B20_EE_ES1 Task1_Edvin Saw Kien Yip_Koay Cheng Chye

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



3

Section cuts X2, 3>X2≥0

FY = 0

R1 – V – 400 = 0

V = 600 – 400

= 200N (CW, SF = +VE)

SF = 200N


FY = 0

V + R1 1000 = 0

V = 1000 600

= 400N (CCW, SF = – VE)

SF = – 400N


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



4

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


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



5


M = 600X3 400(X3 2.5) 600(X3 3)

When X = 3, BM = 1600Nm

When X = 4, BM = 1200Nm


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



8

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



9

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



10

Bending Moment


∑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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∑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

http://wiki.answers.com/Q/What_is_shear_force


http://www.codecogs.com/reference/engineering/materials/shear_force_and_bending_moment.php





http://www.optics.arizona.edu/optomech/references/OPTI_222/OPTI_222_W8.pdf






B20_EE_ES1 Task1_Edvin Saw Kien Yip_Koay Cheng Chye

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