Question

Method 1: Mathematical Approach
Method 2: Engineering Approach

Draw the shear and bending moment diagram for the beam shown below. Use Method 2 30 k 30 k 60k 1 k/ft 25 k-ft 3030 30' 50' 3070 505050

Answer 1

Draw the Free body diagram of the beam and divide the section at internal hinges as follows: 30k 60k 30 k 1 k/ft 25 k-ft BEI 50 50 30 30 30 R IC. 30 50 50 50 7 0 Rc RE 30k 60k 30k R, 1 k/ft 25 k-ft в Е 20 50 50 30 30 20 50 3070 R RC In section EF: 30 k 60 k 30 30 30 Mg = 0 Rzx90–60 x 60–30x30=0 Rx = 50 k F.,=0 Rg+R; -30–60 = 0 Rg = 90 - R Rg = 90 - 50 Ry = 40k

In Section ABE: 25 k-ft A AT 150 * 501 F, = 0 RA+Rz - Rg = 0 RA+R; = R RA+R; = 40 EMA=0 x150-Rx1009–25 = 0 40 x150-25 100 Ry = 59.75 k R2 = 40- Rg = 40-59.75 RA = 19.75 k In section FCD: 30 k 1k/ft FIC 50 | 309 70 Rc EF, = 0 Rc+R, -RF - 30-1x 70=0 Rc+R, = 50+ 30+ 70 Rc+R, = 150 M, = 0 Rş x150 – Rc *100+30x70+1x70x70 50x150+ 2100 + 2450 100 Rc =120.50 k 100 3=0 p

Rc+R, = 150 R, = 150-120.5 R, = 29.45 k For section ABE: Calculate the shear force at different sections by using the engineering method: Calculate the shear force at a section by equating all the vertical forces from the left end: SFA = R=-19.75k SF =R = -19.75k SF3 = RA+R; SF3 = -19.75 +59.75 SF3 = 40 k SF; = SF3 – Rg SFx = 40–40 SF, = 0 Draw the shear force diagram: RE 25 k-ft 50 50 19.75 k 59.75 k 40 k 40 k 19.75 k

Calculate the bending moment at different sections by using the engineering method: Calculate the bending moment at a section by determining the area of the shear force diagram from the left end till that section. BM = 17.95x0 = 0 BM (just let) = A, = 19.75 x 50 = 987.5 k-ft BM (just right) = 987.5+ 25 = 1012.5 k-ft BM3 = BM (just right) + A, BM, =1012.5+19.75x 50= 2000 k-ft BM, =BM3 - A, = 2000 – 40 x 50 = 0 Draw the bending moment: 25 k-ft 19.75 kb 50 MTB 500 180 38 !59.75 k 40 ki 40 k 19.75 k- 987.5 k.ft 1012.5 k.fi 2000 k.ft For section EF: Calculate the shear force at different sections by using the engineering method: Calculate the shear force at a section by equating all the vertical forces from the left end: SF, = R3 = 40 k SFy = 40-30 = 10 k SF. = 10-60 = -50 k SF, =-50+50 = 0k

Draw the shear force diagram: 30 k 60 k F 30 30 30 150k 40k! 40 k 10 k Ok 50k 50k Calculate the bending moment at different sections by using the engineering method: Calculate the bending moment at a section by determining the area of the shear force diagram from the left end till that section. BM, =R x 0 = 40x0 = 0) BMx = 4, = 40 x 30 = 1200 k-ft BM. = 1200+ A2 = 1200+10x 30 = 1500 k-ft BM, = 1500 - Az = 1500- 50 x 30 = 0 Draw the bending moment:

30k 60k No T 40 k! 30 30 30 1 150 k 40 k 4, 10k A 50k - 50k 1500 k-ft 1200 k-ft: For section FCD: Calculate the shear force at different sections by using the engineering method: Calculate the shear force at a section by equating all the vertical forces from the left end: SF:=-R, = -50 k SFS = -50+120.5 = 70.5 k SFR = 70.5 – 30 = 40.5 k SF, = 40.5 - 1x 70 = 29.5 k

Draw the shear force diagram: 30 k RE 1 k/ft 50 7 30 120.5k 70 29.5 k 70.5 k 40.5 k 29.5 40.5 44 29.5 k 50 k Calculate the bending moment at different sections by using the engineering method: Calculate the bending moment at a section by determining the area of the shear force diagram from the left end till that section. BM, = 50x0 = 0 BMc = -A, = 50x50=-2500 k. ft BM, = -2500+ A2 = -2500+ 70.5 x 30 =-385 kºft BM =-385+ Az = -385+2x40.5 x 40.5 = 435.125 k. ft BM= 435.125 - A: = 435.125 – x 29.5 x 29.5 = 0

Draw the bending moment: 30 k 1 k/ft - TCTD 50 | 30 70 ? 120.5k 29.5 k 70.5 k 40.5 k A, 29.5 40.5 29.5 k 50 k 435.125 k.ft 385 k.ft 2500 k.ft

Combine the shear force and bending moment diagram of all the sections: 30 k 30k 60k 1 k/ft 25 k-ft M TE A 19.75 k 50 50 59.75 k NOT F TC 30 30 30 1501 30 30 50 kl 120.5k 70.5 k 40k! 70 :29.5 k 40 ki 40.5 k 29.5 40.5 19.75 k 29.5 k 50k 1500 KROK 1200 k-ft 435.125 k-ft 385 k.ft 987.5 k.ft 1012.5 kft 2000 k-ft 2500 k-ft