Question

If someone could provide solutions to these it would be greatly appreciated!

Question 3: (*) Sketch the show one and Bending Moment s for the show is on 34). Determine the maximus bending moment, and its location on the bean' SIN 40 KN Figure 3(a): The beam is simply supported. (b) Sketch the Shear Force and Bending Moment diagrams for the beam shown in Figure 3(), Determine the maximum bending moment, and its location along the beam's span. Also determine the shear force and bending moment half way along the beam's span. (15) 117 KN 5 KN Figure 3(b): The beam is simply supported, 5 Nm is a uniformly distributed load

Answer 1

9) Given Beam aka 40kn k Gm - to Ac 63m - COIN 9kn Midpoint term TBE EM kam I om R2 = 7612 kall 퀵에 - ! 116.33kW 16.33k i tve c 3m 10 ISED 3m 19 kr -ve TENI 171 kn 7.ka Hi Knim i 22 kalm cs Scanned with -Streigas CamScanner

RB + Ro= uq 76'+ Rp - 49 Shear force calculations i (S-Fo = - =-23.67 KRY S.F) = -71740= 16-33 km (sif eld point = 16.33 kry (S.FR = 16:33 - 7€ =-qlay 8.F) a = - Gent tq kn = 0 Mazinum Shear force = -23-67 ky ) Bending moment Calculations i (Bingon = 0 (Bara)e = x3 = Fikam midpoint -40x3 =22 krom (Bomp Fxø -40% 6 5-27 Kne-m (smartpx12m 40x9 + FLEXA Foto CamScanner

Maximum Bending momentum = Filenam/ narinum Bending moment will occurre at a 3m dis End or qm distance tance from from left . 117 km Right side side end. 5 Em D V Acam to . 6 am - K-549 = 45 45kA 117KW M 4.5 TRC =-16.5 A E Bm - BP 4.5 eg=18.SE t665knd 16.5kal tve A Bm B B 21-5 m mi SEDI - ve 12-EN u ery Im 4-5 BI BMP mi -tours line 1 CS Cams Sqanned with CamScanner 0151 Kalem I 124.835 katm

RBT RC = 117+5x9 = 162 EMC=o -117 x 12+ R3x9-45X4.5 = 0 [Rg = 178.5kry ] RB t RC = 162 Rc = -16.5 kn Shear force Calculations & (S.F = +16.5 kn (S.F, = 16+5+45=Elsky (S.F) 61.5-78.5 =-117 KN 15:5) A = -117 +117 - 0 Maximum Shear force= -117 kn & o it located at Section Rich R 3m distane from left side end.. Scanned with CamScanner it is

sending moment Calculations 5 kind om mmmm (8 me = o Balm =-124. 875 kn -- 16.5X4.5 - sx4.5x4.5 SX -16.5X9 -445 X 4.5 =-35 Italmi = O Rw-m = -16-5712-05 x 7.5 + 178.5X h Maxinum Bonding momeed = -35 l kalam It is located at B sm from left sick end. Mid Point Shear force at take the diagons Shear force diagrams from BC section 16.5 cos u.s from similas triangle c Scanned w20=22-5

Shear force at mid point = 22.5 kalor already Bending moment at midpoint Calculated = 124.875 katm Shear force & Bending moment Sign Conventions from 2t3 St > ↑ > tve Sof a ->-ve Bina - ew >tve Bim cow ove from RHS (o) (2) SF S.F B-m B-m » - - → v tve p > -ve co ove (2) cew > tue con Scanned with CamScanner