm)(16 mm) (b) ULTIMTE LOD IN SHER Shear force on two rivets t thickness of plates 16 mm d diameter of rivets 0 mm 50 kn ULT 180 Ma (for shear in the rivets) (a) MXIMUM ERING STRESS ON THE RIVETS Maximum stress occurs at the middle plate. b bearing area for one rivet dt Shear force on one rivet 4 Let cross-sectional area of one rivet Shear stress t 4 or, d 4 ) d 4( d t the ultimate load: ULT d t ULT (0 mm) (180 Ma) 6 kn roblem bolted connection between a vertical column and a diagonal brace is shown in the figure. The connection consists of three 5 8-in. bolts that join two 1 4-in. end plates welded to the brace and a 5 8-in. gusset plate welded to the column. The compressive load carried by the brace equals 8.0 k. Determine the following quantities: (a) The average shear stress aver in the bolts, and (b) The average bearing stress b between the gusset plate and the bolts. (Disregard friction between the plates.) olumn race End plates for brace Gusset plate

27 SETION 1.6 Shear Stress and Strain 7 Solution Diagonal brace (a) VERGE SHER STRESS IN THE OLTS End plates cross-sectional area of one bolt d in. 4 V shear force acting on one bolt Gusset plate 3 bolts in double shear compressive force in brace 8.0 k d diameter of bolts 5 8 in in. t 1 thickness of gusset plate 5 8 in in. t thickness of end plates 1 4 in. 0.5 in t aver V k 6( in. ) 4350 psi (b) VERGE ERING STRESS GINST GUSSET LTE b bearing area of one bolt t 1 d (0.65 in.)(0.65 in.) in. F bearing force acting on gusset plate from one bolt 3 s b 8.0 k 6830 psi 3 b 3( in. ) roblem hollow box beam of length L is supported at end by a 0-mm diameter pin that passes through the beam and its supporting pedestals (see figure). The roller support at is located at distance L/3 from end. ox beam (a) Determine the average shear stress in the pin due to a load equal to 10 kn. (b) Determine the average bearing stress between the pin and the box beam if the wall thickness of the beam is equal to 1 mm. L L 3 3 ox beam in at support

28 8 HTER 1 Tension, ompression, and Shear Solution Hollow box beam 10 kn d diameter of pin 0 mm t Wall thickness of box beam 1 mm R = L L 3 3 (a) VERGE SHER STRESS IN IN Double shear t aver 4 d Ma d R = R = (b) VERGE ERING STRESS ON IN s b (dt) 41.7 Ma dt roblem The connection shown in the figure consists of five steel plates, each 3 16 in. thick, joined by a single 1 4-in. diameter bolt. The total load transferred between the plates is 100 lb, distributed among the plates as shown. (a) alculate the largest shear stress in the bolt, disregarding friction between the plates. (b) alculate the largest bearing stress acting against the bolt. 360 lb 480 lb 360 lb 600 lb 600 lb Solution lates joined by a bolt d diameter of bolt 1 4 in. t thickness of plates 3 16 in. 360 lb 480 lb 360 lb Free-body diagram of bolt Section : V 360 lb Section : V 40 lb V max max. shear force in bolt 360 lb 600 lb 600 lb (a) MXIMUM SHER STRESS IN OLT t max V max d 4 (b) MXIMUM ERING STRESS F max maximum force applied by a plate against the bolt F max 600 lb s b F max dt 4V max 7330 psi d 1,800 psi

29 SETION 1.6 Shear Stress and Strain 9 roblem steel plate of dimensions m is hoisted by a cable sling that has a clevis at each end (see figure). The pins through the clevises are 18 mm in diameter and are located.0 m apart. Each half of the cable is at an angle of 3 to the vertical. For these conditions, determine the average shear stress aver in the pins and the average bearing stress b between the steel plate and the pins. 3 3 able sling levis.0 m Steel plate ( m) Solution Steel plate hoisted by a sling Dimensions of plate: m Volume of plate: V (.5) (1.) (0.1) m m 3 Weight density of steel: 77.0 kn/m 3 Weight of plate: W V 3.10 kn d diameter of pin through clevis 18 mm t thickness of plate 0.1 m 100 mm FREE-ODY DIGRMS OF SLING ND IN TENSILE FORE T IN LE F vertical 0 T cos 3 W 0 W T cos kn 13.6 kn cos 3 SHER STRESS IN THE INS (DOULE SHER) = W T 3 t aver T 13.6 kn pin ( 4 )(18 mm) able 3 3 H in W 6.8 Ma ERING STRESS ETWEEN LTE ND INS b bearing area td s b T td 13.6 kn (100 mm)(18 mm) 7.57 Ma W H.0 m H W

30 30 HTER 1 Tension, ompression, and Shear roblem special-purpose bolt of shank diameter d 0.50 in. passes through a hole in a steel plate (see figure). The hexagonal head of the bolt bears directly against the steel plate. The radius of the circumscribed circle for the hexagon is r 0.40 in. (which means that each side of the hexagon has length 0.40 in.). lso, the thickness t of the bolt head is 0.5 in. and the tensile force in the bolt is 1000 lb. r Steel plate d (a) Determine the average bearing stress b between the hexagonal head of the bolt and the plate. (b) Determine the average shear stress aver in the head of the bolt. t Solution olt in tension d 0.50 in. (a) ERING STRESS ETWEEN OLT HED ND LTE r t d r r 0.40 in. t 0.5 in lb rea of one equilateral triangle r 3 4 b bearing area b area of hexagon minus area of bolt 3r 3 d 4 3 b (0.40 in.) ( 3) (0.50 in.) in in in. s b 1000 lb 4560 psi b in. rea of hexagon (b) SHER STRESS IN HED OF OLT r 3r 3 s shear area s dt t aver s dt 1000 lb (0.50 in.)(0.5 in.) 550 psi roblem n elastomeric bearing pad consisting of two steel plates bonded to a chloroprene elastomer (an artificial rubber) is subjected to a shear force V during a static loading test (see figure). The pad has dimensions a 150 mm and b 50 mm, and the elastomer has thickness t 50 mm. When the force V equals 1 kn, the top plate is found to have displaced laterally 8.0 mm with respect to the bottom plate. What is the shear modulus of elasticity G of the chloroprene? t a V b

31 SETION 1.6 Shear Stress and Strain 31 Solution earing pad subjected to shear d = 8.0 mm V t = 50 mm b = 50 mm V 1 kn Width of pad: a 150 mm Length of pad: b 50 mm d 8.0 mm t aver V ab g aver d t G t g 1 kn 0.3 Ma (150 mm)(50 mm) 8.0 mm mm 0.3 Ma.0 Ma 0.16 roblem joint between two concrete slabs and is filled with a flexible epoxy that bonds securely to the concrete (see figure). The height of the joint is h 4.0 in., its length is L 40 in., and its thickness is t 0.5 in. Under the action of shear forces V, the slabs displace vertically through the distance d 0.00 in. relative to each other. (a) What is the average shear strain aver in the epoxy? (b) What is the magnitude of the forces V if the shear modulus of elasticity G for the epoxy is 140 ksi? h t L d h V V t Solution Epoxy Joint between concrete slabs V V d h (a) VERGE SHER STRIN g aver d t t (b) SHER FORES V h 4.0 in. t 0.5 in. L 40 in. d 0.00 in. G 140 ksi verage shear stress : aver G aver V aver (hl) G aver (hl) (140 ksi)(0.004)(4.0 in.)(40 in.) 89.6 k