Question

QI A simply supported beam of length L = 10 m carries a uniformly distributed load w of 10 kN/m, as shown in Figure QI (a). The beam is made from a symmetrical I-section and consists of three equal rectangular members of 100 mm x 10 mm (see Figure QI (b)). Self- weight of the beam is neglected. 100 mm w = 10 kN/m A-1 A 100 mm 1/3 L/3 L/3 m [5] Figure Q1 (a) Figure Q1 (b) (a) Find the maximum shear force V in the beam. (b) Using the value of V from (a), derive a general formula for the computation of the shear stress distribution in the cutting line A-A. [15] (c) Apply the formula of (b) to determine the maximum vertical shear stress and also the vertical shear stress in the web at the web/flange intersection. [5]

Answer 1

W = 10 kN/m A RA Watt 1/3 ½ 1/3 X RB L= lom (0) Maximum Shear Force For equilibrium {Fy=O, RATRB Elox 10 = 33.3 KN 3 EMB=0 RAX10 = 10x IORA 166.67 RA 166 67 = 16,667 kn RB = 33.3-16-667 RA = 16 -833 –N Shear Force Calculations Shear Force at A.SA = +16.633 KN SC = 16.633 KN SD = 16.633-10 oxlog Sp -- 6. 657 VN SB = -16.667 KN So maximum Shear force Smax= 16.667 KN Are

(6) General Formula for the computation of shear stress distribution A-A b d II Dh NA. D -A' sly e al y = / first moment of Area Q = Ary+ A262 A = bd gy (%+4) (042) Az = dxy = dy 22 = 8/ Q = ba (pred) + dyx / bd ( 04d) 1 dy? dez shear stress at A-A. Ebel Ib b=d 2 be cond] + d32 v I a 2 Id ta v [b@7+d42 2 I where I = moment of snertia

(c) loomm ITO 1 100 150 To 60 mm 10 I 100mm ☺ - 100+lotlo = 60 mm 2 V= 16.667 KN - 16.667x103N AY+ A2% + 50x10 () = loo xlo (50+40) + 50x10 f [ " Emax Q = 6750 0 mm Moment of hertio T = 100 x 1203 2 P1041503 12 12 T: 12733 333.3 mmų, b= 10 mm Maximum Shear stress Emoxava 16-667x103 x 67500 Exb 12733333.3 x 10 = 8.835 N/mmy And Shear Stress cut Joint and Flangs Q = Aj - l'Ooxlox(50+)-55000mm VQ G (by =100mm I by 16.667x103 x 55000 = 0.72 m Pa 12733333-3x100 b2 (b2=100m) bi T2 = 0.72x100 tub la = 7.2 โอ Ary mm3 NINAS 2 mda