Question

(SE 2ГЕР — - / k/

Answer 1

Ans :

Resultant reaction forces acting at support is given below :

ADS: FB D for Beam: 15 kipf 25 kipf A Xc 714 - 1 4ft 4 ft 4 Ft TRAV 7 RAH S MA * for cantilever beam, at fixed support there are Horizontal & vertical compo. of forces and also a resultant couple moment due rest of the forces on beam acts at pt-A which is counter to them.

1) To find reaction forces at bupport consider Efx = 0 =tve =-ve). RAH=0 Now p=tve Efx=0 ** x=-ve 9 RAV= 15+25 RAV=40 kipf RA Resultant force acting at will be RA= RAH + RAVE TRAVZ RA RAV RA = 40 kipf Now to find to find Resultant couple moment at РЕ. А. consider ) = tve SD = -ve MA 15 X 4 + 25x8 **** MA= 260 kipf oft 4

SFD and BMD calculations and graph is given below :

93 for eg shear force dia. calculations. There are different methods of solving we will make 2 compartments around force left and right and for calculations we will consider always whaterer forces that gre left side of that particular compartment 54 ppose force of io kips, we will make 2 compartments around it as shown. lokies Ha Sos for calculations, we always ce consider forces left to that particular compartment se for left compartment there is no force to the left side SFL = ror left side of right compartment there is lokips force SFR = 10 kips (* = -ve) R O

* SFD calculations (1=tve, v=-ve SFAL= ARE 40 kipf. BL = 490 kipf BR = yo-15 25 kipf. CLE 25 kipf CR = 25 -25 = o kipf o 11 k ipf BMD calculations Také c 2 =-ve, s=tve There is no such compartment method for BM calculations At end points of beam beam B.M is always zero * Bb B. Mo D = o We will take moment about a pt. by considering forces that right side of the particular pt. cantilever beam with support on left side. are on for

?! BM А 2508 - ) 5x4 x BIX) - 260 kipf.ft Now BMB 25X4 -ioo BMB * .ioo kipfo ft There is no force right t to ptoc BMc=0 es

SFD and BMD :

P = 15 P = 25 НА A МА 4 (ft) + 4 (ft) + 4 (ft) * * 1 sec. (X1), 2 sec. (X2) 3 sec. (X3) 40 25 Оy 0 -260 -100 Mx