Fatiga

SECTION 1– DESIGN FOR SIMPLE STRESSES TENSION, COMPRESSION, SHEAR DESIGN PROBLEMS 1.

The link shown, made of AISI C1045 steel, as rolled, is subjected to a tensile load of 8000 lb. Let h = 1.5b . If the load is repeated but not reversed, determine the dimensions of the section with the design based on (a) ultimate strength, (b) yield strength. (c) If this link, which is 15 in. long., must not elongate more than 0.005 in., what should be the dimensions of the cross section?

Problems 1 – 3. Solution: For AISI C1045 steel, as rolled (Table AT 7) su = 96 ksi s y = 59 ksi E = 30× 10 6 psi F A where F = 8000 lb A = bh but h = 1.5b therefore A = 1.5b 2 sd =

(a) Based on ultimate strength N = factor of safety = 6 for repeated but not reversed load (Table 1.1) s F sd = u = N A 96,000 8000 = 6 1.5b 2 5 b = 0.577 in say in . 8

Page 1 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES

h = 1.5b =

15 in 16

(b) Based on yield strength N = factor of safety = 3 for repeated but not reversed load (Table 1.1) s F sd = u = N A 59,000 8000 = 3 1.5b 2 9 in . b = 0.521 in say 16 27 h = 1.5b = in 32

(c) Elongation = δ =

FL AE

where, δ = 0.005 in F = 8000 lb E = 30 × 10 6 psi L = 15 in A = 1.5b 2 then, FL δ= AE

0.005 =

(8000)(15)

(1.5b )(30 ×10 ) 2

b = 0.730 in say

6

3 in . 4

1 h = 1.5b = 1 in 8

2.

The same as 1 except that the material is malleable iron, ASTM A47-52, grade 35 018.

Solution: For malleable iron, ASTM A47-52, grade 35 018(Table AT 6) su = 55 ksi s y = 36.5 ksi

E = 25× 10 6 psi

Page 2 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES F A where F = 8000 lb A = bh but h = 1.5b therefore A = 1.5b 2 sd =

(a) Based on ultimate strength N = factor of safety = 6 for repeated but not reversed load (Table 1.1) s F sd = u = N A 55,000 8000 = 6 1.5b 2 7 b = 0.763 in say in . 8 5 h = 1.5b = 1 in 16

(b) Based on yield strength N = factor of safety = 3 for repeated but not reversed load (Table 1.1) s F sd = u = N A 36,500 8000 = 3 1.5b 2 11 b = 0.622 in say in . 16 1 h = 1.5b = 1 in 32

(c) Elongation = δ = where, δ = 0.005 in F = 8000 lb E = 25× 10 6 psi L = 15 in A = 1.5b 2 then,

Page 3 of 131

FL AE

SECTION 1– DESIGN FOR SIMPLE STRESSES

δ=

FL AE

0.005 =

(8000)(15)

(1.5b )(25 ×10 ) 2

b = 0.8 in say h = 1.5b = 1

3.

6

7 in . 8

5 in 16

The same as 1 except that the material is gray iron, ASTM 30.

Solution: For ASTM 30 (Table AT 6) su = 30 ksi , no s y E = 14.5 × 10 6 psi Note: since there is no s y for brittle materials. Solve only for (a) and (c) F A where F = 8000 lb A = bh but h = 1.5b therefore A = 1.5b 2 sd =

(a) Based on ultimate strength N = factor of safety = 7 ~ 8 say 7.5 (Table 1.1) s F sd = u = N A 30,000 8000 = 7 .5 1.5b 2 3 b = 1.1547 in say 1 in . 16 25 h = 1.5b = 1 in 32 FL (c) Elongation = δ = AE where, δ = 0.005 in F = 8000 lb

E = 14.5 × 10 6 psi

Page 4 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES L = 15 in A = 1.5b 2 then, FL δ= AE

0.005 =

(8000)(15)

(1.5b )(14.5 ×10 ) 2

6

b = 1.050 in say 1 h = 1.5b = 1

4.

1 in . 16

19 in 32

A piston rod, made of AISI 3140 steel, OQT 1000 F (Fig. AF 2), is subjected to a repeated, reversed load. The rod is for a 20-in. air compressor, where the maximum pressure is 125 psig. Compute the diameter of the rod using a design factor based on (a) ultimate strength, (b) yield strength.

Solution: From Fig. AF 2 for AISI 3140, OQT 1000 F su = 152.5 ksi s y = 132.5 ksi F = force =

π

(20)2 (125) = 39,270 lb = 39.27 kips

4 From Table 1.1, page 20 Nu = 8 Ny = 4

(a) Based on ultimate strength N F A= u su π 2 (8)(39.27 ) d = 4 152.5 5 d = 1.62 in say 1 in 8 (b) Based on yield strength NyF A= sy π 2 (4 )(39.27 ) d = 4 132.5

Page 5 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES 1 d = 1.23 in say 1 in 4

5.

A hollow, short compression member, of normalized cast steel (ASTM A27-58, 65 ksi), is to support a load of 1500 kips with a factor of safety of 8 based on the ultimate strength. Determine the outside and inside diameters if Do = 2 Di .

Solution: su = 65 ksi Nu = 8 F = 1500 kips A=

π

(D 4

2 o

)

− Di2 =

π

(4 D 4

2 i

)

− Di2 =

3πDi2 4

3πDi2 N u F (8)(1500 ) A= = = 4 su 65 7 Di = 8.85 in say 8 in 8 3  7 Do = 2 Di = 2 8  = 17 in 4  8 6.

A short compression member with Do = 2 Di is to support a dead load of 25 tons. The material is to be 4130 steel, WQT 1100 F. Calculate the outside and inside diameters on the basis of (a) yield strength, (b) ultimate strength.

Solution: From Table AT 7 for 4130, WQT 1100 F su = 127 ksi s y = 114 ksi From Table 1.1 page 20, for dead load N u = 3 ~ 4 , say 4 N y = 1.5 ~ 2 , say 2 Area, A =

π

(D 4

2 o

)

− Di2 =

π

(4 D 4

F = 25 tons = 50 kips

(a) Based on yield strength 3πDi2 N y F (2 )(50 ) A= = = 4 sy 114

Page 6 of 131

2 i

)

− Di2 =

3πDi2 4

SECTION 1– DESIGN FOR SIMPLE STRESSES 5 in 8 1 5 Do = 2 Di = 2  = 1 in 4 8 (b) Based on ultimate strength 3πDi2 N u F (4 )(50 ) A= = = 4 su 127 7 Di = 0.82 in say in 8 3 7 Do = 2 Di = 2  = 1 in 4 8

Di = 0.61 in say

7.

A round, steel tension member, 55 in. long, is subjected to a maximum load of 7000 lb. (a) What should be its diameter if the total elongation is not to exceed 0.030 in? (b) Choose a steel that would be suitable on the basis of yield strength if the load is gradually applied and repeated (not reversed).

Solution: (a) δ =

FL FL or A = AE δE

where, F = 7000 lb L = 55 in δ = 0.030 in E = 25× 10 6 psi π (7000)(55) A = d2 = (0.030) 30 ×106 4 3 d = 0.74 in say in 4 (b) For gradually applied and repeated (not reversed) load Ny = 3

(

sy =

NyF A

=

)

(3)(7000) = 47,534 psi π (0.75)2 4

s y ≈ 48 ksi

say C1015 normalized condition ( s y = 48 ksi ) 8.

A centrifuge has a small bucket, weighing 0.332 lb. with contents, suspended on a manganese bronze pin (B138-A, ½ hard) at the end of a horizontal arm. If the pin is in double shear under the action of the centrifugal force, determine the diameter

Page 7 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES needed for 10,000 rpm of the arm. The center of gravity of the bucket is 12 in. from the axis of rotation. Solution: From Table AT 3, for B138-A, ½ hard sus = 48 ksi W F = ω 2r g where W = 0.332 lb g = 32.2 fps 2 2π n 2π (10,000 ) ω= = = 1047 rad sec 60 60 r = 12 in W 0.332 F = ω 2r = (1047 )2 (1) = 11,300 lb = 11.3 kips g 32.2 From Table 1.1, page 20 N = 3 ~ 4 , say 4 N F A= u su  π  (4 )(11.3) for double shear 2 d 2  = 48 4  25 d = 0.774 in say in 32

CHECK PROBLEMS 9.

The link shown is made of AISIC1020 annealed steel, with b =

3 in and 4

1 h = 1 in . (a) What force will cause breakage? (b) For a design factor of 4 based 2 on the ultimate strength, what is the maximum allowable load? (c) If N = 2.5 based on the yield strength, what is the allowable load?

Problem 9.

Page 8 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES

Solution: For AISI C1020 annealed steel, from Table AT 7 su = 57 ksi s y = 42 ksi (a) F = su A  3  1  A = bh =  1  = 1.125 in 2  4  2  F = (57 )(1.125) = 64 kips s A (b) F = u Nu Nu = 4  3  1  A = bh =  1  = 1.125 in 2  4  2  (57 )(1.125) = 16 kips F= 4

(c) F =

sy A Ny

N y = 2 .5  3  1  A = bh =  1  = 1.125 in 2  4  2  (42)(1.125) = 18.9 kips F= 2

10.

A ¾-in.bolt, made of cold-finished B1113, has an effective stress area of 0.334 sq. in. and an effective grip length of 5 in. The bolt is to be loaded by tightening until the tensile stress is 80 % of the yield strength, as determined by measuring the total elongation. What should be the total elongation?

Solution: sL δ= E from Table AT 7 for cold-finished B1113 s y = 72 ksi then, s = 0.80 s y = 0.8(72 ) = 57.6 ksi E = 30 ×106 psi = 30,000 ksi sL (57.6)(5) δ= = = 0.0096 in E 30,000

Page 9 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES 11.

A 4-lb. weight is attached by a 3/8-in. bolt to a rotating arm 14-in. from the center of rotation. The axis of the bolts is normal to the plane in which the centrifugal force acts and the bolt is in double shear. At what speed will the bolt shear in two if it is made of AISI B1113, cold finish?

Solution: From Table AT 7, sus = 62 ksi = 62,000 psi 2

 1  3 A = 2 (π )  = 0.2209 in 2  4 8 W F = ω 2 r = sus A g 4 ω 2 (14) = (62,000)(0.2209) 32.2 ω = 88.74 rad sec 2π n = 88.74 ω= 60 n = 847 rpm

12.

How many ¾-in. holes could be punched in one stroke in annealed steel plate of AISI C1040, 3/16-in. thick, by a force of 60 tons?

Solution: For AISI C1040, from Figure AF 1 su = 80 ksi sus = 0.75su = 0.75(80 ) ksi = 60 ksi A = π dt F = 60 tons = 120 kips n = number of holes F 120 n= = = 9 holes Asus (0.2209 )(60 )

13.

What is the length of a bearing for a 4-in. shaft if the load on the bearing is 6400 lb. and the allowable bearing pressure is 200 psi of the projected area?

Solution: pDL = W where p = 200 psi D = 4 in W = 6400 lb

Page 10 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES

(200)(4)L = 6400 L = 8 in

BENDING STRESSES DESIGN PROBLEMS 14.

A lever keyed to a shaft is L = 15 in long and has a rectangular cross section of h = 3t . A 2000-lb load is gradually applied and reversed at the end as shown; the material is AISI C1020, as rolled. Design for both ultimate and yield strengths. (a) What should be the dimensions of a section at a = 13 in ? (b) at b = 4 in ? (c) What should be the size where the load is applied?

Problem 14. Solution: For AISI C1020, as rolled, Table AT 7 su = 65 ksi s y = 49 ksi Design factors for gradually applied and reversed load Nu = 8 Ny = 4

th 3 , moment of inertial 12 but h = 3t h4 I= 36 I=

Moment Diagram (Load Upward)

Page 11 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES

Based on ultimate strength s s= u Nu (a) s =

Mc Fac = I I

h 2 F = 2000 lbs = 2 kips (2)(13) h  65 2 s= = 4 8 h     36  h = 3.86 in h 3.86 t= = = 1.29 in 3 3 say 1 h = 4.5 in = 4 in 2 1 t = 1.5 in = 1 in 2 c=

(b) s =

Mc Fbc = I I

h 2 F = 2000 lbs = 2 kips (2)(4) h  65 2 s= = 4 8 h     36  h = 2.61 in h 2.61 t= = = 0.87 in 3 3 say h = 3 in t = 1 in c=

(c)

Page 12 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES

3 − h 4 .5 − 3 = 4 13 − 4 h = 2.33 in 1 − t 1 .5 − 1 = 4 13 − 4 t = 0.78 in say 5 h = 2.625 in or h = 2 in 8

15.

A simple beam 54 in. long with a load of 4 kips at the center is made of cast steel, SAE 080. The cross section is rectangular (let h ≈ 3b ). (a) Determine the dimensions for N = 3 based on the yield strength. (b) Compute the maximum deflection for these dimensions. (c) What size may be used if the maximum deflection is not to exceed 0.03 in.?

Solution: For cast steel, SAE 080 (Table AT 6) s y = 40 ksi

E = 30×106 psi

Page 13 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES From Table AT 2 FL (4 )(54 ) Max. M = = = 54 kips − in 4 4 bh3 I= 12 but h = 3b h4 I= 36 sy

(a) s = c=

Ny

=

Mc I

h 2

(54) h 

40 2 = 3  h4     36  h = 4.18 in h 4.18 b= = = 1.39 in 3 3 1 h 4 .5 1 say h = 4 in , b = = = 1.5 in = 1 in 2 3 3 2

(b) δ =

(c) δ =

0.03 =

FL3 = 48EI

(4000)(54)3 = 0.0384 in 3   ( )( ) 1 . 5 4 . 5 48(30 ×106 )  

12

 

FL3  h4  48E    36 

(4000)(54)3 (36)

48(30 ×106 )(h 4 ) h = 4.79 in h 4.79 b= = = 1.60 in 3 3

1 h 5.25 3 say h = 5.25 in = 5 in , b = = = 1.75 in = 1 in 4 3 3 4

Page 14 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES 16.

The same as 15, except that the beam is to have a circular cross section.

Solution: s Mc (a) s = y = Ny I

I=

πd4

64 d c= 2 d  M  32M 2 s=  4 = π d  π d3   64   40 32(54 ) = 3 πd3 d = 3.46 in 1 say d = 3 in 2 (b) δ =

I=

FL3 48EI

πd4

64 3 64 FL3 64(4000 )(54 ) δ= = = 0.0594 in 48 E (π d 4 ) 48(30 × 10 6 )(π )(3.5)4 (c) δ =

64 FL3 48E (π d 4 )

64(4000 )(54) 48(30 ×106 )(π )d 4 d = 4.15 in 1 say d = 4 in 4 3

0.03 =

17.

A simple beam, 48 in. long, with a static load of 6000 lb. at the center, is made of C1020 structural steel. (a) Basing your calculations on the ultimate strength, determine the dimensions of the rectangular cross section for h = 2b . (b) Determine the dimensions based on yield strength. (c) Determine the dimensions using the principle of “limit design.”

Page 15 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES Solution:

From Table AT 7 and Table 1.1 su = 65 ksi s y = 48 ksi

N u = 3 ~ 4 , say 4 N y = 1.5 ~ 2 , say 2 FL (6 )(48) = = 72 in − kips 4 4 Mc s= I h c= 2 bh3 I= 12 h but b = 2 4 h I= 24 h M  12 M 2 s = 4  = 3 h h 24 M=

(a) Based on ultimate strength s 12 M s= u = 3 Nu h 65 12(72 ) = 4 h3 h = 3.76 in

Page 16 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES

b=

h 3.76 = = 1.88 in 2 2

h 3.75 3 7 = 1.875 in = 1 in say h = 3.75 in = 3 in , b = = 4 2 2 8

(b) Based on yield strength s y 12 M s= = 3 Ny h 48 12(72 ) = 2 h3 h = 3.30 in h 3.30 b= = = 1.65 in 2 2 h 3 .5 1 3 say h = 3.5 in = 3 in , b = = = 1.75 in = 1 in 2 2 2 4 (c) Limit design (Eq. 1.6)

bh 2 4 h 2  h 2 72 = (48)   4 h = 2.29 in h 2.29 b= = = 1.145 in 2 2 1 h 2 .5 1 say h = 2.5 in = 2 in , b = = = 1.25 in = 1 in 2 2 2 4 M = sy

18.

The bar shown is subjected to two vertical loads, F1 and F2 , of 3000 lb. each, that are L = 10 in apart and 3 in. ( a , d ) from the ends of the bar. The design factor is 4 based on the ultimate strength; h = 3b . Determine the dimensions h and b if the bar is made of (a) gray cast iron, SAE 111; (b) malleable cast iron, ASTM A4752, grade 35 018; (c) AISI C1040, as rolled (Fig. AF 1). Sketch the shear and moment diagrams approximately to scale.

Page 17 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES

Problems18, 19. Solution: F1 = F2 = R1 = R2 = 3000 lb Moment Diagram

M = R1a = (3000)(3) = 9000 lbs − in = 9 kips − in N = factor of safety = 4 based on su bh3 12 h c= 2 h 3  h h4 3  I= = 12 36 I=

(a) For gray cast iron, SAE 111 su = 30 ksi , Table AT 6 h M  s Mc 18M 2 s= u = = 4  = 3 N I h h     36  30 18(9 ) s= = 3 4 h h = 2.78 in h 2.78 b= = = 0.93 in 3 3 say h = 3.5 in , b = 1 in

(b) For malleable cast iron, ASTM A47-52, grade 35 018 Page 18 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES

su = 55 ksi , Table AT 6 h M  s Mc 18M 2 s= u = = 4  = 3 N I h h     36  55 18(9 ) s= = 3 4 h h = 2.28 in h 2.28 b= = = 0.76 in 3 3 1 3 say h = 2 in , b = in 4 4

(c) For AISI C1040, as rolled su = 90 ksi , Fig. AF 1 h M  s Mc 18M 2 s= u = = 4  = 3 N I h h     36  90 18(9 ) s= = 3 4 h h = 1.93 in h 1.93 b= = = 0.64 in 3 3 7 5 say h = 1 in , b = in 8 8

19.

The same as 18, except that F1 acts up ( F2 acts down).

Solution:

[∑ M

A

=0

]

Page 19 of 131

R1 = R2 = 1875 lb

SECTION 1– DESIGN FOR SIMPLE STRESSES Shear Diagram

Moment Diagram

M = maximum moment = 5625 lb-in = 5.625 kips-in

(a) For gray cast iron su 18M = 3 N h 30 18(5.625) = 4 h3 h = 2.38 in h 2.38 b= = = 0.79 in 3 3 1 3 say h = 2 in , b = in 4 4 (b) For malleable cast iron s=

su 18M = 3 N h 55 18(5.625) = 4 h3 h = 1.95 in h 1.95 b= = = 0.65 in 3 3 7 5 say h = 1 in , b = in 8 8 s=

Page 20 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES (c) For AISI C1040, as rolled su 18M = 3 N h 90 18(5.625) = 4 h3 h = 1.65 in h 1.65 b= = = 0.55 in 3 3 1 1 say h = 1 in , b = in 2 2 s=

20.

The bar shown, supported at A and B , is subjected to a static load F of 2500 lb. at θ = 0 . Let d = 3 in , L = 10 in and h = 3b . Determine the dimensions of the section if the bar is made of (a) gray iron, SAE 110; (b) malleable cast iron, ASTM A47-52, grade 32 510; (c) AISI C1035 steel, as rolled. (d) For economic reasons, the pins at A, B, and C are to be the same size. What should be their diameter if the material is AISI C1035, as rolled, and the mounting is such that each is in double shear? Use the basic dimensions from (c) as needed. (e) What sectional dimensions would be used for the C1035 steel if the principle of “limit design” governs in (c)?

Problems 20, 21. Solution:

Page 21 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES

[∑ M [∑ M

A

B

=0 =0

]

3RB = 13(2500)

]

RB = 10,833 lb 3RA = 10(2500 ) RA = 8333 lb

Shear Diagram

Moment Diagram

M = (2500)(10) = 25,000 lb − in = 25 kips − in h = 3b bh3 I= 12 h4 I= 36 h c= 2

h M  Mc 18M 2 s= = 4  = 3 I h h     36 

(a) For gray cast iron, SAE 110 su = 20 ksi , Table AT 6 N = 5 ~ 6 , say 6 for cast iron, dead load s 18M s= u = 3 N h 20 18(25) = 6 h3

Page 22 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES h = 5.13 in h b = = 1.71 in 3 1 3 say h = 5 in , b = 1 in 4 4 (b) For malleable cast iron, ASTM A47-32 grade 32510 su = 52 ksi , s y = 34 ksi N = 3 ~ 4 , say 4 for ductile, dead load s 18M s= u = 3 N h 52 18(25) = 4 h3 h = 3.26 in h b = = 1.09 in 3 3 1 say h = 3 in , b = 1 in 4 4 (c) For AISI C1035, as rolled su = 85 ksi , s y = 55 ksi N = 4 , based on ultimate strength s 18M s= u = 3 N h 85 18(25) = 4 h3 h = 2.77 in h b = = 0.92 in 3 say h = 3 in , b = 1 in

(d) For AISI C1035, as rolled ssu = 64 ksi N = 4 , RB = 10.833 kips s R ss = su = B N A π   π A = 2 D 2  = D 2 4  2 64 10.833 ss = = π 2 4 D 2 D = 0.657 in

Page 23 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES 11 in 16 (e) Limit Design bh 2 M = sy 4 For AISI C1035 steel, s y = 55 ksi

say D =

b=

h 3

h 2  h 3 M = 25 = (55)   4 h = 1.76 in h b = = 0.59 in 3 7 5 say h = 1.875 in = 1 in , b = in 8 8

The same as 20, except that θ = 30o . Pin B takes all the horizontal thrust.

21.

Solution:

FV = F cos θ

[∑ M [∑ M

A

B

=0

=0

] ]

3RB = 13FV 3RB = 13(2500) cos 30 RB = 9382 lb 3RA = 10 FV 3RA = 10(2500 ) cos 30 RA = 7217 lb

Shear Diagram

Page 24 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES

Moment Diagram

M = (2165)(10) = 21,650 lb − in = 21.65 kips − in 18M s= 3 h (a) For gray cast iron, SAE 110 su = 20 ksi , Table AT 6 N = 5 ~ 6 , say 6 for cast iron, dead load s 18M s= u = 3 N h 20 18(21.65) = 6 h3 h = 4.89 in h b = = 1.63 in 3 1 3 say h = 5 in , b = 1 in 4 4 (b) For malleable cast iron, ASTM A47-32 grade 32510 su = 52 ksi , s y = 34 ksi N = 3 ~ 4 , say 4 for ductile, dead load s 18M s= u = 3 N h 52 18(21.65) = 4 h3 h = 3.11 in h b = = 1.04 in 3 say h = 3 in , b = 1 in (c) For AISI C1035, as rolled su = 85 ksi , s y = 55 ksi N = 4 , based on ultimate strength

Page 25 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES su 18M = 3 N h 85 18(21.65) = 4 h3 h = 2.64 in h b = = 0.88 in 3 5 7 say h = 2 in , b = in 8 8 s=

(d) For AISI C1035, as rolled ssu = 64 ksi N = 4 , RBV = 9382 lb RBH = FH = F sin θ = 2500 sin 30 = 1250 lb 2 2 + RBH = (9382) + (1250) RB2 = RBV RB = 9465 lb s R ss = su = B N A π  π A = 2 D 2  = D 2 4  2 64 9.465 ss = = 4 π D2 2 D = 0.614 in 5 say D = in 8 (e) Limit Design bh 2 M = sy 4 For AISI C1035 steel, s y = 55 ksi 2

b=

h 3

h 2  h 3 M = 21.65 = (55)   4 h = 1.68 in h b = = 0.56 in 3 7 5 say h = 1.875 in = 1 in , b = in 8 8

Page 26 of 131

2

SECTION 1– DESIGN FOR SIMPLE STRESSES

22.

A cast-iron beam, ASTM 50, as shown, is 30 in. long and supports two gradually applied, repeated loads (in phase), one of 2000 lb. at e = 10 in from the free end, and one of 1000 lb at the free end. (a) Determine the dimensions of the cross section if b = c ≈ 3a . (b) The same as (a) except that the top of the tee is below.

Problem 22. Solution: For cast iron, ASTM 50 su = 50 ksi , suc = 164 ksi For gradually applied, repeated load N = 7 ~ 8 , say 8 M = F1d + F2 (d + e ) where: F1 = 2000 lb F2 = 1000 lb d = 30 − 10 = 20 in d + e = 30 in M = (2000)(20) + (1000)(30) = 70,000 lb − in = 70 kips − in Mc I Solving for I , moment of inertia s=

(3a )(a ) a  + (3a )(a ) 5a  = [(3a )(a ) + (3a )(a )]y 2

y=

3a 2

Page 27 of 131

 2 

SECTION 1– DESIGN FOR SIMPLE STRESSES

I=

(3a )(a )3 + (3a )(a )(a 2 ) + (a )(3a )3 + (3a )(a )(a 2 ) = 17 a 4 12

12

(a)

3a 2 5a cc = 2 Based on tension s Mct st = u = N I (70) 3a  50  2 = 8  17a 4     2  a = 1.255 in Based on compression s Mcc sc = uc = N I (70) 5a  164  2  = 8  17a 4     2  a = 1.001 in Therefore a = 1.255 in 1 Or say a = 1 in 4 And b = c = 3a = 3(1.25) = 3.75 in ct =

Page 28 of 131

2

SECTION 1– DESIGN FOR SIMPLE STRESSES 3 Or b = c = 3 in 4

(b) If the top of the tee is below

5a 2 3a cc = 2 17a 4 I= 2 M = 70 kips − in ct =

Based on tension s Mct st = u = N I (70) 5a  50  2  = 8  17 a 4     2  a = 1.488 in Based on compression s Mcc sc = uc = N I (70) 3a  164  2  = 8  17a 4     2  a = 0.845 in Therefore a = 1.488 in 1 Or say a = 1 in 2 1 And b = c = 3a = 4 in 2 CHECK PROBLEMS

Page 29 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES 23.

An I-beam is made of structural steel, AISI C1020, as rolled. It has a depth of 3 in. and is subjected to two loads; F1 and F2 = 2F1 ; F1 is 5 in. from one end and F2 is 5 in. from the other ends. The beam is 25 in. long; flange width is b = 2.509 in ; I x = 2.9 in 4 . Determine (a) the approximate values of the load to cause elastic failure, (b) the safe loads for a factor of safety of 3 based on the yield strength, (c) the safe load allowing for flange buckling (i1.24), (f) the maximum deflection caused by the safe loads.

Problems 23 – 25. Solution:

[∑ M [∑ F

V

A

=0

=0

]

]

5F1 + 20(2 F1 ) = 25RB RB = 1.8F1 F1 + 2 F1 = RA + RB RA = 3F1 − 1.8F1 = 1.2 F1

Shear Diagram

Moment Diagram

Page 30 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES M = 9F1 = maximum moment For AISI C1020, as rolled s y = 48 ksi Mc I d 3 where c = = = 1.5 in 2 2 (9 F1 )(1.5) s y = 48 = 2 .9 F1 = 10.31 kips F2 = 2 F1 = 20.62 kips

(a) s y =

sy

Mc N I 48 (9 F1 )(1.5) s= = 3 2 .9 F1 = 3.44 kips F2 = 2 F1 = 6.88 kips

(b) s =

(c)

=

L 25 = = 9.96 < 15 (page 34) b 2.509

sc = 20 ksi ( page 34, i1.24) Mc I (9 F1 )(1.5) 20 = 2 .9 F1 = 4.30 kips F2 = 2 F1 = 8.60 kips sc =

(d) For maximum deflection, by method of superposition, Table AT 2 3

ymax

Fb′  a(L + b′)  2 =  , a > b′ 3EIL  3

or 3

ymax

Fa  b′(L + a )  2 =  , b′ > a 3EIL  3

Page 31 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES ymax caused by F1 3

F a  b′(L + a1 )  2 ymax1 = 1 1  1  , b1′ > a1 3EIL  3 where E = 30,000 ksi a1 = 5 in b1′ = 20 in L = 25 in I = 2.9 in 4 3

ymax1

F1 (5)  20(25 + 5)  2 =  = 0.0022 F1 3(30,000)(2.9)(25)  3

ymax caused by F2 3

F b′  a (L + b2′ )  2 ymax 2 = 2 2  2  , a2 > b2′ 3EIL  3 where b2′ = 5 in a2 = 20 in 3

ymax 2

2 F1 (5)  20(25 + 5)  2 =  = 0.0043F1 3(30,000)(2.9)(25)  3

Total deflection = δ δ = ymax1 + ymax 2 = 0.022 F1 + 0.0043F1 = 0.0065F1 Deflection caused by the safe loads in (a) δ a = 0.0065(10.31) = 0.067 in Deflection caused by the safe loads in (b) δ b = 0.0065(3.44) = 0.022 in Deflection caused by the safe loads in (c) δ c = 0.0065(4.30) = 0.028 in 24.

The same as 23, except that the material is aluminum alloy, 2024-T4, heat treated.

Solution: For aluminum alloy, 2024-T4, heat treated s y = 47 ksi (a) s y =

Mc I

Page 32 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES

s y = 47 =

(9 F1 )(1.5)

2 .9 F1 = 10.10 kips F2 = 2 F1 = 20.20 kips

sy

Mc N I 47 (9 F1 )(1.5) s= = 3 2 .9 F1 = 3.36 kips F2 = 2 F1 = 6.72 kips

(b) s =

(c)

=

L 25 = = 9.96 < 15 (page 34) b 2.509

sc = 20 ksi ( page 34, i1.24) Mc I ( 9 F1 )(1.5) 20 = 2 .9 F1 = 4.30 kips F2 = 2 F1 = 8.60 kips sc =

(d) Total deflection = δ δ = ymax1 + ymax 2 = 0.022 F1 + 0.0043F1 = 0.0065F1 Deflection caused by the safe loads in (a) δ a = 0.0065(10.10) = 0.066 in Deflection caused by the safe loads in (b) δ b = 0.0065(3.36) = 0.022 in Deflection caused by the safe loads in (c) δ c = 0.0065(4.30) = 0.028 in 25.

A light I-beam is 80 in. long, simply supported, and carries a static load at the midpoint. The cross section has a depth of d = 4 in , a flange width of b = 2.66 in , and I x = 6.0 in 4 (see figure). (a) What load will the beam support if it is made of C1020, as-rolled steel, and flange buckling (i1.24) is considered? (b) Consider the stress owing to the weight of the beam, which is 7.7 lb/ft, and decide whether or not the safe load should be less.

Page 33 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES

Solution: (a) For C1020, as rolled, su = 65 ksi Consider flange buckling L 80 = = 30 b 2.66 L since 15 < < 40 b 22.5 22.5 sc = = = 15 ksi 2 2 ( 30 ) L 1 +   1800 1 + 1800 b Mc s= I d 4 c = = = 2 in 2 2

From Table AT 2 FL F (80 ) M= = = 20 F 4 4 Mc s = sc = I ( 20 F )(2 ) 15 = 6 F = 2.25 kips , safe load (b) Considering stress owing to the weight of the beam wL2 (Table AT 2) 8 where w = 7.7 lb ft add’l M =

Page 34 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES wL2  7.7  (80) = = 513 lb − in = 0.513 kips − in  8  12  8 M = 20 F + 0.513 = total moment Mc s = sc = I (20 F + 0.513)(2) 15 = 6 F = 2.224 kips Therefore, the safe load should be less. 2

add’l M =

26.

What is the stress in a band-saw blade due to being bent around a 13 ¾-in. pulley? The blade thickness is 0.0265 in. (Additional stresses arise from the initial tension and forces of sawing.)

Solution:

t = 0.0265 = 0.01325 in 2 r = 13.75 + 0.01325 = 13.76325 in Using Eq. (1.4) page 11 (Text) Ec s= r where E = 30× 106 psi c=

(30 ×10 )(0.01325) = 28,881 psi s= 6

13.76325

27.

A cantilever beam of rectangular cross section is tapered so that the depth varies uniformly from 4 in. at the fixed end to 1 in. at the free end. The width is 2 in. and the length 30 in. What safe load, acting repeated with minor shock, may be applied to the free end? The material is AISI C1020, as rolled.

Solution: For AISI C1020, as rolled su = 65 ksi (Table AT 7) Designing based on ultimate strength, N = 6 , for repeated, minor shock load

Page 35 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES su 65 = = 10.8 ksi N 6 Loading Diagram s=

4 −1 h −1 = 30 x h = 0.10 x + 1 wh 3 I= 12 h c= 2 M = Fx

(Fx ) h 

Mc 3Fx  2  = 6 Fx = 3Fx = = 2 2 3 I 2h h  wh  (0.10 x + 1)2    12  Differentiating with respect to x then equate to zero to solve for x giving maximum stress.  (0.10 x + 1)2 (1) − 2( x )(0.10 x + 1)(0.10 )  ds = 3F  =0 dx (0.10 x + 1)4   0.10 x + 1 − 2(0.10 x ) = 0 x = 10 in h = 0.10(10) + 1 = 2 in s 3Fx s= u = 2 N h 3F (10 ) 10.8 = (2)2 F = 1.44 kips s=

TORSIONAL STRESSES DESIGN PROBLEMS

Page 36 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES 28.

A centrifugal pump is to be driven by a 15-hp electric motor at 1750 rpm. What should be the diameter of the pump shaft if it is made of AISI C1045 as rolled? Consider the load as gradually repeated.

Solution: For C1045 as rolled, s y = 59 ksi sus = 72 ksi Designing based on ultimate strength s s = us , N = 6 (Table 1.1) N 72 s= = 12 ksi 6 33,000hp 33,000(15) = = 45 ft − lb = 540 in − lb = 0.540 in − kips Torque, T = 2π n 2π (1750 ) For diameter, 16T s= π d3 16(0.540 ) 12 = π d3 d = 0.612 in 5 say d = in 8 29.

A shaft in torsion only is to transmit 2500 hp at 570 rpm with medium shocks. Its material is AISI 1137 steel, annealed. (a) What should be the diameter of a solid shaft? (b) If the shaft is hollow, Do = 2 Di , what size is required? (c) What is the weight per foot of length of each of these shafts? Which is the lighter? By what percentage? (d) Which shaft is the more rigid? Compute the torsional deflection of each for a length of 10 ft.

Solution: 33,000hp 33,000(2500 ) T= = = 23,036 ft − lb = 276 in − kips 2π n 2π (570 ) For AISI 1137, annealed s y = 50 ksi (Table AT 8) s ys = 0.6 s y = 30 ksi

Designing based on yield strength N = 3 for medium shock, one direction

Page 37 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES

Design stress s 30 s = ys = = 10 ksi N 3 (a) Let D = shaft diameter Tc J π D4 J= 32 D c= 2 16T s= π D3 16(276 ) 10 = π D3 D = 5.20 in 1 say D = 5 in 4 s=

(b) J =

[

π (Do4 − Di4 ) π (2 Di )4 − Di4 =

32 Do 2 Di c= = = Di 2 2 TDi 32T s= = 4  15π Di  15π Di3    32  32(276 ) 10 = 15π Di3 Di = 2.66 in

32

] = 15π D

4 i

32

Do = 2 Di = 5.32 in say 5 Di = 2 in 8 1 Do = 5 in 4 (c) Density, ρ = 0.284 lb in3 (Table AT 7)

Page 38 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES For solid shaft w = weight per foot of length π  2 w = 12 ρ  D 2  = 3πρD 2 = 3π (0.284 )(5.25) = 73.8 lb ft 4  For hollow shaft π  2 2 w = 12 ρ  (Do2 − Di2 ) = 3πρ (Do2 − Di2 ) = 3π (0.284 ) (5.25) − (2.625) = 55.3 lb ft 4 Therefore hollow shaft is lighter 73.8 − 55.3 Percentage lightness = (100%) = 33.5% 55.3

[

]

(d) Torsional Deflection TL JG where L = 10 ft = 120 in

θ=

G = 11.5 ×103 ksi For solid shaft, J =

θ=

π D4 32

(276)(120)  180  o = 0.039 rad = (0.039 )  = 2 .2 π π  4 3    (5.25) (11.5 × 10 )  32 

For hollow shaft, J =

θ=

π (Do4 − Di4 ) 32

(276)(120)  180  o = 0.041 rad = (0.041)  = 2 .4 π  4 4 3  π   [(5.25) − (2.625) ](11.5 × 10 )

 32  Therefore, solid shaft is more rigid, 2.2o < 2.4o

30.

The same as 29, except that the material is AISI 4340, OQT 1200 F.

Solution: 33,000hp 33,000(2500 ) T= = = 23,036 ft − lb = 276 in − kips 2π n 2π (570 ) For AISI 4340, OQT 1200 F s y = 130 ksi s ys = 0.6 s y = 0.6(130 ) = 78 ksi

Designing based on yield strength

Page 39 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES N = 3 for mild shock

Design stress s 78 = 26 ksi s = ys = N 3 (a) Let D = shaft diameter Tc J π D4 J= 32 D c= 2 16T s= π D3 16(276 ) 26 = π D3 D = 3.78 in 3 say D = 3 in 4 s=

(b) J =

[

π (Do4 − Di4 ) π (2 Di )4 − Di4 =

32 Do 2 Di c= = = Di 2 2 TDi 32T s= = 4  15π Di  15π Di3    32  32(276 ) 26 = 15π Di3 Di = 1.93 in

32

] = 15π D

4 i

32

Do = 2 Di = 3.86 in say Di = 2 in Do = 4 in (c) Density, ρ = 0.284 lb in3 (Table AT 7)

Page 40 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES

For solid shaft w = weight per foot of length π  2 w = 12 ρ  D 2  = 3πρD 2 = 3π (0.284 )(3.75) = 37.6 lb ft 4  For hollow shaft π  2 2 w = 12 ρ  (Do2 − Di2 ) = 3πρ (Do2 − Di2 ) = 3π (0.284 ) (4 ) − (2 ) = 32.1 lb ft 4 Therefore hollow shaft is lighter 37.6 − 32.1 Percentage lightness = (100% ) = 17.1% 32.1

[

]

(d) Torsional Deflection TL JG where L = 10 ft = 120 in

θ=

G = 11.5 ×103 ksi For solid shaft, J =

π D4 32

(276)(120)  180  o = 0.148 rad = (0.148) θ=  = 8.48 π  4 3  π   (3.75) (11.5 ×10 )  32 

For hollow shaft, J =

θ=

π (Do4 − Di4 ) 32

(276)(120 )  180  o = 0.122 rad = (0.122 )  = 6.99 π π  4 4 3    [(4 ) − (2 ) ](11.5 × 10 )

 32  Therefore, hollow shaft is more rigid, 6.99o < 8.48o .

31.

A steel shaft is transmitting 40 hp at 500 rpm with minor shock. (a) What should be its diameter if the deflection is not to exceed 1o in 20 D ? (b) If deflection is primary what kind of steel would be satisfactory?

Solution: 33,000hp 33,000(40 ) (a) T = = = 420 ft − lb = 5.04 in − kips 2π n 2π (500 ) G = 11.5 ×103 ksi L = 20 D

Page 41 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES

θ = 1o = θ= π

π 180

TL JG

rad

(5.04)(20 D )

=

 π D4    11.5 × 103 32   D = 1.72 in 3 say D = 1 in 4 180

(

(b) s =

)

16T 16(5.04 ) = = 4.8 ksi π D 3 π (1.75)3

Based on yield strength N =3 s ys = Ns = (3)(4.8) = 14.4 ksi s ys

14.4 = 24 ksi 0.6 0.6 Use C1117 normalized steel s y = 35 ksi sy =

32.

=

A square shaft of cold-finish AISI 1118 transmits a torsional moment of 1200 inlb. For medium shock, what should be its size?

Solution: For AISI 1118 cold-finish s y = 75 ksi s ys = 0.6 s y = 45 ksi N = 3 for medium shock s T s = ys = N Z′ where, h = b 2b 2 h 2b3 Z′ = = (Table AT 1) 9 9 T = 1200 in − lb = 1.2 in − kips 45 1.2(9 ) s= = 3 2b 3 b = h = 0.71 in 3 say b = h = in 4

Page 42 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES

CHECK PROBLEMS 33.

A punch press is designed to exert a force sufficient to shear a 15/16-in. hole in a ½-in. steel plate, AISI C1020, as rolled. This force is exerted on the shaft at a radius of ¾-in. (a) Compute the torsional stress in the 3.5-in. shaft (bending neglected). (b) What will be the corresponding design factor if the shaft is made of cold-rolled AISI 1035 steel (Table AT 10)? Considering the shock loading that is characteristics of this machine, do you thick the design is safe enough?

Solution: For AISI C1020, as rolled sus = 49 ksi F = sus (π Dt ) 15 where D = in 16 1 t = in 2  15  1  F = 49(π )   = 72.2 kips  16  2  T = Fr 3 where r = in 4 3 T = (72.2 )  = 54.2 in − kips 4 16T π d3 where d = 3.5 in 16(54.2 ) s= = 6.44 ksi π (3.5)3

(a) s =

(b) For AISI 1035 steel, s us = 64 ksi for shock loading, traditional factor of safety, N = 10 ~ 15 Design factor , N = 34.

sus 64 = = 9.94 , the design is safe ( N ≈ 10 ) s 6.44

The same as 33, except that the shaft diameter is 2 ¾ in.

Solution: Page 43 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES d = 2.75 in 16T π d3 16(54.2 ) s= = 13.3 ksi π (2.75)3

(a) s =

(b) For AISI 1035 steel, s us = 64 ksi for shock loading, traditional factor of safety, N = 10 ~ 15 Design factor , N = 35.

sus 64 = = 4.8 , the design is not safe ( N < 10 ) s 13.3

A hollow annealed Monel propeller shaft has an external diameter of 13 ½ in. and an internal diameter of 6 ½ in.; it transmits 10,000 hp at 200 rpm. (a) Compute the torsional stress in the shaft (stress from bending and propeller thrust are not considered). (b) Compute the factor of safety. Does it look risky?

Solution: For Monel shaft, s us = 98 ksi (Table AT 3) N = 3 ~ 4 , for dead load, based on ultimate strength Tc J π Do4 − Di4 π (13.5)4 − (6.5)4 J= = = 3086 in 4 32 32 Do 13.5 c= = = 6.75 in 2 2 33,000hp 33,000(10,000 ) T= = = 262,606 ft − lb = 3152 in − kips 2π n 2π (200 ) (3152)(6.75) = 6.9 ksi s= 3086 (b) Factor of safety,

(a) s =

(

N=

)

[

sus 98 = = 14.2 , not risky s 6 .9

Page 44 of 131

]

SECTION 1– DESIGN FOR SIMPLE STRESSES

STRESS ANALYSIS DESIGN PROBLEMS 36.

A hook is attached to a plate as shown and supports a static load of 12,000 lb. The material is to be AISI C1020, as rolled. (a) Set up strength equations for dimensions d , D , h , and t . Assume that the bending in the plate is negligible. (b) Determine the minimum permissible value of these dimensions. In estimating the strength of the nut, let D1 = 1.2d . (c) Choose standard fractional dimensions which you think would be satisfactory.

Problems 36 – 38. Solution: s = axial stress ss = shear stress (a) s=

F 4F = 2 1 πd2 πd 4

Equation (1) d =

Page 45 of 131

4F πs

SECTION 1– DESIGN FOR SIMPLE STRESSES

s=

F 1 π D 2 − D12 4

(

)

Equation (2) D = ss =

4F 4F 4F = = 2 2 2 2 2 π D − D1 π D − 1.44d 2 π D − (1.2d )

(

)

[

] (

4F + 1.44d 2 πs

F F = π D1h 1.2π dh

Equation (3) h = ss =

=

F 1.2π dss

F π Dt

Equation (4) t =

F π Dss

(b) Designing based on ultimate strength, Table AT 7, AISI C1020, as rolled su = 65 ksi sus = 49 ksi N = 3 ~ 4 say 4, design factor for static load s 65 s= u = = 16 ksi N 4 s 49 ss = us = = 12 ksi N 4 F = 12,000 lb = 12 kips

From Equation (1) 4F 4(12) d= = = 0.98 in πs π (16) From Equation (2) 4F 4(12) 2 D= + 1.44d 2 = + 1.44(0.98) = 1.53 in ( ) πs π 16 From Equation (3) F 12 h= = = 0.27 in 1.2π dss 1.2π (0.98)(12 ) From Equation (4) F 12 t= = = 0.21 in π Dss π (1.53)(12)

Page 46 of 131

)

SECTION 1– DESIGN FOR SIMPLE STRESSES

(c) Standard fractional dimensions d = 1 in 1 D = 1 in 2 1 h = in 4 1 t = in 4

37.

The same as 36, except that a shock load of 4000 lb. is repeatedly applied.

Solution: (a) Same as 36. (b) N = 10 ~ 15 for shock load, based on ultimate strength say N = 15 , others the same. s 65 s= u = = 4 ksi N 15 s 49 ss = us = = 3 ksi N 15 F = 4000 lb = 4 kips

From Equation (1) 4F 4(4) d= = = 1.13 in πs π (4) From Equation (2) 4F 4(4) 2 D= + 1.44d 2 = + 1.44(1.13) = 1.76 in πs π (4) From Equation (3) F 4 h= = = 0.31 in 1.2π ds s 1.2π (1.13)(3) From Equation (4) F 4 t= = = 0.24 in π Dss π (1.76)(3)

Page 47 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES

(c) Standard fractional dimensions 1 d = 1 in 8 3 D = 1 in 4 3 h = in 8 1 t = in 4

38.

The connection between the plate and hook, as shown, is to support a load F . Determine the value of dimensions D , h , and t in terms of d if the connection is to be as strong as the rod of diameter d . Assume that D1 = 1.2d , sus = 0.75su , and that bending in the plate is negligible.

Solution:

s=

F

1 πd2 4 1 F = π d 2s 4 1 s  (1) F = π d 2  u  4 N

Page 48 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES

s=

F

F

=

1 1 π D 2 − D12 π D 2 − 1.44d 2 4 4 1 F = π D 2 − 1.44d 2 s 4 1 s  (2) F = π D 2 − 1.44d 2  u  4 N F F ss = = π D1h 1.2π dh F = 1.2π dhss

(

)

(

(

)

)

(

)

s   0.75su  F = 1.2π dh us  = 1.2π dh  N  N   5s  (3) F = 0.9π dh u   N  F ss = π Dt F = π Dtss s   0.75su  F = π Dt  us  = π Dt   N  N  s  (4) F = 0.75π Dt  u  N

Equate (2) and (1) 1 s  1 s  F = π (D 2 − 1.44d 2 ) u  = π d 2  u  4 N 4 N 2 2 D = 2.44d D = 1.562d Equate (3) and (1) s  1 s  F = 0.9π dh u  = π d 2  u  N 4 N d h= = 0.278d 4(0.9) Equate (4) and (1) s  1 s  F = 0.75π Dt  u  = π d 2  u  N 4 N s  1 s  F = 0.75π (1.562d )(t ) u  = π d 2  u  N 4 N d t= = 0.214d 4(0.75)(1.562)

Page 49 of 131

SECTION 1– DESIGN FOR SIMPLE STRESSES

39.

(a) For the connection shown, set up strength equations representing the various methods by which it might fail. Neglect bending effects. (b) Design this connection for a load of 2500 lb. Both plates and rivets are of AISI C1020, as rolled. The load is repeated and reversed with mild shock. Make the connection equally strong on the basis of yield strengths in tension, shear, and compression.

Problems 39, 40 Solution: (a) ss =

F 1  5 π D 2  4 

4F 5π ss

Equation (1) D = s=

F t (b − 2 D )

Equation (2) b = s=

F + 2D ts

F 5 Dt

Equation (3) t =

F 5 Ds

(b) For AISI C1020, as rolled s y = 48 ksi (Table AT 7) s ys = 0.6 s y = 28 ksi N = 4 for repeated and reversed load (mild shock) based on yield strength 48 s= = 12 ksi 4 28 ss = = 7 ksi 4 From Equation (1)

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SECTION 1– DESIGN FOR SIMPLE STRESSES

D=

4F 5π ss

where F = 2500 lb = 2.5 kips D=

4F 4(2.5) 5 = = 0.30 in say in 5π ss 5π (7 ) 16

From Equation (3) F 2 .5 5 t= = = 0.13 in say in 5 Ds 32 5 5 (12 )  16  From Equation (2) F 2 .5 5 b = + 2D = + 2  = 1.96 in say 2 in ts  5   16   (12 )  32  40.

The same as 39, except that the material is 2024-T4, aluminum alloy.

Solution: (a) Same as 39. (b) ) For 2024-T4, aluminum alloy s y = 47 ksi (Table AT 3) s ys = 0.55s y = 25 ksi N = 4 for repeated and reversed load (mild shock) based on yield strength 47 s= = 12 ksi 4 25 ss = = 6 ksi 4 From Equation (1) 4F D= 5π ss where F = 2500 lb = 2.5 kips

D=

4F 4(2.5) 3 = = 0.33 in say in 5π ss 5π (6) 8

From Equation (3)

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SECTION 1– DESIGN FOR SIMPLE STRESSES

t=

1 F 2 .5 = = 0.11 in say in 8 5 Ds 3 5 (12 ) 8

From Equation (2) 1 F 2 .5 3 b = + 2D = + 2  = 2.42 in say 2 in ts 2 1 8  (12 ) 8 41.

(a) For the connection shown, set up strength equations representing the various methods by which it might fail. (b) Design this connection for a load of 8000 lb. Use AISI C1015, as rolled, for the rivets, and AISI C1020, as rolled, for the plates. Let the load be repeatedly applied with minor shock in one direction and make the connection equally strong on the basis of ultimate strengths in tension, shear, and compression.

Problem 41. Solution: (a)

F sP = t (b − D )

3 F sP = 4 t (b − 2 D )

or

ssR =

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F 1  4 πD 2 (2 ) 4 

Equation (1)

Equation (2)

SECTION 1– DESIGN FOR SIMPLE STRESSES

sR =

F 4 Dt

Equation (3)

(b) For AISI C1015, as rolled suR = 61 ksi , susR = 0.75suR = 45 ksi For AISI C1020, as rolled suP = 65 ksi N = 6 , based on ultimate strength s 65 s P = uP = = 10.8 ksi N 6 s 61 s R = uR = = 10.1 ksi N 6 s 45 ssR = usR = = 7.5 ksi N 6 F = 8000 lb = 8 kips Solving for D F ssR = 2π D 2

F 8 7 = = 0.412 in say in 16 2π ssR 2π (7.5) Solving for t F sR = 4 Dt F 8 1 t= = = 0.453 in say in 4 Ds R 2 7 4 (10.1)  16  Solving for b F Using s P = t (b − D ) F 8 7 b= +D= + = 1.92 in say 2 in ts P 16 1  (10.8) 2 3 F 4 Using s P = t (b − 2 D ) D=

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SECTION 1– DESIGN FOR SIMPLE STRESSES 3F 3(8) 7 + 2D = + 2  = 1.99 in say 2 in 4ts P 16 1 4 (10.8)   2 Therefore b = 2 in 7 D = in 16 1 t = in 2 b=

42.

Give the strength equations for the connection shown, including that for the shear of the plate by the cotter.

Problems 42 – 44. Solution: Axial Stresses

s=

F 1 π D12 4

s=

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=

4F π D12

F (L − D2 )e

Equation (1)

Equation (2)

SECTION 1– DESIGN FOR SIMPLE STRESSES

s=

s=

s=

F D2e

Equation (3)

F 1 π a 2 − D22 4

(

F 1 π D22 − D2 e 4

)

=

=

4F Equation (4) π a 2 − D22

(

4F Equation (5) π D − 4 D2e 2 2

Shear Stresses

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ss =

F 2eb

ss =

F 2(L − D2 + e )t

)

Equation (6)

Equation (7)

SECTION 1– DESIGN FOR SIMPLE STRESSES

43.

ss =

F π at

Equation (8)

ss =

F π D1m

Equation (9)

ss =

F 2 D2 h

Equation (10)

A steel rod, as-rolled AISI C1035, is fastened to a 7/8-in., as-rolled C1020 plate by means of a cotter that is made of as-rolled C1020, in the manner shown. (a) Determine all dimensions of this joint if it is to withstand a reversed shock load F = 10 kips , basing the design on yield strengths. (b) If all fits are free-running fits, decide upon tolerances and allowances.

Solution: (See figure of Prob. 42) 7 t = in = 0.875 in , ssy = 0.6 s y 8 For steel rod, AISI C1035, as rolled s y1 = 55 ksi ssy1 = 33 ksi For plate and cotter, AISI C1020, as rolled s y2 = 48 ksi ssy2 = 28 ksi N = 5 ~ 7 based on yield strength say N = 7

From Equation (1) (Prob. 42) sy 4F s= 1 = N π D12 55 4(10 ) = 7 π D12 D1 = 1.27 in 1 say D1 = 1 in 4

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SECTION 1– DESIGN FOR SIMPLE STRESSES From Equation (9) ssy F ss = 1 = N π D1m 33 10 = 7  1 π 1  m  4 m = 0.54 in 9 say m = in 16 From Equation (3) sy F s= 1 = N D2 e 55 10 s= = 7 D2 e D2e = 1.273 From Equation (5) sy 4F s= 1 = 2 N π D2 − 4 D2e 55 4(10 ) = 2 7 π D2 − 4(1.273) D2 = 1.80 in 3 say D2 = 1 in 4 and D2e = 1.273  3 1 e = 1.273  4 e = 0.73 in 3 say e = in 4 By further adjustment 5 Say D2 = 2 in , e = in 8 From Equation (8) ssy F ss = 2 = N π at 28 10 = 7 π a (0.875) a = 0.91 in say a = 1 in

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SECTION 1– DESIGN FOR SIMPLE STRESSES From Equation (4) sy 4F s= 2 = 2 N π (a − D22 ) 48 4(10) = 7 π (a 2 − 2 2 ) a = 2.42 in 1 say a = 2 in 2 1 use a = 2 in 2 From Equation (7) ssy F ss = 2 = N 2(L − D2 + e )t 28 10 = 5 7  2 L − 2 + (0.875) 8  L = 2.80 in say L = 3 in From Equation (6) ssy F ss = 2 = N 2eb 28 10 = 7 5 2 b 8 b = 2 in From Equation (10) ssy F ss = 2 = N 2 D2 h 28 10 = 7 2(2)h 5 h = 0.625 in = in 8 Summary of Dimensions L = 3 in 5 h = in 8 b = 2 in 7 t = in 8

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SECTION 1– DESIGN FOR SIMPLE STRESSES 9 in 16 1 a = 2 in 2 1 D1 = 1 in 4 D2 = 2 in 5 e = in 8 m=

(b) Tolerances and allowances, No fit, tolerance = ± 0.010 in L = 3 ± 0.010 in h = 0.625 ± 0.010 in t = 0.875 ± 0.010 in m = 0.5625 ± 0.010 in a = 2.500 ± 0.010 in D1 = 1.25 ± 0.010 in For Free Running Fits (RC 7) Table 3.1 Female Male + 0.0030 − 0.0040 b = 2.0 in b = 2.0 in − 0.0000 − 0.0058 allowance = 0.0040 in + 0.0030 − 0.0040 D2 = 2.0 in D2 = 2.0 in − 0.0000 − 0.0058 allowance = 0.0040 in + 0.0016 − 0.0020 e = 0.625 in e = 0.625 in − 0.0000 − 0.0030 allowance = 0.0020 in

44.

A 1-in. ( D1 ) steel rod (as-rolled AISI C1035) is to be anchored to a 1-in. steel plate (as-rolled C1020) by means of a cotter (as rolled C1035) as shown. (a) Determine all the dimensions for this connection so that all parts have the same ultimate strength as the rod. The load F reverses direction. (b) Decide upon tolerances and allowances for loose-running fits.

Solution: (Refer to Prob. 42) (a) For AISI C1035, as rolled su1 = 85 ksi sus1 = 64 ksi For AISI C1020, as rolled

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SECTION 1– DESIGN FOR SIMPLE STRESSES su2 = 65 ksi sus2 = 48 ksi

Ultimate strength Use Equation (1) 1  1  2 Fu = su1  π D12  = (85) π (1) = 66.8 kips 4  4  Equation (9) Fu = sus1 π D1m 66.8 = (64)(π )(1)m m = 0.33 in 3 say m = in 8 From Equation (3) Fu = su1 D2e 66.8 = (85)D2 e D2e = 0.7859 From Equation (5) 1  Fu = su1  π D22 − D2 e  4  1  66.8 = (85) π D22 − 0.7859  4  D2 = 1.42 in 3 say D2 = 1 in 8 3   D2e = 1 e = 0.7859  8 e = 0.57 in 9 say e = in 16 From Equation (4) 1  Fu = su2  π (a 2 − D22 ) 4  2  1   3  66.8 = (65) π  a 2 −1    4    8   a = 1.79 in 3 say a = 1 in 4 From Equation (8)

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SECTION 1– DESIGN FOR SIMPLE STRESSES Fu = sus2 π at

66.8 = (48)(π )(a )(1) a = 0.44 in 1 say a = in 2 3 use a = 1 in 4 From Equation (2) Fu = su2 (L − D2 )e 3  9   66.8 = (65) L − 1   8  16   L = 3.20 in 1 say L = 3 in 4 From Equation (7) Fu = 2 sus2 (L − D2 − e )t 3 9  66.8 = 2(48) L − 1 − (1) 8 16   L = 1.51 in 1 say L = 1 in 2 1 use L = 3 in 4 From Equation (6) Fu = 2 sus1 eb  9 66.8 = 2(64 ) b  16  b = 0.93 in say b = 1 in From Equation (10) Fu = 2 sus1 D2 h  3 66.8 = 2(64 ) 1 h  8 h = 0.38 in 3 say h = in 8 Dimensions 1 L = 3 in 4

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SECTION 1– DESIGN FOR SIMPLE STRESSES 3 in 8 b = 1 in t = 1 in 3 m = in 8 3 a = 1 in 4 D1 = 1 in 3 D2 = 1 in 8 9 e = in 16 h=

(b) Tolerances and allowances, No fit, tolerance = ± 0.010 in L = 3.25 ± 0.010 in h = 0.375 ± 0.010 in t = 1.000 ± 0.010 in m = 0.375 ± 0.010 in a = 1.75 ± 0.010 in D1 = 1.000 ± 0.010 in For Loose Running Fits (RC 8) Table 3.1 Female Male + 0.0035 − 0.0045 b = 1 .0 in b = 1 .0 in − 0.0000 − 0.0065 allowance = 0.0045 in + 0.0040 − 0.0050 D2 = 1.375 in D2 = 1.375 in − 0.0000 − 0.0075 allowance = 0.0050 in + 0.0028 − 0.0035 e = 0.5625 in e = 0.5625 in − 0.0000 − 0.0051 allowance = 0.0035 in 45.

Give all the simple strength equations for the connection shown. (b) Determine the ratio of the dimensions a , b , c , d , m , and n to the dimension D so that the connection will be equally strong in tension, shear, and compression. Base the calculations on ultimate strengths and assume sus = 0.75su .

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SECTION 1– DESIGN FOR SIMPLE STRESSES

Problems 45 – 47. Solution: (a) Neglecting bending 1  Equation (1): F = s π D 2  4   1  Equation (2): F = ss 2 π c 2    4 Equation (3): F = s(2bc ) Equation (4): F = s(ac ) Equation (5): F = s[2(d − c )b] Equation (6): F = ss (4mb ) Equation (7): F = ss (2nb ) Equation (8): F = s(d − c )a su s and ss = us N N Therefore ss = 0.75s Equate (2) and (1)  1   1  F = ss 2 π c 2  = s π D 2    4   4

(b) s =

1  1  0.75s c 2  = s D 2  2  4  c = 0.8165 D Equate (3) and (1) 1  F = s (2bc ) = s π D 2  4  1 2b(0.8165 D ) = π D 2 4 b = 0.4810 D

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SECTION 1– DESIGN FOR SIMPLE STRESSES Equate (4) and (1) 1  F = sac = s π D 2  4  1 a (0.8165 D ) = π D 2 4 a = 0.9619 D Equate (5) and (1) 1  F = s[2(d − c )b ] = s π D 2  4  1 2(d − 0.8165 D )(0.4810 ) = π D 2 4 d = 1.6329 D Equate (6) and (1) 1  F = ss (4mb ) = s π D 2  4  1 0.75(4m )(0.4810 D ) = π D 2 4 m = 0.5443D Equate (7) and (1) 1  F = ss (2nb ) = s π D 2  4  1 0.75(2n )(0.4810 D ) = π D 2 4 n = 1.0886 D Equate (8) and (1) 1  F = s (d − c )a = s π D 2  4  (1.6329 − D − 0.8165D )a = 1 π D 2 4 a = 0.9620 D Summary a = 0.9620 D b = 0.4810 D c = 0.8165 D d = 1.6329 D m = 0.5443D n = 1.0886 D 46.

The same as 45, except that the calculations are to be based on yield strengths. Let ssy = 0.6 s y .

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SECTION 1– DESIGN FOR SIMPLE STRESSES Solution: (Refer to Prob. 45) (a) Neglecting bending 1  Equation (1): F = s π D 2  4   1  Equation (2): F = ss 2 π c 2    4 Equation (3): F = s(2bc ) Equation (4): F = s(ac ) Equation (5): F = s[2(d − c )b] Equation (6): F = ss (4mb ) Equation (7): F = ss (2nb ) Equation (8): F = s(d − c )a

(b) s =

sy

and ss =

s sy

N N Therefore ss = 0.6s Equate (2) and (1)  1   1  F = ss 2 π c 2  = s π D 2     4  4 1  1  0 .6 s  c 2  = s  D 2  2  4  c = 0.9129 D Equate (3) and (1) 1  F = s (2bc ) = s π D 2  4  1 2b(0.9129 D ) = π D 2 4 b = 0.4302 D

Equate (4) and (1) 1  F = sac = s π D 2  4  1 a (0.9129 D ) = π D 2 4 a = 0.8603D Equate (5) and (1) 1  F = s[2(d − c )b ] = s π D 2  4 

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