D. E. C B. -4 m 4 m 15 kN 20 kN 00
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- y''-6y'+9y=t^2e^3t ; y(0)=2,y'(0)=6Solve Subparts b, c and d early I upvote. But only Correctly solution provided.If F1 = 10N, F2 = 15N, F3 = 8N, F4 = 10N, F5 = 18N, and the length of te side of a red square is 1m. Calculate magnltude (+ for CCW, - for CW) of the resuItant moment (N-m) about point A.
- Q) All answers of second moment of area shall be in \times× 106 mm4 Solve it early I upvoteThehorizontalbarshown(Length=15.0 feet)restsonsmoothinclines.Computethedistancexat which load T= 250 lbs should be placed from point B to keep the bar horizontal?Can you check back parts b & c My answers are: using r > 3n method where r= reactions & n= nb of parts b) (16) > (3)(4)statically indeterminate by 4th degreec) (12)> (3)(3) statically indeterminate by 3rd degree let me know where is my mistakethank you
- I'm looking for: B. Moment-area methodsorry，what are these： Step 3 Slope deflection equation In section AB MAB= MFAB+22EIL2θA+θB-3δLMAB= -12.5+4EI5θB MBA= 12.5+4EI52θB In section BC MBC= 0+2EI52θB+θC MCB= 2EI5θB+2θC In section CD MCD= -16.67+4EI52θC MDC= 16.67+4EI5θC arrow_forward Step 4 Equilibrium equations Equilibrium equations MBA+MBC= 0 -(i)MCB+MCD= 0 -(ii) Now, From i12.5+8EIθB5+4EIθB5+2EIθC5= 0EIθB125+EIθC25 = -12.5 -(iii) From ii2EIθB5+4EIθC5+-16.67+8EIθC5=02EIθB5+EIθC125= 16.67 iv Solving equations (iii) and (iv), we get EIθB= -6.548EIθC = 8.037Use fc=14.3N/mm^2 ft=1.43N/mm^2 HRB400 fy=360N/mm^2