Question

3. Refer to Figure 3. A 64-foot steel beam supports a combination of concentrated and distributed loads: Force Fi = 9 K, F2 = 3.65 K, w1 = .525K per lineal foot, and w2 = .212K per lineal foot. In addition, your calculations must include an estimated beam weight of 148 pounds per lineal foot. Part A: What are the magnitudes, in kips, and directions of the vertical components of the reactions at each support? (4 points) Part B: Construct a Load Diagram. [Scale: 1" = 16 feet) (3 points) Part : Construct a Shear Diagram for the loading condition depicted in Part A. (4 points) Part D: What are the extreme values, in kips, of V? (4 points) Part E: Construct a Moment Diagram for the loading condition depicted in Part A. Label the maximum and minimum values on the diagram. Circle (in red pencil) all points of inflection. (6 points) Part F: What are the magnitudes, in Ft-K, and locations of the extreme values of M? Include your calculations. (4 points)

Answer 1

W., 20.525k XI - cilt wery IITTI 0.148K|ft p I R 12ft * 12ft A ft & ICH ** gak 19 ft >k at {FORBx=0 ans. TEMA=O > Rey*(12+6+9) – 9612+6) – 3,65(12+6+9+13) - capa/1 Layne +13 +0,*as knyga - 0.148 (12+6+3+13+1} Self weight between t 0.48 x 12 =0 Self weight betwon PA AQ → Royx 27-9818-3.65×46-0-212 24 0x32 to.525 X 12x6 -0.148X S2² to-148x 12²=0 2 Rey = 27.8854/7 ans {fyzo P. Ray they 0-148x64 -0.525x12 - 0.112*40 --305=0 Ray = 9.017 KT And (0.148x64 -0.525*12 Św wers 22 ask

Part A reactions Row=9.01741, Roy = 27.885kT/ BX0 Ans Parts O.RS to-148 figk 0.212+0.148 0.673k/ft 63-65K 0.36k 1A 0-148 Kft 27.885k 11 A 12ft & 64 $9ft IB ight I oft. Q - 12ft 3.017k Ha or trin for Bm Parte in Q B only. - SF - free Body diagram is drawn in Part B so we will write equations from it Shear force (V) ove-0.673x origin is P ant @ AA, VAC -0.673x12=-8.076 k @ VAR = -8.096+9014 – 0.148 CX—12) Origin ist. @ At Ri VR= 0.941–0-148 X12 =- 0.835K © Ves = -0.835–0.3662-24) © At sGjust left ofs), VSE -0.835-0.36X65–2.995k © Vse =-2.998--0.36(x-30) 8 Just left of B , VB,= -11.998-0.3689 2-15-235k 6 VBT = -15-235-0.36 (x-39) 127.888 Roy © Just left of ti Ya =-15-235+27.885-0.36X19 V = 5.81K

Vre= 5.81–375 -0.36 (X-se) Va0 Shear force diagram 12-657 5-21 oghi ! + R S 2.16 As -2-995 -8.076 Parth -11.995 Tato -15-235 Mane peutive shear= 12-65k man Negative shear =-15-235k Parte Bending moment diagram BM 25 Free Body diagram drawn in Part B so we will swrite equations of BM from it originas P O MPA=-0.673 x ² © ATA ,MA= -0,673 x122= -48.46 ku-ft. © Mar= -0,622x12 (x++ g. 93 (N=12) – 2. 149(812 ③ Mr ==47.82 k-It put 21-24ft in step3, x=24 ® Mes=-0.673X12 x 104-6) +9.01712–129–0.148X126412) -0.36(x-24) 2 0 Ms Pute X=30 in steps Ms=-59.31 Kft.

© MSB = -0.673 X12 (1-6) +9.014(-12)_0.148 x12 (26–18) -0-36 C4-29). – 9.00-30) MB2 -181.85 Kft. X=39ft ¢ Above became bit tedious to for BQ or QB portion we will start origin from Q but note sign Convention ir etgan Mpt = -0.36?. to bite -0.36 X 5 =-6.48 kft. @ MTB P-0-36X2–3.65 (4-6) At B x= 25+ :- MB=- 181.85kft .OK. Check Bending moment diagram Kft. R 6.48 -48.46 KA -47.82 -59.31 Manimum Bending moment =-181-85k-ft at R. Min=0 781.85 Partf Mare BM =181-85k ft at B.