Question

Problems 225 4.7 through 4.9 Two vertical forces are applied to a beam of the cross section shown. Determine the maximum tensile and compressive stresses in portion BC of the beam. 75 mm 200 mm 75 mm PT- 25 mm 150 mm 25 mm- 150 mm 50 mm ne, de- 100 mm 60 KN 60 KN 100 KN 100 KN А в C D 1.5 m- Im 5 m lm Fig. P4.7 0.5 m Fig. P4.8 10 mm 10 mm 10 KN 10 KN VUL 50 mm 10 mm aminum 250 mm 150 mm f safety when it 150 mm --50 mm - Fig. P4.9 log of magnitude M = 25 KN . m are

4.7 and 4.8 how to find the moment and solve !?

Answer 1

I have provided detailed solution of 4.7, kindly follow similar procedure for 4.8 if you find any difficulty let me know.
h. Due to sy monetry CON GON RA R = sokN. I'm 1. 5m | Im Ro w COIN Bending Moment in AB (x from A Mx = RAID (sagging tre Mx = 60x MB = 60 kN-m. Bending Moment in BC (a from B) Mma RA (1+x) - 60x, (SPD) -GOKN GOKN-m. So max. Bending moment Mnax a 60kN-m Centroid A = 45*3)*50 = 11250 mm = 95 mm A 75 X/50 = 11250 mm , - (75+50 ) = 125 mm 75mm. 75mm 150 Y yo- Arh+A24 11950 (25+R 3X1250 ATA2 Moment of Prestia IOG. da a 2, + A CI + 2 + A2 (992 = x (765x3)% 50+ 11.250 (75-35) + xsx150° +11.950 (125-73 - 79687500 munte A the moment is the so sagging will be there, so fibre will be in compression and bottom Abre will be tension. Note: top in

Compressive stress - их Зу = во хто“ N-мт х (200 – 16) 1 96895TO tensile Г = 4ч. 111MPa . stress м у = Geyia“ x 15 15 4800 15 суд во | = 56.40 MP4 |