Answer
45o
failure in ductile materials (A36 steel) due to uniform tensile stress is shear failure because shear strength is less than ultimate strength.
And shear plane will be 450 to that of longitudinal stress.
Shown in the figure below is a solid cylinder. The x-axis is along the longitudinal direction....
Shown in the figure below is a solid cylinder. The x-axis is along the longitudinal direction. The angle o is measured with respect to the x-axis, positive if counterclockwise, i.e. towards the y-axis as shown below; and negative if clockwise. 01 and 02, with are the principal stresses, and Tmax is maximum shear stress. 01202 Fracture surface Normal to Fracture surface Suppose that the cylinder is subjected to a uniform tensile stress of 5000 psi in the longitudinal direction. The...
Shown in the figure below is a solid cylinder. The x-axis is along the longitudinal direction. The angle is measured with respect to the x-axis, positive if counterclockwise, i.e. towards the y-axis as shown below; and negative if clockwise. 01 and 02, with are the principal stresses, and Imax is maximum shear stress. 01202 Fracture surface Normal to Fracture surface Suppose that the cylinder is subjected to a uniform tensile stress of 5000 psi in the longitudinal direction. The orientation...
Question 6 0.5 points Save Answer • Shown in the figure below is a solid cylinder. The x-axis is along the longitudinal direction. The angle is measured with respect to the x-axis, positive if counterclockwise, i.e. towards the y-axis as shown below, and negative if clockwise. 01 and 02, with principal stresses, and Tmax is maximum shear stress. 012021 are the Fractute surface Normal to Fracture surface Suppose that the cylinder is subjected to a uniform tensile stress of 5000...
Shown in the figure below is a solid cylinder. The x-axis is along the longitudinal direction. The angle o is measured with respect to the x-axis, positive if counterclockwise, i.e. towards the y-axis as shown below; and negative if clockwise. 01 and 02, with 01202 , are the principal stresses, and Imax is maximum shear stress. Oz Y Fracture surface Normal to Fracture surface Suppose that the cylinder is subjected to a uniform tensile stress of 5000 psi in the...
Shown in the figure below is a solid cylinder. The x-axis is along the longitudinal direction. The angle o is measured with respect to the x-axis, positive if counterclockwise, i.e. towards the y-axis as shown below; and negative if clockwise. 01 and 02, with are the principal stresses, and Imax is maximum shear stress. 01202 Fracture surface Normal to Fracture surface Suppose that the cylinder is subjected to a uniform tensile stress of 5000 psi in the longitudinal direction. The...
Shown in the figure below is a solid cylinder. The x-axis is along the longitudinal direction. The angle is measured with respect to the x-axis, positive if counterclockwise, i.e. towards the y-axis as shown below; and negative if clockwise. 01 and 02, with 01202 are the principal stresses, and Imax is maximum shear stress. х 8 Fracture surface Normal to Fracture surface Suppose that the cylinder is subjected to a uniform tensile stress of 5000 psi in the longitudinal direction....
Shown in the figure below is a solid cylinder. The x-axis is along the longitudinal direction. The angle o is measured with respect to the x-axis, positive if counterclockwise, i.e. towards the y-axis as shown below; and negative if clockwise. 01 and o2, with are the principal stresses, and Imax is maximum shear stress. 01202 O X Fracture surface Normal to Fracture surface Suppose that the cylinder is subjected to a uniform tensile stress of 5000 psi in the longitudinal...
Shown in the figure below is a solid cylinder. The x-axis is along the longitudinal direction. The angle is measured with respect to the x-axis, positive if counterclockwise, i.e. towards the y-axis as shown below; and negative if clockwise. 01 and 02, with are the principal stresses, and Imax is maximum shear stress. 01202 One Fracture surface Normal to Fracture surface Suppose that the cylinder is subjected to a uniform tensile stress of 5000 psi in the longitudinal direction. Suppose...
Question 13 0.5 points Save Answer Shown in the figure below is a solid cylinder. The x-axis is along the longitudinal direction. The angle is measured with respect to the x-axis, positive if counterclockwise, i.e. towards the y-axis as shown below, and negative if clockwise. 01 and 02, with 01207 are the principal stresses, and Tmax is maximum shear stress. LO Fractute surface Normal to Fracture surface Suppose that the cylinder is subjected to a uniform tensile stress of sooo...
Shown in the figure below is a solid cylinder. The x-axis is along the longitudinal direction. The angle is measured with respect to the x-axis, positive if counterclockwise, i.e. towards the y-axis as shown below; and negative if clockwise. 01 and 02, with are the principal stresses, and Imax is maximum shear stress. 012 02' Fracture surface Normal to Fracture surface Suppose that the cylinder is subjected to a uniform tensile stress of 5000 psi in the longitudinal direction. The...