A cylindrical pressure vessel 2.50 m in diameter is fabricated by shaping two 10-mm-thick steel plates and butt-welding the plates along helical arcs, as shown in Fig. 1. The maximum internal pressure in the pressure vessel is 1200 kPa. For this pressure level, calculate the following quantities: (a) the axial stress and the hoop stress; (b) the absolute maximum shear stress; and (c) the normal stress, σ_n, perpendicular to the weld line, and the shear stress, τ_{nt}, tangent to the weld line.

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(a) In-Plane Stresses: The axial stress and the hoop stress in the cylindrical section of the pressure vessel are given by Eqs. 9.2 and 9.4, respectively. For the given pressure vessel and internal-pressure loading, the axial stress and hoop stress are:

σ_a = \frac{pr}{2t} Axial Stress-Cylinder (9.2)

σ_h = \frac{pr}{t} Hoop Stress-Cylinder (9.4)

σ_a = \frac{pr}{2t}= \frac{(1200 kPa)(1.25 m)}{2(10 mm)} = 75 MPaσ_h = \frac{pr}{t}= \frac{(1200 kPa)(1.25 m)}{(10 mm)} = 150 MPa

Since there is no shear on longitudinal or circumferential cutting surfaces, these stresses are principal in-plane stresses. Therefore,

σ_1 = σ_h = 150 MPa, σ_2 = σ_a = 75 MPa (a) (1)

These stresses are shown on the biaxial stress element in Fig. 2a.

(b) Absolute Maximum Shear Stress: The in-plane Mohr’s circle for stress is shown as the solid-line circle in Fig. 2b. Since both the axial stress and the hoop stress are tension, the absolute maximum shear stress is not the maximum in-plane shear stress. Rather, this is a stress state with σ_1 ≥ σ_2 ≥ 0, as described in Fig. 8.24c. Since r/t = 125, we can neglect the internal pressure, p, in comparison with the in-plane stresses and draw a Mohr’s circle passing through P_1 and the origin, P_3. Thus,

or

τ_{\begin{matrix} abs \\max \end{matrix}} = 75 MPa (b) (2)

(c) Weld-line Stresses: The in-plane Mohr’s circle is redrawn in Fig. 3a. The point N represents the stresses on a face parallel to the weld, since the direction n in Fig. 1 is perpendicular to the weld line. From the Mohr’s circle in Fig. 3, we get

σ_n = \overline{OB} = 112.5 MPa – (37.5 MPa) cos 60° = 93.75 MPa

τ_{nt} = -\overline{NB} = -(37.5 MPa) sin 60° = -32.48 MPa

or

σ_n = 93.8 MPa, τ_{nt} = -32.5 MPa (c) (3)

The normal stress σ_t is given by

σ_t = \overline{OD} = 112.5 MPa + (37.5 MPa) cos 60° = 131.25 MPa

These stresses are shown on the rotated element in Fig. 3b.

The normal stress and shear stress on the weld can be converted to normal-force-per-unit-length and shear-force-per-unit-length (shear flow) as was done in Section 6.8.

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