Problems on simply supported beams
1)
Step 1: Calculation of the reactions
ΣH = 0
ΣV = 0
Ra+ Rb= 4+10+7
Ra+ Rb= 21KN
ΣMa =0
Rb(8)-7(6)-10(4)-4(1.5)=0
Rb = 11 KN ; Ra = 10 KN
Step 2: Calculation of shear force
Shear force at point B = -11KN
Shear force at point C= -11+ 7= -4KN
Shear force at point D=-11+ 7+10 = 6KN
Shear force at point E= -11+ 7+10+4 = 10KN
Shear force at point A=10KN
Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point C= (11X2)= 22KN-m
Bending Moment at point D= (11X4)-(7X2)=30KN-m
Bending Moment at point E= (11X6.5)-(7X4.5)-(10X2.5)= 15KN-m
Bending Moment at fixed point A=(11X8)-(7X6)-(10X4)-(4X1.5) = 0KN-m
Step 1: Calculation of the reactions
ΣH = 0
ΣV = 0
Ra+ Rb= 2+5+4
Ra+ Rb= 11KN
ΣMa =0
Rb(6)-4(4.5)-5(3)-2(1.5)=0
Rb = 6 KN ; Ra = 5 KN
Step 2: Calculation of shear force
Step 2: Calculation of shear force
Shear force at point B = -6KN
Shear force at point E= -6+4= -2KN
Shear force at point D=-6+ 4+5 = 3KN
Shear force at point C= -6+ 4+5+2 = 5KN
Shear force at point A=5KN
Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point E= (6X1.5)= 9KN-m
Bending Moment at point D= (6X3)-(4X1.5)=12KN-m
Bending Moment at point C= (6X4.5)-(4X3)-(5X1.5)= 7.5KN-m
Bending Moment at fixed point A=(6X6)-(4X4.5)-(5X3)-(2X1.5) = 0KN-m
3)
Step 1: Calculation of the reactions
ΣH = 0
ΣV = 0
Ra+ Rb= 3+6
Ra+ Rb= 9 KN
ΣMa =0
Rb(6)-6(4)-3(2)=0
Rb = 5KN ; Ra = 4 KN
Step 2: Calculation of shear force
Step 2: Calculation of shear force
Shear force at point B = -5KN
Shear force at point D= -5+6= 1KN
Shear force at point C= -5+6+3 = 4KN
Shear force at point A= 4KN
Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point D= (5X2)= 10KN-m
Bending Moment at point C= (5X4)-(6X2)=8KN-m
Bending Moment at fixed point A=(5X6)-(6X4)-(3X2) = 0KN-m
4)
Step 1: Calculation of the reactions
ΣH = 0
ΣV = 0
Ra+ Rb= 10(6)
Ra+ Rb= 60 KN
ΣMa =0
Rb(12)-10(6)(3)=0
Rb = 15KN ; Ra = 45 KN
5)
Step 2: Calculation of shear force
Shear force at point B = -15KN
Shear force at point C= -15KN
Shear force at point A= -15+10(6)= 45 KN
Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point C= (15X6)=90KN-m
Bending Moment at fixed point A=(15X12)-(10X6X3)= 0KN-m
5)
Step 1: Calculation of the reactions
ΣH = 0
ΣV = 0
Ra+ Rb= 5(2)+ 10(2)
Ra+ Rb= 30 KN
ΣMa =0
Rb(6)-5(2)(4+1)-10(2)1=0
Rb = 11.67KN ; Ra = 18.33 KN
Step 2: Calculation of shear force
6)
Step 2: Calculation of shear force
Shear force at point B = -11.67KN
Shear force at point D= -11.67+5X2= -1.67KN
Shear force at point C= -1.67 KN
Shear force at point A= -1.67+ 10X2 = 18.33KN
Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point D= (11.67X2)-(5X2X1)= 13.34KN-m
Bending Moment at point C= (11.67X4)-(5X2)(1+2)=16.68KN-m
Bending Moment at fixed point A=(11.67X6)-(5X2)(1+4)-(10X2X1) = 0 KN-m
Step 4: Calculation of point of zero shear force x(point of max BM)
Ra-10(x)=0
18.33-10(x)=0
x=1.833m
Maximum BM corresponding to x
Ra(x)-10(x2/2)
18.33(1.833)-10(1.833x1.833/2)
16.79KN-m = Mmax
6)
Step 1: Calculation of the reactions
ΣH = 0
ΣV = 0
Ra+ Rb= 50+ 10(4)+40
Ra+ Rb= 130 KN
ΣMa =0
Rb(10)-40(6)-10(4)(2+2)-50x2=0
Rb = 50 KN ; Ra = 80 KN
Step 2: Calculation of shear force
Step 2: Calculation of shear force
Shear force at point B = -50KN
Shear force at point D= -50+40= -10KN
Shear force at point C= -50+40+10x4+50= 80 KN
Shear force at point A = 80 KN
Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point D= (50X4)= 200 KN-m
Bending Moment at point C= (50X8)-(40X4)-(10x4x2)=160 KN-m
Bending Moment at fixed point A = 0 KN-m
Step 4: Calculation of point of zero shear force x(point of max BM)
Ra-50-10(x-2)=0
80-50-10(x-2)=0
x=5 m
Maximum BM corresponding to x
Ra(x)-50x3-10x3x1.5
205 KN-m = Mmax
7)
8)
7)
Step 1: Calculation of the reactions
ΣH = 0
ΣV = 0
Ra+ Rb= 4+ 8+8
Ra+ Rb= 20 KN
ΣMa =0
Rb(10)-8(9)-8(5)-4(2)=0
Rb = 12 KN ; Ra = 8 KN
Step 2: Calculation of shear force
Step 2: Calculation of shear force
Shear force at point B = -12 KN
Shear force at point F= -12+8= -4 KN
Shear force at point E= -4 KN
Shear force at point D= -4+8= 4 KN
Shear force at point C= 4+4= 8 KN
Shear force at point A = 8 KN
Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point F= (12X1)= 12 KN-m
Bending Moment at point E= (12X3)-(8x2)=20 KN-m
Bending Moment at point D= (12X7)-(8x6)-(8x2)=20 KN-m
Bending Moment at point C= (12X8)-(8x7)-(8x3)-(4x2)=8 KN-m
Bending Moment at fixed point A = 0 KN-m
Step 4: Calculation of point of zero shear force x(point of max BM)
Ra-4-2(x-3)=0
8-4-2(x-3)=0
x=5 m
Maximum BM corresponding to x
Ra(x)-4x3-4x1
24 KN-m = Mmax
8)
Step 1: Calculation of the reactions
ΣH = 0
ΣV = 0
Ra+ Rb= 25+ 10(6)
Ra+ Rb= 85 KN
ΣMa =0
Rb(6)-25(2)-10(6)(6/2)=0
Rb = 38.33 KN ; Ra = 46.67 KN
Step 2: Calculation of shear force
Shear force at point B = -38.33 KN
Shear force at point C=-38.33+10(4)+25= 26.67 KN
Shear force at point A =-38.33+10(6)+25 =46.67 KN
Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point C= (38.33X4)-(10x4x2)-(25x2)=73.33 KN-m
Bending Moment at fixed point A = 0 KN-m
Step 4: Calculation of point of zero shear force x(point of max BM)
Ra-25-10(x)=0
46.67-25-10(x)=0
x=2.167 m
Maximum BM corresponding to x
Ra(x)-25x0.167-10x2.167x2.167/2
73.48 KN-m = Mmax
Step 1: Calculation of the reactions
ΣH = 0
ΣV = 0
Ra+ Rb= 0
Ra = -Rb
ΣMa =0
Rb(6)-24=0
Rb = 4 KN ; Ra = -4KN
Step 2: Calculation of shear force
Step 2: Calculation of shear force
Shear force at point B = -4 KN
Shear force at point C= -4 KN
Shear force at point A =-4 KN
Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point C = Right side of C=4(4)= 16 KN-m
= Left side of C = 4(2) = -8 KN-m
Bending Moment at fixed point A = 0 KN-m
Step 1: Calculation of the reactions
ΣH = 0
ΣV = 0
Ra+ Rb= 45+ 10(3)
Ra+ Rb= 75 KN
ΣMa =0
Rb(6)-45(3)-10(3)(3/2)-120=0
Rb = 50 KN ; Ra = 25 KN
Step 2: Calculation of shear force
Shear force at point B = -50 KN
Shear force at point D=-50 KN
Shear force at point C=-50+45 = -5 KN
Shear force at point A =-5+10(3) =25KN
Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point D= from B = 50x1.5-120= -45 KN-m
Bending Moment at point C = 50x3-120 =30 KN-m
Bending Moment at fixed point A = 0 KN-m
Step 4: Calculation of point of zero shear force x(point of max BM)
Ra-10(x)=0
25-10(x)=0
x=2.5 m
Maximum BM corresponding to x
Ra(x)-10x2.5x2.5/2
31.25 KN-m = Mmax
Step 1: Calculation of the reactions
ΣH = 0
ΣV = 0
Ra+ Rb= 40+ 5(2)
Ra+ Rb= 50 KN
ΣMa =0
Rb(7)-40(2)-5(2)((2/2)+2))-80=0
Rb = 27.14 KN ; Ra = 22.86 KN
Step 2: Calculation of shear force
Shear force at point B = -27.14 KN
Shear force at point C = -27.14 KN
Shear force at point D= -27.14 KN
Shear force at point E= -27.14+5(2)+40 = 22.86 KN
Shear force at point A = 22.86 KN
Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point C =27.14x1-80=-52.86 KN-m
Bending Moment at point D =27.14x3-80=1.42 KN-m
Bending Moment at point E = 27.14x5-80-5(2)(2/2)=50.7 KN-m
Bending Moment at fixed point A = 0 KN-m
Step 1: Calculation of the reactions
ΣH = 0
Ha = 100cos60+200cos45+300cos30
Ha = 451.23KN
Ha = 100cos60+200cos45+300cos30
Ha = 451.23KN
ΣV = 0
Ra+ Rb= 100sin60+200sin45+300sin30
Ra+ Rb= 378KN
ΣMa =0
Rb(4)-300(3sin30)-200(2)(sin45)-100sin60=0
Rb = 204.86 KN ; Ra = 173.14KN
Step 2: Calculation of shear force
Shear force at point B = -204.86 KN
Shear force at point E= -204.86+300sin30= -54.86 KN
Shear force at point D= -204.86+300sin30+200sin45= 86.56KN
Shear force at point C= 204.86+300sin30+200sin45+100sin60 = 173.14 KN
Shear force at point A =173.14KN
Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point E=204.86x1= 204.86 KN-m
Bending Moment at point D =204.86x2-300sin30 =259.72 KN-m
Bending Moment at point C =204.86x3-300(2)sin30-200(2)sin45=-126.84 KN-m
Bending Moment at fixed point A = 0 KN-m
Step 1: Calculation of the reactions
ΣH = 0
ΣV = 0
ΣV = 0
Ra+ Rb= 0.5(5)(6)
Ra+ Rb= 15KN
ΣMa =0
Rb(5)-0.5(5)(6)(1/3)(5)=0
Rb = 5 KN ; Ra = 10KN
Step 2: Calculation of shear force
ΣMa =0
Step 2: Calculation of shear force
Shear force at point B = -5 KN
Shear force at point B = 10 KN
Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point A= 0KN-m
Step 4: Calculation of point of zero bending moment
-Rb + (0.5)(x)(6x/5) = 0
x = 2.88m
Maximum bending moment = Rb(x) - (0.5)(x)(6x/5)(x/3)
Mmax = 9.62KN-m
Step 1: Calculation of the reactions
ΣH = 0
ΣV = 0
Ra+ Rb= 30+ 0.5(3)(20)
Ra+ Rb= 60KN
ΣMa =0
Rb(6)-30(5)-0.5(3)(20)(1+(2/3(3))=0
Rb = 40KN ; Ra = 20 KN
Step 2: Calculation of shear force
Shear force at point B = -40 KN
Shear force at point E = -40+30 = -10 KN
Shear force at point D= -10 KN
Shear force at point C= -10+(0.5)(3)(20)= 20 KN
Shear force at point A = 20 KN
Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point E= 40X1=40 KN-m
Bending Moment at point D =40x2-30x1=50KN-m
Bending Moment at point C = 40x5-30x4-0.5(20)(3)(2/3(3))=20KN-m
Bending Moment at fixed point A = 0 KN-m
Step 4: Calculation of point of zero bending moment
Ra - (0.5)(x-1)(20/3)(x-1) = 0
x = 1.45m
Maximum bending moment = Ra(x) - (0.5)(x-1)(20/3)(x-1)(2/3)(x-1)
Mmax = 28.79KN-m
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