66 STRENGTH OF MATERIALS 



and consequently stress of only one sign occurs. For a point without 

 the core section the corresponding neutral axis cuts 

 and it is subjected to stresses of both signs. 



Problem 69. Construct the core section for a rectangular cross section of breadth 

 b and height h (Fig. 49). 



Solution. From Problem 35, I, = , I, = W and the correspond!, 

 gyration are t\ - | = ~ and t* = ^. Consequently, the semi-axes of the inertia 

 ellipse are t z = -~ and t u = =- Having constructed the inertia ellipse, the 

 vertices of the core section will be antipoles of the lines PQ, QK. 1: 



P Q 



\Y 



J 



FIG. 49 



FIG. 60 



From Article 59, the antipole of PQ is determined by th<- ivhui.mfM OE = OB 



since OE = - and OH = t, = -A= , OA = - . Similarly, OC = and OB = O7> 



2 2 V3 <* 



Thus the core section is the rhombus ABCD, of which tin- vertices ^1, B, C, D are 

 the antipoles of the lines PQ, PS, Sfl, Qfl respectively, and the sides AB, BC, CD, 

 DA are the antipolars of the points P, S, R, Q respectively. 



Problem 70. Construct the core section for the T-shape in Problem 60. 



Solution. Six lines can be drawn which will luivr two or more points in com- 

 mon with the perimeter of the T-shape without crossing it, namely, PQ. ','/.'. /.' /'. 

 TU, US, and SP (Fig. 50). The vertices A, B, C, D, E of the core section are 

 then the antipoles of these six lines respectively. 



Problem 71. Construct the core section of the I-beam in Problem 61. 



Problem 72. Construct the core section for the chanm-1 in Problt-i 



Problem 73. Construct the inertia ellipse and core section for a circular cross 

 section. 



62. Application to concrete and masonry construction. Since con- 

 crete and masonry are designed to carry only compressive stress 



