A cylindrical tank with flat endplates is constructed from two sections that are welded together circumferentially.The outer diameter of the tank is 1.20, and the wall of the tank is 20.0 mm thick and the mamimum internal pressure is 2.00 MPa.
(a) Calculate the maximum hoop stress in the tank
(b) Calculate the maximum tensile stress in the tank
(c) The tensile strength of the weld is 350 MPa and the shear strength of the weld is 40% of the tensile strength. what is the magnitude of the tensile stress that will cause the weld to burst open
my work:
(a)hoop stress=Pr/t
(b)tensile stress=0.5 hoop stress
(c)how about this question? i really have no idea! please help me!
To answer part (c) of the question, we need to calculate the magnitude of the tensile stress that will cause the weld to burst open.
We know that the tensile strength of the weld is given as 350 MPa. This means that the maximum tensile stress the weld can handle before it fails is 350 MPa.
However, we are also given that the shear strength of the weld is 40% of the tensile strength. So, the shear strength of the weld is 0.4 * 350 MPa = 140 MPa.
Now, during bursting, the weld will fail under a combination of tensile and shear stresses. To calculate the magnitude of the tensile stress that will cause the weld to burst open, we need to consider the principle of superposition.
The maximum tensile stress (σ_t) that will cause the weld to burst open can be calculated using the following equation:
σ_t = √((σ^2_x) + (3 * (τ^2_xy)))
Where σ_x is the tensile stress, and τ_xy is the shear stress.
Since we know the shear strength of the weld (τ_xy = 140 MPa), we can rearrange the equation to solve for σ_x:
σ_x = √((σ_t^2) - (3 * (τ_xy^2)))
Substituting the values, we get:
σ_x = √((350^2) - (3 * (140^2)))
= √((122,500) - (58800))
= √(63700)
= 252.6 MPa
So, the magnitude of the tensile stress that will cause the weld to burst open is approximately 252.6 MPa.