Open Ended Ques

7/27/2019 Open Ended Ques

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BFC 31901

OPEN ENDED QUESTION

1. How to determine the absolute maximum shear and absolute moment using influence

lines variation?

Due to a given system of rolling loads the maximum shear force for every section of the

girder can be worked out by placing the loads in appropriate positions. When these are

plotted for all the sections of the girder, the diagram that we obtain is the maximum shear

force diagram. This diagram yields the ‘design shear’ for each cross section.

While, when a given load system moves from one end to the other end of a girder,

depending upon the position of the load, there will be a maximum bending moment for

every section. The maximum of these bending moments will usually occur near or at the

mid span. The maximum of maximum bending moments is called the absolute maximum

bending moment. When a series of wheel loads crosses a simply supported beam, the

absolute maximum bending moment will occur near mid span under the load Wcr ,

nearest to mid span (or the heaviest load). If Wcr is placed to one side of mid span C, the

resultant of the load system R shall be on the other side of C; and Wcr and R shall be

equidistant from C. Now the absolute maximum bending moment will occur under Wcr .If Wcr and R coincide, the absolute maximum bending moment will occur at mid span. :

When a simply supported beam is subjected to a moving udl longer than the span, the

absolute maximum bending moment occurs when the whole span is loaded. Mmax max =



Procedure:

i. Allow a unit load (either 1b, 1N, 1kip, or 1 tonne) to move over beam from left to

right.

ii. Find the values of shear force or bending moment, at the point under

consideration, as the unit load moves over the beam from left to right.

iii. Plot the values of the shear force or bending moment, over the length of the beam,

computed for the point under consideration



2. By using the data in Table 2 from the experiment moment influence lines, explain how to

determine the absolute maximum shear?

Location Position of hanger from

left hand support (m)

Digital

Force

Reading

(N)

Experiment

Moment

(Nm)

Theoretical

Moment (Nm)

100 g 200 g 300 g

1 0.04 0.20 0.32 3.0 0.375 0.814

2 0.06 0.22 0.34 2.8 0.350 0.873

3 0.08 0.26 0.36 2.8 0.350 0.942

4 0.10 0.30 0.38 2.7 0.338 1.010

TOTAL 1.413 3.639

AVERAGE 0.3533 0.9098

Notes:

1. Experimental Moment = Digital Force Reading x 0.125

2. Theoretical Moment is calculated using Equation (3)

CALCULATION EXAMPLE FOR PART 2

1. Experiment Moment = Digital Force Reading x 0.125

= 3.0 x 0.125

= 0.375 N



2. Theoretical Moment is calculated using Equation (3)

Moment at ‘cut’ section = F1 y1 + F2 y2 + F3 y3……..

F1 (100g) = 0.981N

F2 (200g) = 1.962 N

F3 (300g) = 2.943 N

x

0.3m 0.14m

‘Cut’

0.981N

0.0955

1.962N 2.943N

x

x



EXAMPLE

x1 = 0.06mx2 = 0.22m

x3

= 0.3m

Influence lines x

0.075 = _y1_

0.3 0.04

y1 = 0.0127m

0.075 = _y2_

0.3 0.16

y2 = 0.0509m

0.075 = y3 __

0.14 (0.44-0.34)

y3 = 0.0682m

Moment at ‘cut’ section = F1y1+ F2y2 + F3y3

= 0.981(0.0127) + 1.962(0.0509) + 2.943(0.0682)= 0.0125 + 0.0999 + 0.2007

= 0.3131Nm



12.5 kN 37.5 kN

35.23 kN 39.77 kN

0.04 m 0.16 m 0.12 m 0.12 m

35.23 kN

-2.27 kN

-39.77kN

+ve

1.4092 kN.m

5.046 kN.m

4.7736kN.m

39.00 kN.m

22.73 kN

Shear Force Diagram and Moment Deflection Diagram

25 kN



As we know the absolute maximum shear occurred at one of the support of simply supported

beam. So when we calculated shear using influence lines method, we can get the value of shear

when we inserted the load at each point of the beam. Therefore, we can sketch the SFD or named

as Shear Force Diagram. From the graph above, we found that the absolute maximum shear is

35.23kN happened at supported A.

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