Question

Q1 A cantilever steel beam of length L = 7.5 m carries both a uniformly distributed load w of 20 kN/m throughout its length and a point load P of 10 kN at its free end, as shown in Figure Q1 (a). The beam is made from a rectangular hollow box section with a width of 300 mm and a depth of 450 mm (refer to Figure Q1 (b)). The wall thickness of the box section is constant throughout which is 20 mm. Self-weight of the beam is neglected. 300 mm P= 10 kN w = 20 kN/m 450 mm 7.5 m 1 20 mm Figure Q1 (a) Figure Q1 (b) (a) Determine the maximum bending moment of the beam. [5] (b) Calculate the maximum bending stress of the beam and draw the corresponding bending stress distribution across the section. [10] (c) If the beam is subject to an additional axial tensile load of 500 kN at its free end, calculate the maximum longitudinal stresses and draw the resulting stress distribution across the section. [10]

Answer 1

IOKN 2011/m B e 7.5 T X Consider a at a - ; distance of Section 'X-X from free end. .x OKN 120klyn 2- X S.F- - Shear force at 10+ (20xx). Litve gave = 10 + 20% AF I so > SES 107.2010) = 10KN At X375 sf = 10+2017.5) - 160KN. Bending moments At x = (10x x) – (26x2x3) 10X-100 At x=0 i) BM=0 2 At x=75., B.M = - 10(7,5).10(7.5) -637.5 kNam alme 2

2 IOKN 20 KN m 7.5 160 SED 10 120 CM 21IUHAOSFD 0 0 BMD LILLLLL -637.5 a) Maximum. bending bending Moment g the beam : 637.5km 6). Bending Bending shen distribution 11 Elim cela '] Here I- Ixx Y = distance from neatral axes Ime 2

3 300 BB bat 그자 12 ( 150 20) ( 3 3 3COX 1503 260 6 12 12 1 20mmo U --7,84 xro9 mm At top fibres 10 150 로 225 g H13 이 이 M .xy x 225 엄 637.5X168 168 1.8 68 ) tension Hilaman 2 182.95 엄 1. AH Neutral axis Seo .. JE0 At bottom fible! As y'is Same | 05 -182.95 Compression

14 182.865W/ y F Neutral axis -18295/mmy Bending stress distribution c) Additional 500 KN (Axial)! 500X18 Area Area? (2x450x 20) + (2x20x260) ✓ = 28400 mm 500x10B ON 28400 (Gension H/ mm =1706

5 Resultant Strom distribution ! all top ! 182,95+:17.6 op > 200.55 Nm At Pony Neutral Arus! R=ot 17.6 = = 17.6 N/mm Bottom ! all PR=-182.95 + 1746 = -165-35 N/mm Stress distributions 200.5 N/mmy 11 7 Neutral äris $17.6.1/my -165.35 N/mma