DETAILED EXPLANATION PLEASE © Solved: For the beam and lc | + v mygreatlakes org
ID: 1712860 • Letter: D
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DETAILED EXPLANATION PLEASE
© Solved: For the beam and lc | + v mygreatlakes orgmyLSU-Single Sign Nelnet Sign in to your Real Property 1 Career 2Geaux Sign UREC Portal M Sign ln Zy Home -zyBooksChase Bank- Credit For the beam and loading shown, conslder section n-n and take a-108 mm, b-20 mm and c 15 mm. Determine the shear stress at a 160 mm 180 kN I n 500 mm 500 mm 30 mm 30 mm 20 mm 025 points The magnitude of the shear stress at a is given as 13.93 (MPa) 2-16 PM O Type here to search 10/21/2017 2Explanation / Answer
From beam analysis,since the beam is symmetrically loaded and supported, the reaction forces at both dupports will be equal
Reaction at each support=180/2=90 kN
Shear force at section a-a =V=90 kN=90000N
Let us now determine the deprh of neutral axis of the section from the topmost surface
Let the depth of neutral axis be d
d=((160*20*10)+2*(88*20*64))/((160*20)+(2*88*20))=38.29 mm
Moment of inertia of section about neutral axis=I=
(160*20³/12)+(160*20*28.29²)+(2*20*88³/12)+(2*20*88*25.71²)=7.27*10^6 mm^4
Static moment of area above point a about netral axis=Q=160*20*28.29=90528 mm³
Width of section at a=160mm
Shear stress at just top of a=(90000*90528)/(7.27*10^6*160)=7 Mpa
Shear stress at just bottom of a=(90000*90528)/(7.27*10^6*40)=28 Mpa
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