Question

please use the following diagram to answer the questions in the picture below. feel free to make random dimensions that you choose to convey the process.

Thank you very much

2 Given the hollow box beam shown below, do the following: 1. Determine the state of stress, expressing the magnitude and direction of normal stress, shear stress, and the internal moment, at point C, Section C-C. Point C is at the juncture of the web and the top flange. Section C-C is located 0.5 meters to the right of the left end (the beam's midpoint). Draw the free body diagram of the beam at section C-C. Draw Mohr's Circle and use it to determine the principle stresses, 01 and o2, and the angle 0 that the plane of principle stress makes with the x-axis. Draw Mohr's Circle and use it to determine the absolute maximum shear stress, TAbs Max a. b. C. d. 2. Given the same beam, do the following: Sketch the elastic curve a. b. Draw the free body diagram for a section of the beam at a point where 0

Answer 1

Heve they Values ave am te auming Since Yandom not and use you Said dimensions 3iven F500N F800N L = 1 m, de h= 6omm So, de 0.5m O.2m y=1Smm b= 30 mm neutal axis to distance brom ;e *b Poiut y=15 mm. C

0.5m 800N F2 ーう500N C 1-15mm .0 2m Ax Im b 30 mm x0Ax -97 500N MAO s +ve *o )Ay+B Aru=640N +B = 80ON 800X0.2 - X B,-16ON 3trem shear stres at ) Novmal Secton C-C BooN C 500 0.5 640 +800-640 o C-C P-c 500N フーク」 C-C > Vc-C = -160 N n ternal memend at (1) e, -c = 160 N + 800x 0.3-640x0-5-0 MC 80 N.m Mc-C 240320

axiad load: Pe-c A= hxb A 60 x30 500 N/mm 1800 mm 1800 Bending moment bending we knoo Hat Ir 60 x30 L4 - 540000omm 2 2 4=15mm 8OX10 (N.mn x 15 mtn . 540000 mm 0 2.22 MPa stvem, o O28 +2.22 '. 7otal novmal Ox2.5 MPa Sheav stvesi g& ven by tVansveyse shear VQ Hat 30 15 mm we know Tayy 30 V=-160N (A x g) VE160N (1) Paytiag avea. Q=AX Paytial avea tEthicknes ok the section - (e0x15)x(1515) Q=30X15 x22.5 030mm =b mm2 Q=10125 mm3 160x10125 : 2 xy 540000 x 30 O.) MPa at C-Ci b) F.B.D B00N O.2m Ve-c C AX K 0.5m Ay

(P centve, c C = Cx 2 .5 Radius, R= 7Tmax RE254MPa /C 15 2.5 2 O.5 2.5MPA 02 OMPA Zmax .254M Pa .574 20 bi

Elastic CUYVE Smax b/w 02x21m- F-800 N o.2m where, o 21 m O.5 m Ax A4 foyce Fs the deflectan acking whe Ye the At Point w:99 be maximum Hhat F. we knouo 40.2 2 M6X) h P ET Ay M)F-O.2) A a Agx F (X-0.2) on both sides Integvation get we 0 = 7 2 = e= ET dx 2 get On both S%des we again Integyation 3. 3 -0.2) C, +C2 E1 YX) 4 at Conditiony Boundovy Ag 1 ET O f06.67-68 27+C, -38. 4

6440 Co.2) 6 2m - 200 mm 38 4 EI 3 6나0 (200) 38.4 3 200X10X 540000 85333329 49 3 200X 10 54000 O -3 7. 90 X10 -0.2-38. 3 640 g00 d) 6 6 EI