7.0 RESULT

PART 1

Location Of LoadFrom Left Hand

Support (M)

Digital ForceDisplay

Reading (N)

Moment AtCut Section

(Nm)

ExperimentalInfluence Line

Value (Nm)

TheoreticalInfluence

Lines Value(Nm)

0.04 0.2 0.025 0.013 0.0130.06 0.3 0.038 0.019 0.0190.08 0.4 0.050 0.025 0.0250.10 0.5 0.063 0.032 0.0320.12 0.6 0.075 0.038 0.0380.14 0.7 0.088 0.045 0.0450.16 0.8 0.100 0.051 0.0510.18 0.9 0.113 0.058 0.0570.20 1.0 0.125 0.064 0.0640.22 1.2 0.150 0.076 0.0700.24 1.3 0.163 0.083 0.0760.26 1.6 0.200 0.102 0.0830.30 1.2 0.150 0.076 0.0960.32 0.9 0.113 0.058 0.0820.34 0.7 0.088 0.045 0.0680.36 0.5 0.063 0.032 0.0550.40 0.1 0.013 0.007 0.027

8.0 CALCULATION

PART 1

Moment at cut section = Digital force reading x 0.125

a. Moment at cut section = 0.2 x 0.125= 0.025 Nm

b. Moment at cut section = 0.3 x 0.125= 0.038 Nm

c. Moment at cut section = 0.4 x 0.125= 0.050 Nm

d. Moment at cut section = 0.5 x 0.125= 0.063 Nm

e. Moment at cut section = 0.6 x 0.125= 0.075 Nm

f. Moment at cut section = 0.7 x 0.125= 0.088 Nm

g. Moment at cut section = 0.8 x 0.125= 0.100 Nm

h. Moment at cut section = 0.9 x 0.125= 0.113 Nm

i. Moment at cut section = 1.0 x 0.125= 0.125 Nm

j. Moment at cut section = 1.2 x 0.125= 0.150 Nm

k. Moment at cut section = 1.3 x 0.125= 0.163 Nm

l. Moment at cut section = 1.6 x 0.125= 0.200 Nm

m. Moment at cut section = 1.2 x 0.125= 0.150 Nm

n. Moment at cut section = 0.9 x 0.125= 0.113 Nm

o. Moment at cut section = 0.7 x 0.125= 0.088 Nm

p. Moment at cut section = 0.5 x 0.125= 0.063 Nm

q. Moment at cut section = 0.1 x 0.125= 0.013 Nm

Experimental Influence line values = Moment (Nm) Load (N)

a. Experimental Influence line values = 0.025 1.962

= 0.013 Nmb. Experimental Influence line values = 0.038

1.962= 0.019 Nm

c. Experimental Influence line values = 0.050 1.962

= 0.025 Nmd. Experimental Influence line values = 0.063

1.962= 0.032 Nm

e. Experimental Influence line values = 0.075 1.962

= 0.038 Nmf. Experimental Influence line values = 0.088

1.962= 0.045 Nm

g. Experimental Influence line values = 0.100 1.962

= 0.051 Nmh. Experimental Influence line values = 0.113

1.962= 0.058 Nm

i. Experimental Influence line values = 0.125 1.962

= 0.064 Nmj. Experimental Influence line values = 0.150

1.962= 0.076 Nm

k. Experimental Influence line values = 0.163 1.962

= 0.083 Nml. Experimental Influence line values = 0.200

1.962= 0.102 Nm

m. Experimental Influence line values = 0.150 1.962

= 0.076 Nmn. Experimental Influence line values = 0.113

1.962= 0.058 Nm

o. Experimental Influence line values = 0.088 1.962

= 0.045 Nmp. Experimental Influence line values = 0.063

1.962= 0.032 Nm

q. Experimental Influence line values = 0.013 1.962

= 0.007 Nm

Theoretical Influence lines valueEquation 1 for load position 40 to 260 mm

MX = (L-x)a – 1(a-x) L

a. MX = (0.44 – 0.04)0.30 – 1(0.30-0.04) 0.44

= 0.013 Nmb. MX = (0.44 – 0.06)0.30 – 1(0.30-0.06)

0.44= 0.019 Nm

c. MX = (0.44 – 0.08)0.30 – 1(0.30-0.08) 0.44

= 0.025 Nmd. MX = (0.44 – 0.10)0.30 – 1(0.30-0.10)

0.44= 0.032 Nm

e. MX = (0.44 – 0.12)0.30 – 1(0.30-0.12) 0.44

= 0.038 Nmf. MX = (0.44 – 0.14)0.30 – 1(0.30-0.14)

0.44= 0.045 Nm

g. MX = (0.44 – 0.16)0.30 – 1(0.30-0.16) 0.44

= 0.051 Nmh. MX = (0.44 – 0.18)0.30 – 1(0.30-0.18)

0.44= 0.057 Nm

i. MX = (0.44 – 0.20)0.30 – 1(0.30-0.20) 0.44

= 0.064 Nmj. MX = (0.44 – 0.22)0.30 – 1(0.30-0.22)

0.44= 0.070 Nm

k. MX = (0.44 – 0.24)0.30 – 1(0.30-0.24) 0.44

= 0.076 Nml. MX = (0.44 – 0.26)0.30 – 1(0.30-0.26)

0.44= 0.083 Nm

Theoretical Influence lines valueEquation 2 for load position 320 to 400 mm

MX = xb – (x-a) L

a. MX = (0.32)(0.14) – (0.32-0.30) 0.44

= 0.082 Nmb. MX = (0.34)(0.14) – (0.34-0.30)

0.44= 0.068 Nm

c. MX = (0.36)(0.14) – (0.36-0.30) 0.44

= 0.055 Nmd. MX = (0.40)(0.14) – (0.40-0.30)

0.44= 0.027 Nm

9.0 DISCUSSIONS

PART 1

1. Derive equation 1 and 2

∑Fx = 0∑Fy = RA + RB – 1

= 0

RA + RB = 1RA(L) – 1(L - x) = 0

RA(L) = 1(L - x)RA = 1(L - x)

L= 1 – x

L

RB = 1 –(1 – x) = x

L L

Equation 1, 0≤x≤a

-Mx + RA(a) – 1(a-x) = 0

Mx = (1 – x)a – 1(a – x) L

= (L – x)a – 1(a – x) L

Equation 2, a≤x≤b

Mx – RB(b) + 1(x - a) = 0

Mx = RB(b) – 1(x – a)= x(b) – 1(x – a)

L2. On the graph paper given, plot the theoretical and experimental values against distance from

left hand support.Comment on the shape of the graph. What does it tell you about how moment varies at the cut section as a load moved on the beam.

0.040.06

0.08 0.10.12

0.140.16

0.18 0.20.22

0.240.26 0.3

0.320.34

0.36 0.40

0.02

0.04

0.06

0.08

0.1

0.12

Experimental ValueTheoretical Value

From the graph, a peak shaped graph can be obtained. The peak is the weakest point of the beam where there is a hinge in the beam. As load is being moved on the beam, the influence line which was constructed can be used to obtain the value of the moment. As load is moved across near to it, the moment will increase. So does the other way round when load is moving further than the hinge, the value of moment will decrease as the load is moving towards the

support at the end. As the load is moving along towards the hinge from both side of support, it will come to a peak where the value of moment is the same.

3. Comment on the experimental results and compare it to the theoretical results.The experimental results that we obtained are only slightly different with the theoretical results. During conducted the experiment, we tried to minimize the error by ensuring the Digital Force is zero before we place the hangers.

10.0 CONCLUSIONS

As a conclusions, both objectives were achieved. Moment influence line could be plot and the influence line which was constructed can be used to obtain the value of the moment