9780203009512.ch5

9780203009512.ch5 - 0749_Frame_C05 Page 73 Wednesday,...

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73 Mechanical Equilibrium and the Principle of Virtual Work 5.1 TRACTION AND STRESS 5.1.1 C AUCHY S TRESS Consider a differential tetrahedron enclosing the point x in the deformed configura- tion. The area of the inclined face is dS , and dS i is the area of the face whose exterior normal vector is e i . Simple vector analysis serves to derive that n i = dS i / dS (see Exercise 1 in Chapter 1). Next, let d P denote the force on a surface element dS , and let d P ( i ) denote the force on area dS i . The traction vector is introduced by τ = d P / dS . As the tetrahedron shrinks to a point, the contribution of volume forces, such as inertia, decays faster than surface forces. Balance of forces on the tetrahedron now requires that . (5.1) The traction vector acting on the inclined face is defined by , (5.2) from which (5.3)± 5 dP dP jj i i = τ= d dS P τ j j i i j i i i i ij i dP dS dP dS dS dS Tn = = = () . © 2003 by CRC CRC Press LLC
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74 Finite Element Analysis: Thermomechanics of Solids with . (5.4) It is readily seen that T ij can be interpreted as the intensity of the force acting in the j direction on the facet pointing in the i direction, and is recognized as the ij th entry of the Cauchy stress T . In matrix-vector notation, the stress-traction relation is written as (5.5) The next section will show that T is symmetric by virtue of the balance of angular momentum. Equation 5.5 implies that T T is a tensor, thus, it follows that T is a tensor. To visualize T , consider a differential cube. Positive stresses are shown on faces pointing in positive directions, as shown in Figure 5.2. In traditional depictions, the stresses on the back faces are represented by arrows pointing in negative directions. However, this depiction can be confusing—the arrows actually represent the directions of the traction components. Consider the one-dimensional bar in Figure 5.3. The traction vector t e 1 acts at x = L , while the traction vector t e 1 acts at x = 0. At x = L , the corresponding stress is t 11 = t e 1 e 1 = t . At x = 0, the stress is given by ( t e 1 ) ( e 1 ) = t . Clearly, the stress at both ends, and in fact throughout the bar, is positive (tensile). We will see later that the stress tensor is symmetric by virtue of the balance of angular momentum. FIGURE 5.1 Equilibrium of a tetrahedron. 1 3 dF (3) (2) n (1) dS 2 T dP dS ij j i i = () τ= T T n © 2003 by CRC CRC Press LLC
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Mechanical Equilibrium and the Principle of Virtual Work 75 5.1.2 1 ST P IOLA -K IRCHHOFF S TRESS The transformation to undeformed coordinates is now considered. In Chapter 3, we saw that (5.6) is known as the 1st Piola-Kirchhoff stress tensor, and it is not symmetric. FIGURE 5.2 Illustration of the stress tensor. FIGURE 5.3 Tractions on a bar experiencing uniaxial tension. 1 2 3 T 33 32 23 21 12 11 13 31 22 Y Z t X τ Τ Τ dS dS dS dS 00 = = == T T SS T n JF n nJ F T T 1 ,.
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This note was uploaded on 05/18/2011 for the course MAE 269A taught by Professor Ju during the Spring '11 term at UCLA.

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9780203009512.ch5 - 0749_Frame_C05 Page 73 Wednesday,...

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