2023年全國碩士研究生考試考研英語一試題真題(含答案詳解+作文范文)_第1頁
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1、Apago PDF EnhancerApago PDF Enhancer695 11.2 Strain EnergyLet us now consider the work dU done by the load P as the rod elongates by a small amount dx. This elementary work is equal to the product of the magnitude P of

2、 the load and of the small elon- gation dx. We writedU 5 P dx (11.1)and note that the expression obtained is equal to the element of area of width dx located under the load-deformation diagram (Fig. 11.3). The total w

3、ork U done by the load as the rod undergoes a deforma- tion x1 is thusU 5 #x10P dxand is equal to the area under the load-deformation diagram between x 5 0 and x 5 x1.The work done by the load P as it is slowly applied

4、to the rod must result in the increase of some energy associated with the defor- mation of the rod. This energy is referred to as the strain energy of the rod. We have, by definition,Strain energy 5 U 5 #x10P dx (11.2

5、)We recall that work and energy should be expressed in units obtained by multiplying units of length by units of force. Thus, if SI metric units are used, work and energy are expressed in N ? m; this unit is called a

6、joule (J). If U.S. customary units are used, work and energy are expressed in ft ? lb or in in ? lb.In the case of a linear and elastic deformation, the portion of the load-deformation diagram involved can be represent

7、ed by a straight line of equation P 5 kx (Fig. 11.4). Substituting for P in Eq. (11.2), we haveU 5 #x10kx dx 5 1 2 kx2 1orU 5 1 2P1x1 (11.3)where P1 is the value of the load corresponding to the deformation x1.PO xFig

8、. 11.2 Load-deformation diagram.PP U ? AreaOx x x1dxFig. 11.3 Work due to load P.PP ? kxU ? P1x1x1 xP1O1 2Fig. 11.4 Work due to linear, elastic deformation.bee80288_ch11_692-758.indd Page 695 11/12/10 5:12:23 PM user-

9、f499 bee80288_ch11_692-758.indd Page 695 11/12/10 5:12:23 PM user-f499 /Users/user-f499/Desktop/Temp Work/Don't Delete Job/MHDQ251:Beer:201/ch11 /Users/user-f499/Desktop/Temp Work/Don't Delete Job/MHDQ251:Beer:

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