Page 1 Strength of Materials Prof. M. S. Sivakumar Indian Institute of Technology Madras Stresses Stress at a point Stress Tensor Equations of Equilibrium Different states of stress Transformation of plane stress Principal stresses and maximum shear stress Mohr's circle for plane stress Page 2 Strength of Materials Prof. M. S. Sivakumar Indian Institute of Technology Madras Stresses Stress at a point Stress Tensor Equations of Equilibrium Different states of stress Transformation of plane stress Principal stresses and maximum shear stress Mohr's circle for plane stress Strength of Materials Prof. M. S. Sivakumar Indian Institute of Technology Madras Introduction 2.1 stress at a point Figure 2.1 Consider a body in equilibrium under point and traction loads as shown in figure 2.1. After cutting the body along section AA, take an infinitesimal area ?A lying on the surface consisting a point C. The interaction force between the cut sections 1 & 2, through ?A is ?F. Stress at the point C can be defined, A0 F lim A ?? ? s= ? 2.1 ?F is resolved into ?F n and ?F s that are acting normal and tangent to ?A. Normal stress, n n A0 F lim A ?? ? s= ? 2.2 Shear Stress, ?? ? s= ? s s A0 F lim A 2.3 Top Page 3 Strength of Materials Prof. M. S. Sivakumar Indian Institute of Technology Madras Stresses Stress at a point Stress Tensor Equations of Equilibrium Different states of stress Transformation of plane stress Principal stresses and maximum shear stress Mohr's circle for plane stress Strength of Materials Prof. M. S. Sivakumar Indian Institute of Technology Madras Introduction 2.1 stress at a point Figure 2.1 Consider a body in equilibrium under point and traction loads as shown in figure 2.1. After cutting the body along section AA, take an infinitesimal area ?A lying on the surface consisting a point C. The interaction force between the cut sections 1 & 2, through ?A is ?F. Stress at the point C can be defined, A0 F lim A ?? ? s= ? 2.1 ?F is resolved into ?F n and ?F s that are acting normal and tangent to ?A. Normal stress, n n A0 F lim A ?? ? s= ? 2.2 Shear Stress, ?? ? s= ? s s A0 F lim A 2.3 Top Strength of Materials Prof. M. S. Sivakumar Indian Institute of Technology Madras 2.2 stress Tensor Figure 2.2 Consider the free body diagram of an infinitesimally small cube inside the continuum as shown in figure 2.2. Stress on an arbitrary plane can be resolved into two shear stress components parallel to the plane and one normal stress component perpendicular to the plane. Thus, stresses acting on the cube can be represented as a second order tensor with nine components. xx xy xz yx yy yz zx zy zz ? ? ss s ? ? s= s s s ? ? ? ? ss s ? ? ? ? 2.4 Is stress tensor symmetric? Figure 2.3 Page 4 Strength of Materials Prof. M. S. Sivakumar Indian Institute of Technology Madras Stresses Stress at a point Stress Tensor Equations of Equilibrium Different states of stress Transformation of plane stress Principal stresses and maximum shear stress Mohr's circle for plane stress Strength of Materials Prof. M. S. Sivakumar Indian Institute of Technology Madras Introduction 2.1 stress at a point Figure 2.1 Consider a body in equilibrium under point and traction loads as shown in figure 2.1. After cutting the body along section AA, take an infinitesimal area ?A lying on the surface consisting a point C. The interaction force between the cut sections 1 & 2, through ?A is ?F. Stress at the point C can be defined, A0 F lim A ?? ? s= ? 2.1 ?F is resolved into ?F n and ?F s that are acting normal and tangent to ?A. Normal stress, n n A0 F lim A ?? ? s= ? 2.2 Shear Stress, ?? ? s= ? s s A0 F lim A 2.3 Top Strength of Materials Prof. M. S. Sivakumar Indian Institute of Technology Madras 2.2 stress Tensor Figure 2.2 Consider the free body diagram of an infinitesimally small cube inside the continuum as shown in figure 2.2. Stress on an arbitrary plane can be resolved into two shear stress components parallel to the plane and one normal stress component perpendicular to the plane. Thus, stresses acting on the cube can be represented as a second order tensor with nine components. xx xy xz yx yy yz zx zy zz ? ? ss s ? ? s= s s s ? ? ? ? ss s ? ? ? ? 2.4 Is stress tensor symmetric? Figure 2.3 Strength of Materials Prof. M. S. Sivakumar Indian Institute of Technology Madras Consider a body under equilibrium with simple shear as shown in figure 2.3. Taking moment about z axis, () ( ) zyxxz y xyyz x Mddd ddd =t - t =0 xy yx t =t Similarly, . xz zx yz zy and t=t t =t Hence, the stress tensor is symmetric and it can be represented with six components, , instead of nine components. xx yy zz xy xz yz , , , , and ss s t t t xx xy xz xx xy xz yx yy yz xy yy yz zx zy zz xz yz zz ??? ss s s t t ??? s= s s s = t s t ??? ??? ss s t t s ??? ??? ? ? ? ? ? ? 2.5 Top Page 5 Strength of Materials Prof. M. S. Sivakumar Indian Institute of Technology Madras Stresses Stress at a point Stress Tensor Equations of Equilibrium Different states of stress Transformation of plane stress Principal stresses and maximum shear stress Mohr's circle for plane stress Strength of Materials Prof. M. S. Sivakumar Indian Institute of Technology Madras Introduction 2.1 stress at a point Figure 2.1 Consider a body in equilibrium under point and traction loads as shown in figure 2.1. After cutting the body along section AA, take an infinitesimal area ?A lying on the surface consisting a point C. The interaction force between the cut sections 1 & 2, through ?A is ?F. Stress at the point C can be defined, A0 F lim A ?? ? s= ? 2.1 ?F is resolved into ?F n and ?F s that are acting normal and tangent to ?A. Normal stress, n n A0 F lim A ?? ? s= ? 2.2 Shear Stress, ?? ? s= ? s s A0 F lim A 2.3 Top Strength of Materials Prof. M. S. Sivakumar Indian Institute of Technology Madras 2.2 stress Tensor Figure 2.2 Consider the free body diagram of an infinitesimally small cube inside the continuum as shown in figure 2.2. Stress on an arbitrary plane can be resolved into two shear stress components parallel to the plane and one normal stress component perpendicular to the plane. Thus, stresses acting on the cube can be represented as a second order tensor with nine components. xx xy xz yx yy yz zx zy zz ? ? ss s ? ? s= s s s ? ? ? ? ss s ? ? ? ? 2.4 Is stress tensor symmetric? Figure 2.3 Strength of Materials Prof. M. S. Sivakumar Indian Institute of Technology Madras Consider a body under equilibrium with simple shear as shown in figure 2.3. Taking moment about z axis, () ( ) zyxxz y xyyz x Mddd ddd =t - t =0 xy yx t =t Similarly, . xz zx yz zy and t=t t =t Hence, the stress tensor is symmetric and it can be represented with six components, , instead of nine components. xx yy zz xy xz yz , , , , and ss s t t t xx xy xz xx xy xz yx yy yz xy yy yz zx zy zz xz yz zz ??? ss s s t t ??? s= s s s = t s t ??? ??? ss s t t s ??? ??? ? ? ? ? ? ? 2.5 Top Strength of Materials Prof. M. S. Sivakumar Indian Institute of Technology Madras 2.3 Equations of Equilibrium Figure 2.4 Consider an infinitesimal element of a body under equilibrium with sides dx as shown in figure 2.4. dy 1 × × B x , B y are the body forces like gravitational, inertia, magnetic, etc., acting on the element through its centre of gravity. x F0 = ? , () () () () ( ) yx xx xx y xx yx yx x dx d 1 dy 1 dy dx 1 dx 1 B dx dy 1 0 xdy ?t ?? ?s ?? s+ × -s × +t + × -t × + × × = ?? ?? ? ?? ?? Similarly taking and simplifying, equilibrium equations of the element in differential form are obtained as, y F 0 = ? yx xx x xy yy y B0 xy B0 xy ?t ?s + += ?? ?t ?s + += ?? 2.6 Extending this derivation to a three dimensional case, the differential equations of equilibrium become,Read More

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