Question #205929

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.

Expert's answer

"\\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"

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