神马品牌网去他个小饼干截图:哪位能够帮助我翻译一下机械专业英语,
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When developing the flexure formula we imposed a condition that the cross-sectional area be symmetric about an axis perpendicular to the neutral axis; furthermore, the resultant internal moment M acts along the neutral axis. Such is the case for the “T” or channel sections shown in Fig. 6-31. These conditions, however, are unnecessary, and in this section we will show that the flexure formula can also be applied either to a beam having a cross-sectional area of any shape or to a beam having a resultant internal moment that acts in any direction.
Moment Applied Along Principal Axis. Consider the beam’s cross section to have the unsymmetrical shape shown in Fig.6-32a. as in Sec.6.4, the right-handed x,y,z coordinate system is established such that the origin is located at the centroid C of the cross section, and the resultant internal moment M acts along the +z axis. We require the stress distribution acting over the entire cross-sectional area to have a zero force resultant, the resultant internal moment about the y axis to be zero, and the resultant internal moment about the z axis to equal M.* These three conditions can be expressed mathematically by considering the force acting on the differential element dA located at (0,y,z), Fig.6-32a. this force s dF＝σdA, and therefore we have
(6-14)
(6-15)
(6-16)
Moment Applied Along Principal Axis. Consider the beam’s cross section to have the unsymmetrical shape shown in Fig.6-32a. as in Sec.6.4, the right-handed x,y,z coordinate system is established such that the origin is located at the centroid C of the cross section, and the resultant internal moment M acts along the +z axis. We require the stress distribution acting over the entire cross-sectional area to have a zero force resultant, the resultant internal moment about the y axis to be zero, and the resultant internal moment about the z axis to equal M.* These three conditions can be expressed mathematically by considering the force acting on the differential element dA located at (0,y,z), Fig.6-32a. this force s dF＝σdA, and therefore we have
(6-14)
(6-15)
(6-16)
当发展我们强加了一种情况代表性的区域是对称的有关对中立的轴轴垂直线的事屈曲公式的时候; 此外, 合量内在的片刻 M 沿着中立的轴行动。 如此的是 " T" 或频道在图 6-31 中被显示的区段情形. 这些情况,然而, 是不必要的,而且在这一个区段中我们将会表示屈曲公式也能被应用或到光线有一个任何形状的代表性区域或对光线有一个合量内在的片刻以在任何的方向中行动。
片刻沿着主要的轴应用。 考虑光线的交叉区段有在 Fig.6-32 中被显示的非对称的形状一。当在 Sec.6.4 ，惯用右手的 x ， y 中， z 同等的人物系统是确定的以致于起源位于交叉的区段图心 C ，而且合量内在的片刻 M 行动之时沿着那 +z 轴。 我们需要遍及整个的代表性的区域行动有一个零的力量合量的压迫力分配, 合量内在的片刻有关 y 轴的事是零的, 和合量内在的片刻有关 z 轴的事等于 M。*这三种情况能被藉由考虑对位于的差别元素 dA 有所反应的力量， Fig.6-32 算术地表达一。 这力量 s dF ＝σ dA, 因此我们有
(6-14)
(6-15)
(6-16)