Question

6 kN/m 15 kN 100 mm E NA - B D 100 mm 0.8 m For the truss system shown, a. (10 pts) Draw the shear and moment diagrams for the rectangular wooden beam shown. b. (6 pts)Determine the magnitude of the absolute maximum moment (Mmax) and its location c. (Z pts) Find the maximum bending stress in the member (max) d. (Z pts)Compute the Bending Stress at a point on the section B that is 25 mm below the top of the beam. 120 mm 2 m 1.0 m- IRA = 14 kN Rg = 19 kN

Answer 1

08 | Part a. 15 KN 6 kN/m (E C D Kim 0.8 m 2m- RA-14 KN RE=19 KN Shear force calculation :- at section (В SFB = Ra-6*0.g SFB = 14-4.8 SFB 9.2 KN Left 1 to the section e (SFO) (Sfo), - RA-6*2 (SFD), - 2 KN - 14 - 12 = 2 KN Right to the section @ (SFO)R (SFO). Ra - 6*2-15 = 14-12 -15 = -13 VN (SFOR R = -13 KN

Po Bending Bending moment calculaten Moment at Section ③ MB MB = RA*0.8 - 6*0.8*0.8 2 MB= 14*0.8 - 1.92 = 9.28 KN-M MB B = 9.28 KN-M at Section ③ MO MD=RA*2 - 6*2* 2 / 2 Mo - 14* 2-12 = 16 KN-M Mo = 16 KN-M

(Pg. 3) Shear force & Bending moment diagram :- 115 KN 6 KNM E D Kim 0.8 m 2 m. RA- 34 KN RE=19 KN Shear force diagram 14 KN 9.2 KN 2KN B) -13 KN -19KN Bending moment diagram :- + 9.28 KN-M 16 KN-M

Po 4 Part bl from Bending moment diagram Absolute maximum bending moment is 16 KN-M at section (D) Part C Given cross section T 100mm 100 mm K 120 mm 不 Since cross section is same througout the span maximum bending stres will be at the location of maximum bending moment max = EN = Section modulus , z bd² I 20*2002 6 6 z = 800000 mm2 = 8+104m3 16 KN-M =2*10 KN 8*10*4 m3 15 may DY or 20 MPa

Pa 5 Part D 25mm 100 mm 75 mm Neutrat Axis 100 mm t K 120 mm > Bending Stress at 25 mm below the top at B M = Bending moment at OBE the section MY I Y = Distance of point from neutral axis Mg= 9.28 KN-M I= Moment of Inertia of Section Y= 75 mm=75-10² m I= bd? 120*200* 12 12 since I = 8x10 mm" - -12 :8*10 m Imm= 10 m 10?m? so, og = 9.28* MY I 75*103 8*105 - 8700 KN or 8.7 mpa ma