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8/3/2019 B20_EE_ES1 Task1_Edvin Saw Kien Yip_Koay Cheng Chye
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A) Shear force is generally related to tearing force which means,an external force are
applied perpendicularly first on any material and then it causes internal forces in the
specified material.
For example,put a ruler between two table then apply a force perpendicularly downwards
on the middle of the ruler. As the force applies,it will create a shear force inside the ruler.
Since you are creating a force that's perpendicular to the material,the bigger force you apply
to the ruler, the higher the shear force the material is going to experience in general. Please
also note that shear force is an internal force, and the ruler in this case, the shear force can
vary at different point in the material.
A free body diagram can be drawn which represent how much shear force a material is
experiencing at different point.
Shear force are classified to two types which are the positive shear force and negative
shear force. The types of shear force can be identified by knowing the direction of the shear
turns. It is a positive shear force when it turns clockwise direction and negative shear force
when it turn the vice versa way.
Based on the example given where a force are applied perpendicularly downwards to the
ruler,bending will occur even the least force applied other than shear force. Thus,this
‘bending’ are what we mean for Bending Moment.
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Case 1
Fy = 0
R1 + R2 = 400 + 600 + 200
= 1200N – (1)
MR2 = 0
MCW = MCCW
6R1 = 3.5(400) + 600(3) + 200(2)
R1 = 600N – (2)
(2) substitute into (1)
600 + R2 = 1200N
R2 = 600N
Shear Force
Section cuts X1, 2.5> X1≥0
FY = 0
V = R1
= 600N (CW, SF = +VE)
SF = 600N
R2
400N 600N200N
2.5m 0.5m 1.0m 2.0m
L= 6.0m
X1 X2 X3 X4
X
X1
R1=600N
X1
X2
X2
R1=600N
V400N
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Section cuts X2, 3>X2≥0
FY = 0
R1 – V – 400 = 0
V = 600 – 400
= 200N (CW, SF = +VE)
SF = 200N
Section cuts X3, 4>X3≥0
FY = 0
V + R1 1000 = 0
V = 1000 600
= 400N (CCW, SF = – VE)
SF = – 400N
Section cuts X4, 6>X4≥0
FY = 0
V+ R1 – 400 – 600 – 200 = 0V+600 – 400 – 600 – 200 = 0
V = 600N (CCW, SF = -VE)
SF = – 600N
X3
400N 600N
X3
V
R1=600N
X4 X4
400N 600N 200NV
R1=600N
R1=600N R2=600N
400N 600N 200N V
X5
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Section cuts X5, L=6
FY = 0 V=0N
V +600 – 400 – 600 – 200 + 600 = 0 SF=0NV + R1 – 400 – 600 – 200 + 600 = 0
Bending Moment
Section cuts X1, 2.5>X1≥0
M = 600X1
When X = 0, BM = 0Nm
When X = 2.5, BM = 1500Nm
Section cuts X2, 3>X2≥0
M = 600 X2 – 400 (X2 – 2.5)
When X = 2.5, BM = 1500Nm
When X = 3, BM = 1600Nm
R1=600N
X1 X1
R1=600N X2
X2
400N(X2- 2.5)
X3 X3
400N
(X3- 2.5)
(X3- 3)
600N
R1=600N
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Section cuts X3, 4>X3≥0
M = 600X3 400(X3 2.5) 600(X3 3)
When X = 3, BM = 1600Nm
When X = 4, BM = 1200Nm
Section cuts X4, 6>X4≥0
M = 600 X4 – 400(X4 - 2.5) – 600(X4 – 3) – 200(X4 – 4)
When X = 5, BM = 600Nm
When X=6, BM = 0Nm
-800
-600
-400
-200
0
200
400
600
800
0 1 2 3 4 5 6 7
SF (N)
x (m)
Shear Force Diagram
R1=600NX
4
400N600N
200N
(X4- 2.5)(X4- 3)
(X4- 4)
(X4)
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F) When shear force is decreasing, bending moment is increasing. The length of bending
moment will increases as the shear force increases from negative reading value. Maximum
length of the bending force are 3m long. Thus,it means that after this maximum point the
potential to bend will depleted as the shear force increases.
G) Based on the graph,the maximum shear force are 600N which starts from 0 and remain the
same amount at 600N until 2.5m at x-axis.The maximum bending moment are 1600Nm
which produces at the length of 3m long. The coordinate of the point are (3,1600).
Case 2
w=100N/m
L= 6m
0
200
400
600
800
1000
1200
1400
1600
1800
0 1 2 3 4 5 6 7
BM (Nm)
x (m)
Bending Moment Diagram
3.5m2m
RA RB X X2 X3
X4200N 400N
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∑Fy=0 ∑M=0
∑Fup=∑Fdown W=100N/m ∑Mcw=∑Mccw
RA + RB = 200N + 400N + W (6m) 200N (2m) + 400N (3.5m) + 600N (3m) = RB (6m)
RA + RB = 200N + 400N + 100(6m) 3600Nm = RB (6m)
RA + RB = 1200N RB = 600N
RT = 1200N RA = RT - RB
RA= 1200N - 600N
RA = 600N
Shear Force
Section cuts X1, 2> X1 ≥ 0
∑Fy = 0
V + W(X1) RA=0
V = 600N W(X1) (CW, SF= +Ve)
SF = 600N – 100X1
When x = 0, SF = 600N
When x = 1, SF = 500N
When x = 1.999, SF ≈ 400N
X RA=600N
X
V
w=100N/m
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Section cuts X2, 3.5 >X2 ≥ 0
∑Fy = 0
V + 200N + W(X2) RA = 0
V= RA – 200N – W(X2)
V = 600N – 200N – 100(X2)
V = 400N – 100(X2) (CW, SF= +Ve)
SF = 400N – 100(X2)
When x = 2, SF = 200N
When x = 3, SF = 100N
When x = 3.49, SF ≈ 50N
Section cuts X3, 6 > X3 ≥ 0
∑Fy = 0
V + 200N + 400N + W(X3) RA=0
V= 600N 200N 400N 100(X3)
RA=600N X2
200N
200N
RA=600N X3
400N
X2
X3
w=100N/m
w=100N/M
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V=100(X3) (CCW, SF= - Ve)
SF= -100X3
When x = 3.5, SF = -350N
When x = 4, SF = -400N
When x = 4.5, SF = -450N
When x = 5, SF = -500N
When x = 6, SF = -600N
Section cuts X4, when L=6
∑Fy=0
V + 200N + 400N + W(X4) RA RB = 0
V = 600N + 600N 200N 400N 100(6)
V = 0N
SF = 0N
RA=600N
RA
200N 400N
V
X4
RB
X4
w=100N/M
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Bending Moment
Section cuts X1, 2>X1≥0
∑M = RA (X1) – F (X1 /2)
= RA (X1) – W (X1)2 /2
=600(X1) 100(X1)2 /2
When x = 0, BM = 0Nm
When x = 1, BM = 550Nm
When x = 2, BM = 1000Nm
Section cuts X2 3.5>X2≥0
∑M = RA (X2) – 200 (X2 – 2) – F (X2 /2)
= RA (X2) – 200 (X2 – 2) – W (X2)2 /2
=600 X2 – 200 (X2 – 2) – 100(X2)2 /2
When x = 2m, BM = 1000Nm
When x = 3m, BM = 1150Nm
When x = 3.5m, BM = 1187.5Nm
RA=600N X1
X1
X1 /2
RA=600N X2
X2
X2 /2
200N
X2 - 2
F=W(X2)
F=W(X1)
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Section cuts X3, 6>X3≥0
∑M= RA (X3) – 200(X3 – 2) – F (X3 /2) – 400(X3 – 3.5)
= RA (X3) – 200(X3 – 2) – W (X3)2 /2 – 400(X3 – 3.5)
= 600X3 – 200(X3 – 2) – 100(X3)2 /2 – 400(X3- 3.5)
When x=4m, BM= 1000Nm
When x= 4.5m, BM =787.5Nm
When x= 5m, BM= 550Nm
Section cuts X4, when L=6
∑M= RA X4 – 200(X4 – 2) – F(X4 /2) – 400(X4 – 3.5) + RB (X4)
= RA X4 – 200(X4 – 2) – W(X4)2 /2 – 400(X4 – 3.5) + RB (X4)
=600(6) – 200(4) – 100(36)/2 – 400(2.5) + 600(0)
=0Nm
When x=6m, BM= 0Nm
F=W(X3)
RA=600N
X3 /2
X3 - 2
X3
X3
X3 – 3.5200N
400N
RA=600N
F=W(X4)
X4 /2
200N
400N
X4 – 3.5
X4 - 2
X4 RB=600N
X4
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G) The maximum shear force are 600N when x=0. Whereas,the maximum bending moment are
1200Nm at 3.5m.
h) It is found that in case 2,Uniformly distributed load(UDL) are applied along the beam. Thus,this
causes the graph gained has slanted line at the beginning. (According to the comparison of the shear
force diagram). Also,the peak point in bending moment diagram for case 1 are much more sharp than
the peak point in case 2.
Conclusion,the strength of a material can be determine by finding the amount of shear force and
bending moment. The highest reading for each diagram (which are the bending moment diagram and
shear force diagram) are the maximum shear force and bending moment which the material or object
can withstand and exert. To add on it,both diagram are related to each other. Thus,to make a strong
-800
-600
-400
-200
0
200
400
600
800
0 1 2 3 4 5 6 7
SF(N)
x (m)
Shear Force Diagram
0
200
400
600
800
1000
1200
1400
0 1 2 3 4 5 6 7
BM (Nm)
x (m)
Bending Moment Diagram
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material,we need to take in notice about both amount which is the shear force and bending moment
amount. Supportive item can be added into or to support the material in order to strengthen it and to
avoid it from exceeding its maximum point.
References:
1. http://wiki.answers.com/Q/What_is_shear_force
2. http://www.codecogs.com/reference/engineering/materials/shear_force_and_bending_mo
ment.php
3. http://www.optics.arizona.edu/optomech/references/OPTI_222/OPTI_222_W8.pdf