Question

Show all work and steps please.

*6-80. If the beam is subjected to a moment of M 100 kN m, determine the bending stress at points A B, and C. Sketch the bending stress distribution on the cross section

Answer 1

solution: - Juo mm criven as M- IOU KNM. TM loux.lo3 N.m] 30 mm. CO's B150 mm * - I po isomm X - [fig (a) [STEP-I – First we will consider the x-x' as - reference line as shown in fig (a). - Now, we will find out the centre of geanity (CG) af the given section. The given section is symmetrical about the axis y-y so that the ca y the given section will lie on y_y axis. - Now this section is split up into two rectangles PORS & uuwt as shown's Let y = Distance of the c.Go of the section from - the bottom line (reference line xx') PS. a = Aread rectangle PORS = (150 tiso) x 30 - 300 X30 = gooo mom Y = Distance of c.o.of the ares a fogom the bottom line (x-x') = 30 = 15 mm, az = Are y rectangle CUMWT) = 30x 300 = gooo many Y2 = Distance of c.a. of rectangle (Urwt) from the bottom line (-2') = 30 + 300 = 30 +150 = ldoman

a, y + Q₂Y2 9,+Q2 put all values in equation to ③ - ý = gooox ist 9000 groux 180 9000 13 SOU - + 1 620000 18000 175 5000 18000 J = 97.5 mm. 4 Neutralaxts 97.5 y fig (6) STEP-II/ In next step, we will calculate the moment of inestia of the cross-section about the neutral axis - Let In = moment of inestia of rectangle (Cors) about the hulizental mis and passing through its cia. Ia = moment of inentia of Rectangle (rowt) about the horizontal anys and passing through the C.G.

hi= The distance between the coco of the a given section and the c.a. af the rectangle (PQRS). h-(9,-5) = 5-y, = (97.5-15) The = 82.5 mm. h2 = The distance between the car of the given Section & the coc. of the rectangle (vw) hp = 9, -9 = 180 - 97.5 = 82 Tnh. m = 82.5 mm. Now- 300 x(30)3 = 6,74000 mm NOW- • 6.750 12 12 12 Ia, = 675000 mma l Il = 30 x (300) 3 = 67500000 mm" Ia = 675,00,000 mm" flom the parallel axis theorem, the moment of inertia of the rectangle (PORs) about the holigental axis passing through the carry the given section- = In + a, hy2 = 675000 + 9000 ( 82.5) 2 6193 1250 mm 4

similarly, the m. It of the rectangle 4 (UUwT) about the holizontal are's passing through the car of the given section. = In + a2 hz ? = 67500000 + 90oo (82.5)? = 128756250 mmy - Thus the m.Ir of the cross-section about the newhal anis is, Ixx = 61931250 + 128756250 Ixx = 190687500 mm = 0.190685 Ixx = 0.1907X10-3 m47 STEP- II In next step, we will find out - the bending stress- - The distance from the neutral anis to points A, B, & cis calculaed as follows: YA = 330 – 97.5 = 232.5 mm = 0.233 m YB = 97.5 mm = 0.0975m. y = 975-30 = 67.5 mm = 0.0675m.

- [STER TV] – - 7 now using the heading equation. 5 - - ramy MYA = (100 x103) * 0.233 0.1907810-3 = 0.233 x 10 5x103 0.1907 = 122-18 X106 Nl. m2 VA = 122.18 MPa. Ol wmm2 / to complessive.) Ane Similarly, TB = MYB = = (100x103) x 0.0975 0.1967 x 10-3 0.0975% 10 5x103 0.1907 51.13 x 106 N/m2 51.13 MPa (N) mm2) / Ane = 1 TB = c = Mye - (100x103) X(0.0675) 0.19078103 = 0.0675 X 105 x 103 0.1507 = 35.4 x 106 N/m2 To = 35.4 MPa od (N/mm2) | Ais.

STEP-T Now using these values , we will construct the lending stress distribution across the cous- section is shown in fig (c) -NA (Netral axis) fig (c) - x -