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Solution : Е, 200(10°) E 11(10° The width of steel must be increased to an equivalent width for wood: Page 1 -18.182 mm 1 width (A |(10³) m² Failure of wood: Mc 0.5MA 20(10°) = I M = kNm Failure of steet I 130(10) – (18.182 (0.5MA). М- Ans

Solution : Е, 200(10°) E 11(10° The width of steel must be increased to an equivalent width for wood: Page 1 -18.182 mm 1 width (A |(10³) m² Failure of wood: Mc 0.5MA 20(10°) = I M = kNm Failure of steet I 130(10) – (18.182 (0.5MA). М- Ans

Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Barry J. Goodno, James M. Gere
Chapter6: Stresses In Beams (advanced Topics)
Section: Chapter Questions
Problem 6.10.17P
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Solution : E 200(10°) E 1110° ) The width of steel must be increased to an equivalent width for wood: Page 1 =18.182 W. mm 1 I = width (A |(10*) m² Failure of wood: Mc (6. I 0.5MA 20(10° ) = M = ANm I Failure of steel: nMc st /max I (18.182)0.5MA). 130(10° ) – Nm M = Ans I
Transcribed Image Text:Solution : E 200(10°) E 1110° ) The width of steel must be increased to an equivalent width for wood: Page 1 =18.182 W. mm 1 I = width (A |(10*) m² Failure of wood: Mc (6. I 0.5MA 20(10° ) = M = ANm I Failure of steel: nMc st /max I (18.182)0.5MA). 130(10° ) – Nm M = Ans I
Question 2 : The composite beam consists of a wood core and two plates of steel. If the allowable bend ng stress for the wood is (oaiow )y = 20MPa and for the s(o aienl =130MPA zilow determine the maximum moment that can be applied to the beam. E, =11GPA , E= 200GPA Take: A = 125 mm A mm B = 75 mm M 20 mm- B mm 20 mm Solution : E 200(10" E 1110° The width of steel must be increased to an equivalent width for wood: Page 1 18.182 W. mm
Transcribed Image Text:Question 2 : The composite beam consists of a wood core and two plates of steel. If the allowable bend ng stress for the wood is (oaiow )y = 20MPa and for the s(o aienl =130MPA zilow determine the maximum moment that can be applied to the beam. E, =11GPA , E= 200GPA Take: A = 125 mm A mm B = 75 mm M 20 mm- B mm 20 mm Solution : E 200(10" E 1110° The width of steel must be increased to an equivalent width for wood: Page 1 18.182 W. mm
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