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,
a.) Calculate the allowable tensile
force Pallow considering tension in
b.) Recompute Pallow for shear in
the diameter of pin, dp = 11 mm
the outer diameter of tubes dAB =40 mm
the outer diameter of tube dBC= 28 mm
the wall thicknes tAB = 6 mm
the wall thickness tBC = 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
First, we calculate the allowable stress in yield and ultimate strength criteria.
Factor of safety is defined as:
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.