Question

A beam of length 12 m has overhanging of 3 m on left and right leaving the span between the supports of 6 m. It carries UDL of 8 KN/m over the entire length and a concentrated load of 10 KN at the right extreme end. Draw SF and BM diagrams and find the point of contra flexure point.                     

Answer 1

$\sum M_B=0$

$16kN(3m)-wL( \frac{L}2-3)+R_c(6m)-10kN(L-3)=0$

$30-8*12*3+6R_c-90=0$

$R_c=58kN$

$ZM_c=0$

$8kN(9m)+wL(9m-\frac{L}2)-10*3-R_B*6=0$

$72+8*12*3-30-6R_B=0$

$R_B=55kN$

$V_x=-10kN-w$

$At \space x=0_m,V_A=-10-w(o)=V_A=-10kN$

Just before B $At \space x=3m,V_B=-10-w(3)=V_B=-10-24=-34kN$

Just after B $At \space x=3m,V_B=-10-wx+R_B=V_B=-10-8(3)+55=21kN$

Just before C ,$x=9m,V_c=-10-wx+R_B=V_c=-10-8(9)+55=-27kN$

Point zero shear between , $B \space and \space C=0=-10kN-wx+R_B$

$0=-10-8x+55$

$x=5.625m$

Just after C , $V_c=-10-wx+R_B+R_c=V_C=-10-8*9+55+58=31kN$

Just Before, D, $V_D=-10-wx+R_B+R_c=V_D=-10-8*12+55+58,=V_D=7kN$

Just after D, $V_D=0$

Bending moment at x from point A

$M_x=-10kN(x)- \frac{1}2wx^2$

$At \space x=0,BM \space at \space A \space is \space M_A=-10(0)- \frac{1}2w(o)^2=0$

$M_B=-10(x)- \frac{1}2wx^2+R_B(x-3)$

$M_B=-10(3)- \frac{1}2(8)*3^2+R_B(3-3)$

$M_B=-66kNm$

Bending moment at point zero shear

$M=-10(x)- \frac{1}2 wx^2+R_B(x-3)$

$M= -10 (5.625)- \frac{1}2*8*5.625^2+55(5.625-3)$

$M=-38.4375kNm$

Bending moment just after C

$M_c=-10(x)- \frac{1}2wx^2+R_B(x-3)+R_c(x-9)$

$M_c=-10(9)- \frac{1}2 *8*9^2+55(9.3)+58(9-9)$

$M_c=-84kNm$

Bending moment at $D=0$