B -strength Of Materials - Sp - Solution

5. The diameter of a brass rod is 6 mm. What force will stretch it by 0.2% of its length. Ebrass = 9 x 1010 Pa. a. 5090 N b. 5060 N c. 9050 N d. 6050 N Solution: FL δ= AE FL 0.002L = π  2 10  (0.006) 9 ×10 4   F = 5089.4 N (a)

STRENGTH OF MATERIALS – PROBLEMS 1. A steel tie rod on bridge must be made to withstand a pull of 5000 lbs. Find the diameter of the rod assuming a factor of safety of 5 and ultimate stress of 64,000 lb/in2. a. 0.75 in b. 0.71in c. 0.84 in d. 0.79 in Solution: σu 4F = 2 N πd 64,000 4(5000) = 5 πd 2 d = 0.7052 in (b)

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6. A wire length 2.50 m has a percentage strain of 0.012% when loaded with tensile force. Determine the extension of the wire. a. 0.1 mm b. 0.3 mm c. 0.4 mm d. 0.2 mm Solution: δ 0.012 δ ε= = L , 100 2.5

2. If the ultimate strength of a steel plate is 42,000 lb/in2, what force is necessary to punch a 0.75 in diameter hole in a 0.625 in thick plate? a. 63,000 lbs b. 68,080 lbs c. 61,850 lbs d. 66,800 lbs Solution: F = σ uπdt = (42,000)(π )(0.75)(0.625) = 61,850 lb (c) 3. What modulus of elasticity in tension is required to obtain a unit deformation of 0.00105 m/m from a load producing a unit tensile stress of 44,000 psi? a. 42.300 x 106 psi b. 41.202 x 106 psi c. 43.101 x 106 psi d. 41.905 x 106 psi Solution: stress 44,000 E= = = 41.9048 ×106 psi (d) strain 0.00105 4. How many 5/6 inch holes can be punch in one motion in a steel plate made of SAE 1010 steel, 7/16 inch thick using a force of 55 tons. The ultimate strength for shear is 50 ksi and use 2 factor of safety. a. 5.8 b. 3.7 c. 5 d. 6.2 Solution: Punching don’t need factor of safety. F τu = πdtn 55(2000) 50,000 = 5 7 π   n  16  16  n = 5.12 say 5 holes (c)

δ = 3×10−4 m = 0.3 mm (b) 7. Determine the polar section modulus, Zp (in3) of a shaft delivering 10 hp at 150 rpm. The diameter of the shaft is 3 in dia. and an allowable shear stress of 6,000 psi. a. 5.8 b. 6.4 c. 7.0 d. 5.3 Solution: Zp =

πd 3

Zp =

π (3)3

= 5.3 in 3

16 , 16 8. Find the polar section modulus of a hollow shaft with OD = 6 in and ID = 3 in. a. 28.97 cu in b. 39.76 cu in c. 45.45 cu in d. 51.98 cu in Solution:

Zp =

π  do4 − di4 

 16 

do

  

π  (6)4 − (3)4 

3   = 39.76 in 16  6  9. A horizontal beam 16 ft long is subjected to a load of 500 lb located to its center. The dimension of the beam is 2 x 4 inches respectively and its unit weight is 100 lb/ft. Find its flexural stress.

Zp =

1

a. b. c. d. Solution:

Solution: TL 32TL θ= = 4 JG πd G

11,696.34 psi 10,233.2 psi 15,677.2 psi 15,388.1 psi

θ=

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32 3 ×106 (1400) = 0.0035205 rad π (110 )4 (83,000)  180 deg   = 0.2017 deg (c)   π rad 

θ = (0.0035205 rad )

12. Determine the outside diameter if a hollow shaft whose inside diameter is 1 inch if it is to replace a 1.5 inch diameter solid shaft for power transmission. The shafts have equal torsional strengths. a. 2.25 in b. 1.5 in c. 2.0 in d. 1.75 in Solution: 16Tdo 16T τ= 3= πd π do4 − d i4

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d o4 − di4 = do d 3 do4 − (1)4 = do (1.5)3

100(16 ) + 500 = 1050 lb 2 Shear and moment diagram 1 M = (800)(8) + (250)(8) = 5200 ft − lb = 62,400 in − lb 2 h 4 Mc c = = = 2 in sf = I , 2 2 R1 = R2 =

d o = 1.588 in

Use next higher standard size of 1.75 in (d) 13. A circular bar solid cast iron 60-in long carries a solid circular head 60-in diameter. The bar is subjected to a torsional moment of 60,000 in-lb which is supplied at one end. It is desired to keep the torsional deflection of the circular head below 1/32-inch when the bar is transmitting power over its entire length in order to prevent the chattering of the piece. What would be the diameter of the bar, if the working stress is taken as 3000 psi and the transverse modulus of elasticity is 6 million psi? a. 4.7 inch b. 8.8 inch c. 6.7 inch d. 9.5 inch Solution: Solving for diameter using shear stress formula: 16T τ= 3 πd 16(60,000) 3000 = πd 3 d = 4.67 in

bh 3 2(4)3 = = 10.67 in 4 12 12 (62,400)(2 ) sf = = 11,696.34 lb in 2 (a) 10.67 10. What is the bending moment in ft-lb at the fixed end of a 10 ft truss with a uniform weight of 100 lb/ft and a concentrated vertical load at the free end of 1200 lbs. a. 15,000 b. 1800 c. 1500 d. 17,000 Solution: I=

 10  M = (10)(100)  + (1200)(10 ) = 17,000 ft − lb (d)  2 11. Compute the torsional deflection in degrees of a 110 mm diameter, 1.4 meter long shaft subjected to a twist moment of 3 x 106 N-mm. The torsional modulus of elasticity is 83,000 N/mm2. a. 0.27 b. 0.31 c. 0.20 d. 0.24

Solving for diameter using torsional deflection δ = rθ 1  60  =  θ 32  2 

θ = 0.00104 rad θ= 2

TL JG

0.00104 =

60,000(60)

π 32

(

d 4 6 ×106

b. 9,434 in-lb c. 8,000 in-lb d. 5,000 in-lb Solution:

)

d = 8.75 in Therefore, the safest diameter of the bar is 8.8 in. 14. Compute the induced/compressive stress, in kPa, of a steel solid shafting of 50 mm diameter and 800 mm in length that is subjected to an increase of temperature by 80 deg C. a. 196,530 kPa b. 181,445 kPa c. 162,256 kPa d. 112,187 kPa Solution:

Te = T 2 + M 2 Te = 5000 2 + 8000 2 = 9434 in − lb (b) 18. The maximum shear stress induced on a 1.5 in diameter solid shaft loaded with a torque of 8,000 in-lb and a bending moment of 12,000 inlb is: a. 21,763 psi b. 62,237 psi c. 27,220 psi d. 14,422 psi Solution:

σ = αE∆t , α = 6.6 ×10−6 F for steel E = 30×106 psi for steel

Te = T 2 + M 2

 9F   = 144 F ∆t = (80 C )   5C 

Te = 8000 2 + 120002 = 14,422.2 in − lb

σ = (6.6 ×10−6 )(30×106 )(144) = 28,512 psi

τ=

 101.325 kPa   = 196,529 kPa (a)   14.7 psi 

19. The shaft whose torque varies from 2000 to 6000 in-lbs has 1 ½ inches in diameter and 60,000 psi yield strength. Compute for the shaft mean average stress. a. 6036 psi b. 6810 psi c. 5162 psi d. 5550 psi Solution: 16T 16(6000 ) τ max = max = = 9054 psi 3 πd π (1.5)3

σ = 28,512 psi 

15. A steel train rails is 400 meters long, in March it is at -30 C and 40 C in July. What is the change in length in mm? K of steel = 11.7 x 10-6 m/m-C. a. 421 mm b. 303 mm c. 328 mm d. 503 mm Solution:

δ = αL∆t , δ = (11.7 ×10−6 )(400)[(40) − (− 30)] δ = 0.3276 mm = 327.6 mm (c)

τ min =

16. A 5.1 cm diameter solid steel shaft has a maximum bending moment of 677.9 N-m and an applied torque of 339 N-m. What is the equivalent bending stress in MPa? a. 35.14 b. 55.13 c. 45.45 d. 65.48 Solution: 1 Me = M + T 2 + M 2 2 1 Me = 677.9 + 339 2 + 677.9 2 = 717.918 N − m 2 32Me 32(717,919 N − mm ) σ= = πd 3 π (51 mm )3

( (

16Te 16(14,422.2) = = 21,763 psi (a) πd 3 π (1.5)3

16Tmin 16(2000) = = 3018 psi πd 3 π (1.5)3

τ max + τ min

9054 + 3018 = 6036 psi (a) 2 2 20. A body weighting 1000 lbs falls 6 inches and strikes a 2000 lbs (per inch) spring. What is the deformation of the spring? a. 3 inches b. 100 mm c. 6 inches d. 2 inches Solution: 1 W (h + δ ) = kδ 2 2 1 1000(6 + δ ) = (2000)δ 2 2 (δ − 3)(δ + 2 ) = 0

τm =

)

)

σ = 55.127 MPa (b)

=

δ = 3 in (a)

17. The equivalent twisting moment of a 2-in diameter shaft under a torque of 5000 in-lb and a bending moment of 8000 in-lb is: a. 13,000 in-lb

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