Question

Duration: 100 min Q1- Draw the shear and moment diagrams for the beam shown in figure and calculate the maximum bending stress. (70%) 4KN 2KN V B cross-section 3m + 1m Im 3m

Answer 1

Ans :

Ans: FBD : 4X103N 2x103N RAH A B sm RAV Rc RA and RAC to find consider Efx=0 IRAH=0 ** {fy=0 TRAV+ Rc = 6000 ** Z X EMA=0 (5 = tve, 2-ve) 4X 1 + 2x 4= Rc X 4 (All in KN RC= 4+ 8 =12 4 4 TRC = 3KN * from egn RAV=3KN Hence RA= RAV2+RAH = RAV RA= 3KN

* оге shear force dia calculations There different methods of solving we will make a compartments around force left and right and for calculations we will consider always. Whatever forces that are left side of that particular compartment suppose force of 10 kips. we will make 2 compartments around it as shown. lokios *** for e-g. th R sos for calculations, we always consider forces left to that particular compartment sae. for left compartment there is no force to the left side SFL = O for left side of right compartment there is lokips force SFR = lo kips Car-ve) 19

calculations for SFD ** (4=tve, * = -ve) SFAL=0 SFAR= 3_KN SF BL = 3 KN SF BR = 3 - 4 = - 1kN SF CL =-/ KN SECRE -1 -2+ 3 OKN ** BMD calculations: Take Q-ve, 5 = tve) There are no compartment method for BM Calculations, At end points of beam B. M is always zero BMA =0] Knim BMC KN.m We will take moment about a pt by considering forces that are on LHS of pt. - 3X) (in kN andm) BMB 80% - -3 BM . )

P = 4 P = 2 R RE A НА B TO + 1 (m) + 3 (m) 1 sec. (XF) 2 sec. (X2) 3 Qy -1 Mx 0 3

B4 the Bending equation Max. Bending Stress is given by 6 = M I stress. here f = Bending y = distance of extreme fiber from neutral axis. M = Moment OP Resistance moment of inertia about X-30 axis (Here we will take all values in N and mm) I = 3 m = 3000mm - 1000mm I = bd3 1000 X (3000) 3 12 2.25 x 1012 mm 4 I = 2.25 XI0 12 mm4 , - d 3000 - 1500 mm 2 2 y = 1500 mm end of simply is zero * Bending moment at the supported beam - Hence moment of resistance due to that joad at end points is also zero.

max . Bending stress = 2000 N / m²  

-L And also me due to point lood applied on simply supported beam is given by ** M = Woo.b M due to 4 KN Hence is given by 48103 X 1000 X 3000 400o Nomm ) 6. S 3x10 & due to 2KN is given by M= 2x103 X 4000 X o 4o0o Hence M = 0 Total M = 3x106 Nomm so by given equation M.y = 3x10 6 X 1500 2.25 81012 Tob 2X10-3 N N/mm2 Toba 2000 N/m2]