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Mechanical EngineeringQ&A LibraryProblem 4:Two steel tubes are joined at B byfour pins (dp = 11 mm), as shownin the cross-section a-a in thefigure. The outer diameters of thePintubes are daB = 40 mm and dec = 28mm. The wall thicknesses are taB =6 mm and tBc = 7 mm. The yieldstress in tension for the steel is oy= 200 MPa and the ultimate stressin tension is ou = 340 MPa. Thecorresponding ultimate value inshear of the pin is 140 MPa.Assume that the factors of safetywith respect to the yield andultimate stress are 4 and 5,respectively.a.) Calculate the allowable tensileforce Pallow Considering tension indpTAB1всthe tubes.dABdpcb.) Recompute Pallow for shear inthe pins.Question

Asked Feb 24, 2020

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Two steel tubes are joined at B by

four pins (dp = 11 mm), as shown

in the cross-section a-a in the

figure. The outer diameters of the

tubes are dAB = 40 mm and dBC = 28

mm. The wall thicknesses are tAB =

6 mm and tBC = 7 mm. The yield

stress in tension for the steel is y

= 200 MPa and the ultimate stress

in tension is U = 340 MPa. The

corresponding ultimate value in

shear of the pin is 140 MPa.

Assume that the factors of safety

with respect to the yield and

ultimate stress are 4 and 5,

respectively.

a.) Calculate the allowable tensile

force Pallow considering tension in

the tubes.

b.) Recompute Pallow for shear in

the pins.

Step 1

given:

the diameter of pin, d_{p} = 11 mm

the outer diameter of tubes d_{AB} =40 mm

the outer diameter of tube d_{BC}= 28 mm

the wall thicknes t_{AB} = 6 mm

the wall thickness t_{BC} = 7 mm

The yield strength ,ultimate tensile strength & ultimate shear strength is given as:

The factor of safety for ultimate strength = 5

The factor of safety for yield strength = 4

Step 2

(a)

First, we calculate the allowable stress in yield and ultimate strength criteria.

Factor of safety is defined as:

Step 3

Now we calculate the effective area of the tube as across a-a given in problem.

Aeffective = the effective area of the tube for calculation of tensile noramal stress.

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