Question

Use the graphical method to construct the shear-force and bending moment diagrams for the beam shown. Let a=4.0 ft, b=5.0 ft, c=5.0 ft,d=3.0 ft. w = 7.5 kips/ft and P = 30 kips. Construct the shear-force and bending-moment diagrams on paper and use the results to answer the questions in the subsequent parts of this GO exercise. с D E b For this loading, calculate the reaction forces A, and E, acting on the beam. Positive values for the reactions are indicated by the directions of the red arrows shown on the free-body diagram below. (Note: Since A, O, it has been omitted from the free-body diagram.) B с D b d Answers: Ay kips, Ey kips.

Answer 1

3 okips 7.5 kips/ft u OE po TOT 'Ey 3ft SH 5f 4++ Ay Eîyao » Ay+ Ey 5-7,5x5+30 Aytty = -7.5 EMATO Ey x 17 5-7:545 X 6.5 + 30414 Ey = 10.3676 kips. Answer put the value of Ey in Eq" 0 Aya -17.8676 kips. Ansub Past B) Eqh of Shear force along the Beam Vin) = -17.8676 <)' + 7.542-97 - 7.5<2-99 -30 <-1439 a) V(ot) = -17.8676 kips. Answer b) IV (A) = -17:8676 kips. answer e) V() = 19.6324 kips. Answer (V(14) =19.6324 kips answer

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e> N(14+) = -10.3676 kips Lanswer $> [V(17) = -10.3676 Icipe . Answer Part 3) Equ of Bending moment in Beam M(W) 3717.8676<u) + 7,5 <4-412-7.504-937-3084-19) IM(4) = -71.4704 kips-ft Answer Part 4) at u=hthen (16)=0 Ülo 4 + 17.8676 4 = 6.38235 ft 7.5 Answer Maxinun Bending moment. (Mo) Parts) Mo = -(7.867686.39235) + 7,5x19:58235) M0--92.753868 kips. ft. Answer part by a) M (9) = -67.0584 kips-ft | Answer by M(19) = 31.1036 kips-ft I answer

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