1. A cylindrical steel pressure vessel 400 mm in diameter with a wall thickness of 20 mm, is subjected to an internal pressure of 4.5 MN/m2. (a) Calculate the tangential and longitudinal stresses in the steel. (b) To what value may the internal pressure be increased if the stress in the steel is limited to 120 MN/m2? (c) If the internal pressure were increased until the vessel burst, sketch the type of fracture that would occur. (d) Illustrate to justify your solution

Q: 4) Explain and illustrate variations of normal stresses everywhere on the bar and state why it is…

A: Given: A bar carrying tensile load and having two holes on the bar. Tensile load=P Due to this load,…

Q: A punch of 20 mm diameter is used to punch a hole in 8 mm thick plate. If the force required to…

A: Given data- Diameter of the punch = 20 mm, thickness = 8 mm, force = 110 kN, To Find- The value of…

Q: P- 3P 2P Steel Bronze Al A = 200 mm A = 400 mm A = 500 mm = 150MPA e= 100MPA c = 80MPA

A:

Q: A circular aluminum alloy [E = 70 GPa; a = 22.5 x 10-6/°C; v = 0.33] pipe has an outside diameter of…

A: Given:Outside diameter of pipe, D=240 mmWall thickness=10 mmInside diameter of pipe, d=240-2×10…

Q: Problem # 4 A 4 in x 3 in steel block follows an elasto-perfectly plastic stress strain relation…

A: Calculate the modular ratio: m=modulus of elasticity of steelmodulus of elasticity of…

Q: Select the types of losses are both common for post tensioning and pre tensioning. a. relaxation,…

A: To determine: Types of losses are both common for post tesnsioning and pretensioning.

Q: A steel band 8mm thick is to be pulled on to a cylinder 30cm in diameter. The internal diameter of…

A: Given, Diameter of cylinder = 30 cm Thickness of steel band = 8 mm Internal diameter of the hoop…

Q: Strain, Deformation and Thermal Stress Steel Rod A = 600 mm L=4.0 m E= 200 GPa a = 11.70 um/(m°C)…

A: Given- Aluminum Rod Area=900mm2 Length=3.0m Modulus of elasticity =70Gpa Coefficient of thermal…

Q: A mild steel specimen having Young’s modulus of 200 GPa and original length of 300 mm is tested for…

A: Given: Young’s Modulus, E = 200 GPa Original length, l = 300 mm Stress = 500 MPa

Q: A high strength steel band saw of rectangular cross-section of size 20 x 0.8 mm thick runs over…

A:

Q: An applied torque results in a shear strain (y) of 3.75 x 10-3 in a solid circular steel shaft. At…

A:

Q: A circular shaft of diameter 120 mm is welded to a rigid plate by a fillet weld of size 6 mm. If a…

A: Given data- diameter=120 mmfillet weld=6 mmtorque=8 kNmTo Find-The maximum stress in the weld.

Q: The failure stress of a homogeneous steel bar having a diameter of 10 mm is 250 MPa. Using a factor…

A:

Q: In a fillet weld, the direct shear stress and bending tensile stress are 50 MPa and 150 MPa…

A: Given data- direct shear stress q=50 MPabending tensile stress σb=150 MPaTo Find-The value of…

Q: On the free surface of an aluminum[E- 10000 ksi, v= 0.35) component, the strain rosette shown was…

A: Solution: 1) Normal strain at the point associated with axis x and y: We have,…

Q: 1.3 (A). Define the terms shear stress and shear strain, illustrating your answer by means of a…

A: SHEAR STRESS: It is defined as the force, internal resisting force per unit area.The force acting on…

Q: At an axial load of 24 kN, a 40-mm-wide by 20-mm-thick polyimide polymer bar elongates 2.7 mm while…

A:

Q: A load applied to a machine component results in the state of plane stress ?x=80 MPa, ?y=100 MPa,…

A:

Q: The state of two-dimensional stresses acting on a concrete lamina consists of a direct tensile…

A: given- Direct tensile stress-σx=1.5 N/mm2σy=0Shear stress-τxy=1.2 MPa

Q: A typical aluminum-alloy scuba diving tank is shown. The outside diameter of the tank is 200 mm and…

A:

Q: A circular aluminum alloy [E = 70 GPa; a = 22.5 x 10-6/°C; v = 0.33] pipe has an outside diameter of…

A: Given: Modulus of elasticity =E  Coefficient of thermal expansion Outside diameter Thickness  Length…

Q: 1) The plastic polymer cylindrical pipe is carrying flowing water at 0.5 N/mm pressure. The pipe has…

A:

Q: 2. Which of the below stresses are normal stresses? a) Tensile and compressive stresses b) Tensile…

A: Which of the below stresses are normal stresses? a) Tensile and compressive stresses

Q: ".If the diameter of circular bar is (15mm) while the diameter of the bar at the section where…

A:

Q: The thin-walled vessel shown is fabricated by joining two cylinders with thickness 26 mm using 20…

A: Given, Thickness of the vessel, t= 26mm Number of bolts= 20 Diameter of the bolts= 20mm Outer…

Q: A circular aluminum alloy [E = 70 GPa; a = 22.5 × 10-6/°C; v = 0.33] pipe has an outside diameter of…

A: Given data  E = 70 GPA α = 22.5 × 10-6 /°C  μ  = 0.33  Drop temperature (∆T) = -43° C  Length of…

Q: The point loads are placed at the fixed positions shown in the figure and they are live loads. A K a…

A: Since the condition for each subpart is varying, the first part will be calculated. Please submit…

Q: In a tensile test experiment of carbon steel. a standard specimen (D= 0.505 in. Gauge length=2.0 in…

A: Given DataDiameter of the specimen=0.505 inGauge length=2 inTotal length=8 inYield strength=80000…

Q: Consider the compound system consisting of three sections with cross-sectional areas 3A, 2A, and A…

A: The correct answer to the given questions are as under

Q: 2. Shown below are pieces of unsaid material. They are combined and locked in each other and a force…

A:

Q: Problem Two: Consider the cross-sectional shown in the figure below. The cross-section is reinforced…

A: Free body diagram -  Moment of inertia -  Ig = 300×500312 - 180×380312⇒Ig = 2301920000 mm4

Q: A 42 meter solid shaft is under torsional loading 53 Nm. If in this application, the allowable shear…

A: Given: Torsional loading, T = 53N-m Allowable shear stress = 369MPa Required: Minimum diameter of…

Q: Q5) The wooden block shown in Figure (4) will fail if the shear stress acting along the grain is…

A: In the given plane only stresses acting along X and Y direction and no shear stress (Txy =0) is…

Q: A circular aluminum alloy [E = 70 GPa; a = 22.5 x 10-6/°C; v=0.33] pipe has an outside diameter of…

A:

Methods to Determine Vertical Stress

The load applied to the soil is transferred to the underground. The presence of water at soil pores and solid grains above the soil are the primary components that are responsible for the effective load transfer. Due to the load acting, the internal component of the soil or the underground formations undergoes deformations or a tectonic shift. The amount of deformation depends upon the amount of load acting on the soil surface, which indeed, depends on the weight of everything present above the soil. Due to this, the underground soil element develops internal stresses which try to resist the tectonic changes.