Stress and strain

[匿名]

2013-07-12 2:16 am

請問有人能幫忙解答一下以下的題目嗎？(只須答案)

A wire of dia. 25mm has a tensile strength of 139MPa and E=170GPa, find
(1) the maximum allowable pull on the wire (kN) ;
(2) the corresponding elongation (mm) on a span of 38mm

A hollow cylinder of outer dia. 50mm and inner dia. 21mm is 3.3m long, the max allowable stress is 165N/mm^2
(3) Find the max allowable load (kN) ;
(4) Find the corresponding change in length (mm), E=116GPa.

(5) A tie bar of gross cross section 34mm x 17mm has a 20mm hole drilled in it to receive bolts. Calculate the safe load (kN) if the allowable tensile stress is 118N/mm^2

(6) A tie of rectangular section is to transmit an axial load of 106kN. The member is connected by at each end by 2 lines of rivets in 20mm dia. holes. Find the breadth (mm) required if the member is 13mm thick and the allowable stress is 136N/mm^2

(7) A cube of side dimension 75mm failed in a compression test at a load of 111kN. Find the allowable stress if the factor of safety if 5.4

(8) A rod of dia. 17mm sustained a max load of 75kN. Find the breadth (mm) of flat bar of the same material with thickness of 16mm that would transmit an load of 77kN. Factor of safety=5.9

(9) A cable of dia. 25mm and length 298mm hangs freely down a shaft. Find the elongation (mm). Density of cable = 7530kg/m^3. E=135kN/mm^2

A short concrete column of 326mm square section is reinforced with 5 dia. 20mm bars and carries an axial load of 1690kN. E for steel and concrete = 242 and 19kN/mm^2
(10) Find the stress in the steel (MPa) ;
(11) Find the stress in the concrete (MPa)

回答 (1)

Shu Fun

2013-07-15 12:40 am

✔ 最佳答案

(1) Stress = Force/Area
Maximum allowable pull
= Maximum Allowable Stress * Area
= 139*10^6[=139E6]*(π/4)*0.025^2
= 68200 N = 68.2kN

(2) Strain = Elongation / Original length = Stress / Young's Modulus
Corresponding elongation = 139E6/170E9 * 0.038 = 3.11E-5 m = 0.0311mm

(3) Maximum allowable load
= Max Stress * Area
= 165/(10^-3)^2*(π/4)*(0.05^2-0.021^2)
= 165E6*5.1475E-4π
= 267000 N = 267 kN

(4) From(2), Elongation = Stress / Young's Modulus * Original length
= 165E6/116E9*3.3
= 4.69E-3 m = 4.69 mm

(5) Smallest Cross-section = (34-20)*17 = 238 mm^2
Safe load = 118*238 = 28 084 N

(6) Let the breadth required is x mm
Smallest Section = 13(x-20*2) = 13x-520 mm^2

Allowable stress = 106E3/(13x-520) = 136
Required breadth = x = 100 mm

(7) Allowable stress
= Ultimate strength/Safety factor
= 111E3/(0.075^2)/5.4 = 3.65E6 N/m^2

(8) Ultimate strength = 75E3/(0.017^2*π/4) = 3.30E7 Pa

Let the required breadth be x mm
Then 77E3/(0.016*x*10^-3) = 3.30E7/5.9
Required breadth = x = 85.9 mm

(9) Mass = Density*Volume = 7530*0.025^2*π/4*0.298 = 1.10kg
Weight = mg = 1.10*9.81 = 10.8 N
Elongation = 10.8/(25^2*π/4)/135E3*298 = 4.86E-5 mm

(10)(11) Area of column = 0.326^2 = 0.106 mm^2, Area of bars = 5*(0.02)^2*π/4 = 0.001571 m^2

Since the bar and the concrete extend to the same extent, the stress in the concrete is 19/242 of that in the steel.

Let the stress in the steel be x Pa
Total stress
=[0.001571x+0.106(19x/242)]/(0.106+0.001571)
=1.69E6/(0.106+0.001571) = 1.57E7 Pa
Therefore 0.00991x = 1.96E6, x = 198E6

Stress in the steel = 198 MPa
Stress in the concrete = 19/242 = 15.5 MPa

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