Question

100 kN/m 9-22. The T-beam is subjected to the distributed loading that is applied along its centerline. Determine the principal stress at point A and show the results on an element located at this point. 1 m 0.5 m 200 mm 20 mm 200 mm 75 mm

Answer 1

20 mm The location of the centroid c of the T cross-section, Fig. a, is 2A 0.1(0.2) (0.02) 0.21(0.02)(0.2) 0.155 m 0.2(0.02) 0.02(0.2) (0.02)(0.23) + 0.02(0.2)(0.155-0.1 2 12 l 0.2m (02)0.02)0.20.02)0.21 0.15 37.6667 (106) m 0.02m. Referring to Fig. b, QA A'0.1175(0.075)(0.02) 0.17625(103) m Using the method of sections and considering the FBD of the left cut segment of the beam, Fig. c V 100(1-0 V 100 kN 2Mc0 100(1)(0.5)-M = 0 M = 50 kN·m The normal stress developed is contributed by bending stress only. For point A, y = 0.155-0.075 = 0.08 m. Thus My 50(10) (0.08) I37.666710) = 106 MPa The shear stress is contributed by the transverse shear stress only. Thus, Ve 100(0 10.17625(0 23 40(106) pa = 23 40 MPa

The state of stress of point A can be represented by the element shown in Fig, c. Here, σχ--106. 19 MPa, σ,-0 and τΧ,-23.40 MPa σ1,2 ry -106.19 0 106.1923.40 =-53.10 ± 58.02 arī 4.93 MPa σ2 =-111 MPa Ans.

tan 2θ,-(ox-dry)/2 23.40 (-10619-0)/2 -0.4406 (o - o,)/2 (106.19 011.89 as 78.11 Substitute θ =-11 .89", cos 2θ + Txy sin 2 -106.19 + 0-106. 19-0 cos (-23.78) 23.40 5m (-23.78°) =-111.12 MPa = σ Thus, p78.1 p21.9 Ans. The state of principal stress can be represented by the element shown in Fig. e #0-55m ト0.02 m Cb)