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4.47 The reinforced concrete beam shown is subjected to a positive bending moment of 175 kN • m. knowing that the modulus of elasticity is 25 GPa for the concrete and 200 GPa for the steel, determine (a) the stress in the steel, (b) the maximum stress in the concrete.

4.47 The reinforced concrete beam shown is subjected to a positive bending moment of 175 kN • m. knowing that the modulus of elasticity is 25 GPa for the concrete and 200 GPa for the steel, determine (a) the stress in the steel, (b) the maximum stress in the concrete.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Problem 4.47 4.47 The reinforced concrete beam shown is subjected to a positive bending moment of 175 kN · m. knowing that the modulus af elasticity is 25 GPa for the concrete and 200 GPa for the steel, determine (a) the stress in the steel, (b) the maximum stress in the concrete. 450 mm 22-mm diameter E: 200 GPa 25 GPa 8.0 50 mm - 250 mm A, = 4. Td - (4>(E)(22}* = 1.5205 *10* mm %3D 250 + nAs = 12. 164 x10 mm 'Locate the neutral axis Qxis 250 x - (12.164 */0*)(00 -x ) 400 125 x? + 12. 164 x10 x 4, 3657 x10“ = Solving for x - 12.164 x10 +(12.164x t) + (4}(125){4. 8657 * 106 X = (2)(125) 154.55 mm, 400 - x 245. 45 mm I = 3 250 x + (12.164 x10° )(400 - x )* * (250)(|54.55)+ (12. 164 *10*)(2+5.4s ,* = 1.0404 x jo" mm 1. 0404 x l0 m* %3D n My 6 = (a) Steel : 245.45 mm - 0.245 45m 6 = - 8.0)(175 xl0*)(-0.24545) 1.0404 x 103 330x10* Pa - 330 MPa %3D (6) Concrete : : 154.55 mm O.15455 m y = (1.01(175x10)(0.15455) 1.0404 x l0-3 6:- - -26,0 x 10° Pa = -26.0 MPa
Transcribed Image Text:Problem 4.47 4.47 The reinforced concrete beam shown is subjected to a positive bending moment of 175 kN · m. knowing that the modulus af elasticity is 25 GPa for the concrete and 200 GPa for the steel, determine (a) the stress in the steel, (b) the maximum stress in the concrete. 450 mm 22-mm diameter E: 200 GPa 25 GPa 8.0 50 mm - 250 mm A, = 4. Td - (4>(E)(22}* = 1.5205 *10* mm %3D 250 + nAs = 12. 164 x10 mm 'Locate the neutral axis Qxis 250 x - (12.164 */0*)(00 -x ) 400 125 x? + 12. 164 x10 x 4, 3657 x10“ = Solving for x - 12.164 x10 +(12.164x t) + (4}(125){4. 8657 * 106 X = (2)(125) 154.55 mm, 400 - x 245. 45 mm I = 3 250 x + (12.164 x10° )(400 - x )* * (250)(|54.55)+ (12. 164 *10*)(2+5.4s ,* = 1.0404 x jo" mm 1. 0404 x l0 m* %3D n My 6 = (a) Steel : 245.45 mm - 0.245 45m 6 = - 8.0)(175 xl0*)(-0.24545) 1.0404 x 103 330x10* Pa - 330 MPa %3D (6) Concrete : : 154.55 mm O.15455 m y = (1.01(175x10)(0.15455) 1.0404 x l0-3 6:- - -26,0 x 10° Pa = -26.0 MPa
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