Question

4. A T-shaped cross-sectional beam is loaded as shown in the figure. Determine the following a. Sketch the internal shear force and bending moment diagrams for the beam. b. Calculate the maximum magnitude of the bending stress. Indicate where this occurs on the cross-section and along the length of the beam. c. Calculate the transverse shearing stress at the centroid of the cross-section using the maximum magnitude of the transverse shear force. - 200 mm 8 KN 1.5 kN/m 20 kN.m 1 75 mm LIV . 1187 250 mm 2m >> im > 2m > 3m 100 mm

Answer 1

solution 200mm 1.5KN/m 20kalom Sk 250m RAN PRAY de k k 20 IM 29 30 Efy to (RAY) + (RB) y = 8+4.5 - 12-5kN 100mm MATO - 20 -(883) + PB) yx5-(4:5) x6.5) 70 | Le B) y = 14.65 kN) 14.65 2.15KN RAY = 12-5- 200mm 75mm 250mm k roomm SED & BRD

1.5 kilm 20kn-m SYN RY) = 2.15KN (Ry) A 2m 2 m 3m 14.65KN 4.50KN) SEN u o - 2.15KN 0-5 7-10.15KN mansman shear at r=3 to In m) - 6.75 IN-B - 4.3 kN/m BmD 113.55KN 115-7 kalah maur mums bending at a=25

Now mom cut of of t seetion Inertice 2.00 mm) 10W 75mm mm 250mm 185.93 Yoo R mm It is Symmetrie 4 about y-anis A 250x 100 - 25000mm A2 = 75x200 = 15000 mm VIF 250 - 125mm Yn 250+ 75 = 287.5mm y - Aly Ary - 185.93mm. AITA. Tag = bh + Ad? 100x250)3 - 100X150) X(185, 932 n 9944 57455.8 mm 4 I rege, = bb3 + And² – 200 x 75 3+ 200x95 x Qol. 5772 (101.5922 - 161778223.5 man I 1156 23.5 679mmt.

A # G # (200*75) X(64+75) |$22500 mm b) maamum bendung moment = (m) 15-7kalom at n = 20 manim bending stress (on) - Mz xy) = 15.7 x 64.67475 1156235679 x 10 mg 15.7x10 3 x (101.57) x 10 (m) 11562 356 79 & 10-12 1379.17 2.974 N/m2 - 1.37 mpa manimum shearing stress, (7) - va – 10.15 x103 x 1522500 X10 9 m3 TXt = 115623567981012 10.15X10° X 1822500X10-9c lookm4 11$6235679x10512 X 100X10-3 133652.4662 N/m2 1.13652 mpa