122 CHAPTER 2 Axially Loaded Numbers. Stresses on Inclined Sections

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1 1 CHATER Aiall Loaded Numbers Stresses on Inclined Sections roblem.6-1 A steel bar of rectangular cross section (1.5 in..0 in.) carries a tensile load (see figure). The allowable stresses in tension and shear are 15,000 psi and 7,000 psi, respectivel. Determine the maimum permissible load ma. 1.5 in..0 in. Solution.6-1 Rectangular bar in tension.0 in. 1.5 in. A 1.5 in..0 in. 3.0 in. Maimum Normal Stress: Maimum shear stress: t ma s A allow 15,000 psi allow 7,000 psi Because allow is less than one-half of allow, the shear stress governs. ma allow A (7,000 psi) (3.0 in. ) 4,000 lb s A roblem.6- A circular steel rod of diameter d is subjected to a tensile force 3.0 kn (see figure). The allowable stresses in tension and shear are 10 Ma and 50 Ma, respectivel. What is the minimum permissible diameter d min of the rod? d = 3.0 kn Solution.6- Steel rod in tension d 3.0 kna d 4 Maimum normal stress: s A Maimum shear stress: t ma s A allow 10 Ma allow 50 Ma Because allow is less than one-half of allow, the shear stress governs. t ma or50 Ma A 3.0 kn () d 4 Solve for d: d min 6.18 mm

2 SECTION.6 Stresses on Inclined Sections 13 roblem.6-3 A standard brick (dimensions 8 in. 4 in..5 in.) is compressed lengthwise b a force, as shown in the figure. If the ultimate shear stress for brick is 100 psi and the ultimate compressive stress is 3600 psi, what force ma is required to break the brick? 8 in. 4 in..5 in. Solution.6-3 Standard brick in compression Maimum shear stress: 8 in. 4 in..5 in. t ma s A ult 3600 psi ult 100 psi Because ULT is less than one-half of ULT, the shear stress governs. A.5 in. 4.0 in in. Maimum normal stress: s A t ma A or ma At ult ma (10.0 in. )(100 psi) 4,000 lb roblem.6-4 A brass wire of diameter d.4 mm is stretched tightl between rigid supports so that the tensile force is T 9 N (see figure). What is the maimum permissible temperature drop T if the allowable shear stress in the wire is 60 Ma? (The coefficient of thermal epansion for the wire is / C and the modulus of elasticit is 100 Ga.) T d T robs..6-4 and.6-5 Solution.6-4 Brass wire in tension T d T d.4 mm A d mm /C E 100 Ga allow 60 Ma Initial tensile force: T 9 N Stress due to initial tension: s T A Stress due to temperature drop: E(T) (see Eq. -18 of Section.5) Total stress: s T E( T ) A MAXIMUM SHEAR STRESS t ma s 1 B T E( T )R A Solve for temperature drop T: T t ma TA t E ma t allow SUBSTITUTE NUMERICAL VALUES: T (60 Ma) (9 N)(4.60 mm ) (100 Ga)( C) 10 Ma 0 Ma 50C MaC

3 14 CHATER Aiall Loaded Numbers roblem.6-5 A brass wire of diameter d 1/16 in. is stretched between rigid supports with an initial tension T of 3 lb (see figure). (a) If the temperature is lowered b 50 F, what is the maimum shear stress ma in the wire? (b) If the allowable shear stress is 10,000 psi, what is the maimum permissible temperature drop? (Assume that the coefficient of thermal epansion is / F and the modulus of elasticit is psi.) Solution.6-5 Brass wire in tension d 1 16 in. A d 4 T d T in /F E psi Initial tensile force: T 3 lb Stress due to initial tension: s T A Stress due to temperature drop: E(T ) (see Eq. -18 of Section.5) Total stress: s T E( T ) A (a) MAXIMUM SHEAR STRESS WHEN TEMERATURE DROS 50F t ma s 1 B T E( T )R A Substitute numerical values: t ma 9,190 psi (b) MAXIMUM ERMISSIBLE TEMERATURE DRO IF allow 10,000 psi Solve Eq. (1) for T: T t ma TA t E ma t allow Substitute numerical values: T 60.F (Eq. 1) roblem.6-6 A steel bar with diameter d 1 mm is subjected to a tensile load 9.5 kn (see figure). (a) What is the maimum normal stress ma in the bar? (b) What is the maimum shear stress ma? (c) Draw a stress element oriented at 45 to the ais of the bar and show all stresses acting on the faces of this element. d = 1 mm = 9.5 kn

4 SECTION.6 Stresses on Inclined Sections 15 Solution.6-6 Steel bar in tension 9.5 kn d = 1 mm = 9.5 kn (c) STRESS ELEMENT AT 45 = 45 0 (a) MAXIMUM NORMAL STRESS s A 9.5 kn 84.0 Ma 4 (1 mm) NOTE: All stresses have units of Ma. s ma 84.0 Ma (b) MAXIMUM SHEAR STRESS The maimum shear stress is on a 45 plane and equals /. t ma s Ma roblem.6-7 During a tension test of a mild-steel specimen (see figure), the etensometer shows an elongation of in. with a gage length of in. Assume that the steel is stressed below the proportional limit and that the modulus of elasticit E psi. T in. T (a) What is the maimum normal stress ma in the specimen? (b) What is the maimum shear stress ma? (c) Draw a stress element oriented at an angle of 45 to the ais of the bar and show all stresses acting on the faces of this element.

5 16 CHATER Aiall Loaded Numbers Solution.6-7 Tension test T in. T Elongation: in. ( in. gage length) Strain: e L in in. Hooke s law : Ee ( psi)( ) 18,000 psi (a) MAXIMUM NORMAL STRESS is the maimum normal stress. s ma 18,000 psi (b) MAXIMUM SHEAR STRESS The maimum shear stress is on a 45 plane and equals /. t ma s 9,000 psi (c) STRESS ELEMENT AT 45 Note: All stresses have units of psi. 0 = 45 roblem.6-8 A copper bar with a rectangular cross section is held without stress between rigid supports (see figure). Subsequentl, the temperature of the bar is raised 50 C. Determine the stresses on all faces of the elements A and B, and show these stresses on sketches of the elements. (Assume / C and E 10 Ga.) A B 45 Solution.6-8 Copper bar with rigid supports 45 A B T 50C (Increase) /C E 10 Ga STRESS DUE TO TEMERATURE INCREASE E (T) (See Eq. -18 of Section.5) 105 Ma (Compression) STRESSES ON ELEMENTS A AND B 105 A = 45 B NOTE: All stresses have units of Ma. MAXIMUM SHEAR STRESS t ma s 5.5 Ma

6 SECTION.6 Stresses on Inclined Sections 17 roblem.6-9 A compression member in a bridge truss is fabricated from a wide-flange steel section (see figure). The cross-sectional area A 7.5 in. and the aial load 90 k. Determine the normal and shear stresses acting on all faces of stress elements located in the web of the beam and oriented at (a) an angle 0, (b) an angle 30, and (c) an angle 45. In each case, show the stresses on a sketch of a properl oriented element. Solution.6-9 Truss member in compression sin cos (1.0 ksi)(sin 10) (cos 10) 5. ksi 90 k A 7.5 in. s A 90 k 7.5 in. 1.0 ksi (Compression) = (a) ksi 1.0 ksi 0 (b) 30 Use Eqs. (-9a) and (-9b): cos (1.0 ksi)(cos 30) 9.0 ksi sin cos (1.0 ksi)(sin 30)(cos 30) 5. ksi cos (1.0 ksi)(cos 10) 3.0 ksi NOTE: All stresses have units of ksi. (c) 45 cos (1.0 ksi)(cos 45) 6.0 ksi sin cos (1.0 ksi)(sin 45) (cos 45) 6.0 ksi = NOTE: All stresses have units of ksi.

7 18 CHATER Aiall Loaded Numbers roblem.6-10 A plastic bar of diameter d 30 mm is compressed in a testing device b a force 170 N applied as shown in the figure. Determine the normal and shear stresses acting on all faces of stress elements oriented at (a) an angle 0, (b) an angle.5, and (c) an angle 45. In each case, show the stresses on a sketch of a properl oriented element. 100 mm lastic bar 300 mm = 170 N d = 30 mm Solution.6-10 lastic bar in compression 100 mm 300 mm = 170 N lastic bar d = 30 mm FREE-BODY DIAGRAM sin cos (96.0 ka)(sin.5)(cos.5) 340 ka cos (96.0 ka)(cos 11.5) 141 ka sin cos (96.0 ka)(sin 11.5)(cos 11.5) 340 ka 100 mm F Compressive force in plastic bar F 4 4(170 N)680 N LASTIC BAR (ROTATED TO THE HORIZONTAL) s F A 680 N 4 (30 mm) 96.0 ka (Compression) (a) 0 F F (b).5 Use Eqs. (-9a) and (-9b) cos (96.0 ka)(cos.5) 81 ka 96 ka 300 mm 0 d = 30 mm 0 F 96 ka = 170 N = NOTE: All stresses have units of ka. (c) 45 cos (96.0 ka)(cos 45) 481 ka sin cos (96.0 ka)(sin 45)(cos 45) 481 ka = NOTE: All stresses have units of ka.

8 SECTION.6 Stresses on Inclined Sections 19 roblem.6-11 A plastic bar fits snugl between rigid supports at room temperature (68 F) but with no initial stress (see figure). When the temperature of the bar is raised to 160 F, the compressive stress on an inclined plane pq becomes 1700 psi. (a) What is the shear stress on plane pq? (Assume / F and E psi.) (b) Draw a stress element oriented to plane pq and show the stresses acting on all faces of this element. p q robs and.6-1 Solution.6-11 lastic bar between rigid supports p q /F E psi Temperature increase: T 160F 68F 9F NORMAL STRESS IN THE BAR E(T ) (See Eq. -18 in Section.5) ( psi)( /F)(9F) 484 psi (Compression) ANGLE TO LANE pq cos For plane pq: 1700 psi Therefore, 1700 psi (484 psi)(cos ) 1700 psi cos u psi cos (b) STRESS ELEMENT ORIENTED TO LANE pq psi 1150 psi cos (484 psi)(cos 14.18) 784 psi sin cos (484 psi)(sin 14.18)(cos 14.18) 1150 psi = NOTE: All stresses have units of psi. (a) SHEAR STRESS ON LANE pq sin cos (484 psi)(sin 34.18)(cos 34.18) 1150 psi (Counter clockwise)

9 130 CHATER Aiall Loaded Numbers roblem.6-1 A copper bar is held snugl (but without an initial stress) between rigid supports (see figure). The allowable stresses on the inclined plane pq, for which 55, are specified as 60 Ma in compression and 30 Ma in shear. (a) What is the maimum permissible temperature rise T if the allowable stresses on plane pq are not to be eceeded? (Assume / C and E 10 Ga.) (b) If the temperature increases b the maimum permissible amount, what are the stresses on plane pq? Solution.6-1 Copper bar between rigid supports p q /C E 10 Ga lane pq: 55 Allowable stresses on plane pq: allow 60 Ma (Compression) allow 30 Ma (Shear) (a) MAXIMUM ERMISSIBLE TEMERATURE RISE T cos 60 Ma (cos 55) 18.4 Ma sin cos 30 Ma (sin 55)(cos 55) Ma Shear stress governs Ma Due to temperature increase T: E(T) (See Eq. -18 in Section.5) Ma (10 Ga)( /C)(T) T 31.3C (b) STRESSES ON LANE pq Ma cos (63.85 Ma)(cos 55) 1.0 Ma (Compression) sin cos (63.85 Ma)(sin 55)(cos 55) 30.0 Ma (Counter clockwise) roblem.6-13 A circular brass bar of diameter d is composed of two segments brazed together on a plane pq making an angle 36 with the ais of the bar (see figure). The allowable stresses in the brass are 13,500 psi in tension and 6500 psi in shear. On the brazed joint, the allowable stresses are 6000 psi in tension and 3000 psi in shear. If the bar must resist a tensile force 6000 lb, what is the minimum required diameter d min of the bar? p q d

10 SECTION.6 Stresses on Inclined Sections 131 Solution.6-13 Brass bar in tension d p q n = 54 Tensile stress: cos s s allow 6000 psi cos u (cos 54) 36 17,370 psi (3) Shear stress: sin cos 6000 lb A d 4 s ` t allow sin ucos u ` 3,000 psi (sin 54)(cos 54) 6,310 psi (4) STRESS BASED UON ALLOWABLE STRESSES IN THE BRASS Tensile stress ( 0): allow 13,500 psi 13,500 psi (1) Shear stress ( 45): allow 6500 psi t ma s allow 13,000 psi () STRESS BASED UON ALLOWABLE STRESSES ON THE BRAZED JOINT ( 54) ALLOWABLE STRESS Compare (1), (), (3), and (4). Shear stress on the brazed joint governs psi DIAMETER OF BAR A 6000 lb in. s 6310 psi A d 4 d 4A d 4A min B 1.10 in. allow 6000 psi (tension) allow 3000 psi (shear) roblem.6-14 Two boards are joined b gluing along a scarf joint, as shown in the figure. For purposes of cutting and gluing, the angle between the plane of the joint and the faces of the boards must be between 10 and 40. Under a tensile load, the normal stress in the boards is 4.9 Ma. (a) What are the normal and shear stresses acting on the glued joint if 0? (b) If the allowable shear stress on the joint is.5 Ma, what is the largest permissible value of the angle? (c) For what angle will the shear stress on the glued joint be numericall equal to twice the normal stress on the joint?

11 13 CHATER Aiall Loaded Numbers Solution.6-14 Two boards joined b a scarf joint or Due to load : 4.9 Ma or (a) STRESSES ON JOINT WHEN 0 Since must be between 10 and 40, we select 33.3 n = Note: If is between 10 and 33.3, 90 a.5 Ma. If is between 33.3 and 40, 9070 cos (4.9 Ma)(cos 70).5 Ma. (c) WHAT IS if? 0.57 Ma Numerical values onl: sin cos sin cos cos (4.9 Ma)(sin 70)(cos 70) t u ` ` 1.58 Ma s u (b) LARGEST ANGLE IF allow.5 Ma sin cos cos sin cos or tan allow sin cos The shear stress on the joint has a negative sign. Its numerical value cannot eceed allow.5 Ma. Therefore,.5 Ma (4.9 Ma)(sin )(cos ) or 6.6 NOTE: For 6.6 and 63.4, we find 0.98 Ma and 1.96 Ma. sin cos From trigonometr: sin u cos u 1 sin u Therefore: sin (0.459) Solving : or Thus, ` t u s u ` as required.

12 SECTION.6 Stresses on Inclined Sections 133 roblem.6-15 Acting on the sides of a stress element cut from a bar in uniaial stress are tensile stresses of 10,000 psi and 5,000 psi, as shown in the figure. (a) Determine the angle and the shear stress and show all stresses on a sketch of the element. (b) Determine the maimum normal stress ma and the maimum shear stress ma in the material. 5,000 psi = 10,000 psi 10,000 psi 5,000 psi Solution.6-15 Bar in uniaial stress 5,000 psi 10,000 psi From Eq. (1) or (): s 15,000 psi tan u s sin u cos u (15,000 psi)(sin 35.6)(cos 35.6) 7,070 psi Minus sign means that acts clockwise on the plane for which ,000 psi 5,000 psi 5,000 (a) ANGLE AND SHEAR STRESS cos 10,000 psi 7, ,000 7,070 = 35.6 s u s cos u 10,000 psi cos u (1) 10,000 5,000 LANE AT ANGLE 90 s u90 s [cos(u 90)] s [sin u] s sin u s u90 5,000 psi s s u90 5,000 psi sin u sin u Equate (1) and (): 10,000 psi cos u 5,000 psi sin u tan u 1 tan u 1 u 35.6 () NOTE: All stresses have units of psi. (b) MAXIMUM NORMAL AND SHEAR STRESS s ma s t ma s 15,000 psi 7,500 psi