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The below wooden double overhanging beam is under a uniformly distributed load w. The wood is weak along the orientation of tPLESAE CONTINUE FROM C,D IF you consider the whole question to be too long thanks

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(A SSignmenot , AN M. Grmax 2 SMR? RA M=? N, 25 MA NA 2SMPa Zog= Eayles G=Gy) Rings Co DING Lll matee ecuerAssignmait It NA 200 6omm V Go x(L00[y + Ix 60 = (92S/ Ealyy 160-4 120 -Sn 20 2 26 2 Simple cedin) tanMattab er txceNG! COLUMN BUcteLi NG Sfrensth Gyveld A TE Ger Bucleling A- Eyreld hen C reld 2 CA-1/COLUMN BuckeLiNG Bacleleny load Per Buchdeg Stoers Ger Por A A A A Cradius of gyhathan) Ger ehere A- 12 12 A 2 hh V12 CerXam ple CLUMN BUCiLING 31.4 x10 mm mt = T. 4.& x ib mmf L = 7m le 2 L = 14 m 4.9m 11 エユ_L 316, 23 teN Fer 316.23 TEIy 337.67

The below wooden double overhanging beam is under a uniformly distributed load w. The wood is weak along the orientation of the grain (or wood cell fibres) that makes an angle of 30° with the horizontal (see figure). The maximum shear stress on a plane parallel to the grain that the wood can sustain is tmax = 5 MPa, and the maximum normal stress of wood is amax = 25 MPa. The Young modulus of this wood is E=15 GPa (a) Using w 18 kN/m, check if the beam is safe against failure & Determine the deflections at free ends and at the middle of the beam (b) Determine the value of w so that the maximum deflection of the beam will not exceed 2 mm AND the beam is safe against failure (c) If the uniformly distributed load w on the overhanging parts of the beam is removed (or w is applied over the 2-m span only), will the deflections calculated in question (a) increase or decrease. Explain with details (e.g. sketch) but no calculations. Sketch the bending moment and shear diagrams (d) If an additional concentrated force (vertical & downward direction) is applied in the middle of the beam (w is still applied over its entire length of the beam), will the deflections calculated in question (a) increase or decrease. Explain with details (e.g. sketch) but no calculations. Sketch the bending moment and shear diagrams W 200 mm 300 700 mm 700 mm 2000 mm 60 mm
(A SSignmenot , AN M. Grmax 2 SMR? RA M=? N, 25 MA NA 2SMPa Zog= Eayles G=Gy) Rings Co DING Lll matee ecuer
Assignmait It NA 200 6omm V Go x(L00[y + Ix 60 = (92S/ Ealyy 160-4 120 -Sn 20 2 26 2 Simple cedin) tanMattab er txce
NG! COLUMN BUcteLi NG Sfrensth Gyveld A TE Ger Bucleling A- Eyreld hen C reld 2 CA-1
/COLUMN BuckeLiNG Bacleleny load Per Buchdeg Stoers Ger Por A A A A Cradius of gyhathan) Ger ehere A- 12 12 A 2 hh V12 Cer
Xam ple CLUMN BUCiLING 31.4 x10 mm mt = T. 4.& x ib mmf L = 7m le 2 L = 14 m 4.9m 11 エユ_L 316, 23 teN Fer 316.23 TEIy 337.67 tN ?16.23 143.4にN。 2.2
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Answer #1

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