Question

Calculate the maximum deflection of the steel beam (E=200 GPa), if a=0.6 m and P=14 kN. Take I = 102 x 100 mm4. Write your answer in mm to 2 decimal places. P 0 |Α C IB + a a a a

Answer 1

Given that, Elastic modulus (El: 200 G Pa Moncent of Inertia (I): 102x106mma since MS-Eldy dur leted 14kw 14N Maximum deflection IC B OCCENS at midpoint A 0.6 06 0.6 0.6

Me = [RA XX – 14 (N=0.6) - 1 uba-rol) – 14(n=12) Dret redan Eidy = 210 – 14(2-0.6)-14(usive) - 14{ rs du El dy 211 + (-0.6) f14fac12)-14 (n=1-8) da * ( 24 7 q (u + 3)="nx(2-407 if nios if WAL ifurtos +ch tecto -Y = (1 - (oceli da sole tght CL At n=0; y=0 =) C=0 At *24; 920 0- * ( (6a2_14_0039 _ty (ms) - tea (out) EL + C (2.0) G 14 (3+6.176.612 zalewa २५ 4:-1.26

Maximum deflection occurs at c (4312) SU 1.2 g1 EI 21 avna 1-26k 14 (60.612_1-26 (1927-44100 12602] EL 3 Ely 2-9.032 xló N-m 3 , y: -9.03L810 200x109 x 102 x 10°x x 10% 10% ge -9.032 m V-0.469 2008 IOL Therefore deflection atc (41= -0.469 mm il Manimum deflection

-Demand V = 21 kN ction 8 = -0.469 mm Reactions Uit 21 Uit 21 00 0.5 1.0 2.0 15 Distance from left of Beam (m) . Shear 20 10 Shear (N) D 0.5 2 .10 Moment 15 10 Moment (1) 5 D 0.5 1.5 2 Deflection 0.5 1.5 0 -0.1 -02 Dellection (arn) -03 x 1.19 Detection - 269 patch 2 14 12 KN 21 kN 21 KN 0.5 20 1.0 1.5 Distance from Le of Beam (m)

So maximum deflection = 0.469 mm

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