Structural Design Principles: A Comprehensive Report

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20 principals of structural design
1

Table of Contents
LO1..............................................................................................................................................................3
Task 1..........................................................................................................................................................3
(A)............................................................................................................................................................3
a...............................................................................................................................................................3
(b)............................................................................................................................................................5
(C):...........................................................................................................................................................7
(D)..........................................................................................................................................................10
(E)..........................................................................................................................................................13
B................................................................................................................................................................17
C................................................................................................................................................................17
D................................................................................................................................................................20
LO2............................................................................................................................................................21
A................................................................................................................................................................21
B................................................................................................................................................................23
C................................................................................................................................................................24
D................................................................................................................................................................26
LO3............................................................................................................................................................28
Task A:.......................................................................................................................................................28
Task B........................................................................................................................................................30
(b)..........................................................................................................................................................30
C................................................................................................................................................................30
LO4............................................................................................................................................................31
(A)..............................................................................................................................................................31
Task B:.......................................................................................................................................................32
Task C........................................................................................................................................................32
Task D........................................................................................................................................................34
References.................................................................................................................................................36
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Table of Figures
Figure 1 Acting of point load at mid............................................................................................................4
Figure 2 Force of shear and bending moment.............................................................................................6
Figure 3 Acting of point load at 2 m............................................................................................................7
Figure 4 shear and bending force................................................................................................................8
Figure 5 Load acting uniformly....................................................................................................................9
Figure 6 Solution figure.............................................................................................................................11
Figure 7 Point and uniformly acting...........................................................................................................12
Figure 8 Solution image.............................................................................................................................14
Figure 9 point and uniformly acting load...................................................................................................15
Figure 10 Solution image...........................................................................................................................17
Figure 11 Image of given loads..................................................................................................................20
Figure 12 Solution for LFRD.......................................................................................................................20
Figure 13 Solution for LRFD.......................................................................................................................21
Figure 14 Image for task 2.........................................................................................................................22
Figure 15 Deflection in beam.....................................................................................................................24
Figure 16 Task 2 figure...............................................................................................................................25
Figure 17 Point load figure........................................................................................................................27
Figure 18 Structural diagram.....................................................................................................................33
Figure 19 Straw bale..................................................................................................................................33
Figure 20 Wood.........................................................................................................................................34
Figure 21 Bamboo.....................................................................................................................................34
Figure 22 Timber.......................................................................................................................................35
Figure 23 Structure of BIM........................................................................................................................35
3

LO1
Task 1
(A)
a.
Given parameters:
Length of span = 144 m
A load of point ac on center = 150KN
From these parameters, the arrangement can be shown as:
Figure 1 Acting of point load at mid
1. Calculation of the reaction force by the above arrangements
A calculation can be made by using the above diagram:
Submission of the reacting forces that can act in upward direction = submission of the
reacting forces that can act in a downward direction
That means, Ra + Rc = Rb (a)
It is known that the sum of moments acting is zero = 0
Now, consider the point A:
Rc*14 = 150*7
Rc = (150*7/14) or
= 75 KN
Now, consider the point C:
Ra + Rc = 150
4

Ra = 150 – Rc
Ra = 150 – 75 or
= 75 KN
2. The calculation for the shear forces that act based on the above arrangements
Consider point A:
Force of shear at point A = (acting force is 150 KN/2) or
= 75 KN
Consider point C:
A force of shear at point C= (force acting is 150 KN/2) – (force acting is 150
KN/2)
Force of shear at point C = 0 KN
Now, consider point B
A force of shear at point B = (force acting is 150 KN/2) - acting force is 150 KN
= (acting force is 150 KN- 2*acting force is 150
KN)/2
= -75 KN
3. The calculation for the moment of bending based in the upward diagram
Consider point A:
Moment of bending at point A = zero (0)
Consider point C:
Moment of bending at point C = zero (0)
Consider point B;
Moment of bending at point B = Ra*(half length of the span)
=75 KN * 7 m or
=55 KN m
So, now create the image for a force of shear along with the moment of bending that can
be activated according to the above diagram and on the origin of calculations, so we get:
5

Figure 2 Force of shear and bending moment
(b)
from the given data in question:
Length of span = 8m
Load f point act on the center = 65 KN
6

Load position = (from left) 2m
The arrangement will be based on the given data:
Figure 3 Acting of point load at 2 m
a. Calculation of the force of reacting based on the above diagram
Consider point A:
Rc*8 = 65*2 or
Rc = (65*2)/8 or
= 16.25 KN
b. The calculation for a force of shear that acts according to the above diagram
Consider point A:
A force of shear at point A = (acting force of 65KN*6)/8 or
= 48.75 KN
Consider point C:
Force of shear at point C = (acting force of 65KN*)/8 - (acting force of
65KN*)/8
Force of shear at point C = 0KN
Consider point B:
Force of shear at point B = (acting force of 65KN*)/8 - (acting force of
65KN*)
= [(acting force of 65KN*) - 2(acting force of
65KN*)]/8
= -16.25KN
c. Calculation for moment of bending act according to above diagram
Consider point A:
Moment of bending = zero (0)
Consider point C:
Moment of bending = zero (0)
Consider point B:
Moment of bending = (65*6/8) *2 or
= 97.50KN
So, now create the image for shear force along with the bending moment that acts
Following the result of the above calculations. So, we get,
7

Figure 4 shear and bending force
(C):
As a basis on the given data in the question:
Length of the span = 25m
Uniform acting load = 25KN/m
8

The arrangement can be based on the upward data:
Figure 5 Load acting uniformly
a. The calculation for a force of reaction that acts on the basis on the diagram:
Form this setup, we calculate for the equilibrium condition:
Submission of the reacting force in the above direction = submission of the reacting force
in the downward direction
That is Ra + Rc = 25*25 = 625 (1)
As we know the mean of acting of the moments = zero (0)
Consider point A:
Rc*25 = 625*12.5 or
Rc = (625*12.5)/25 or
= 312.50 KN
Consider point C and use the equation (1):
Ra + Rr = 625
Ra= 625 – 312.50 or
Ra = 312.50 KN
b. The calculation for shear forces act based on the diagram:
Consider point A:
A force of shear at point A = - (acting force of 25KN*25)/2 or
= - 312.50 KN
Consider point B:
Force of shear at point B + (acting force of 25KN*25)/2 = (acting force of
25KN*25)/2
Force of shear at point B =zero (0) KN
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Consider point C:
A force of shear at point C = (acting force of 25KN*25)/2 or
= 312.50 KN
c. The calculation for the moment of bending that based on the given arrangement
Consider point A:
Moment of bending at point A = zero (0)
Consider point C:
Moment of bending at point C = zero (0)
Consider point B:
Moment of bending at point B = force of sheer * half-length for half span
= (312.50 KN) *12.5 / 2 or
= 1953.12 KN
So, now create the image for the moment of bending along with the force of shear that
can be based on the result of upward calculations. So, we get:
10

Figure 6 Solution figure
(D)
Based on the question:
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Length of span = 14m
Load point act on the center =150 KN
Uniformly acting load = 20 KN/m
The set u for this will be shown as:
Figure 7 Point and uniformly acting
a. The calculation for the reaction force based on the arrangement
The equilibrium condition can be calculated using the above diagram:
Submission of forces that act in upward direction = submission of forces that act in the
downward direction
That is Ra + Rc = 150 + (20*14) or
= 430 KN (1)
As we know that the mean for acting moment = zero (0)
Consider point P:
Rc * 14 = (150*7) + (20*14*7) or
= 1050 + 1960
Rc = 1050 + 1960/14 or
Rc = 215 KN
Consider point R:
Ra + Rc = 430
Ra = 430 – Rr or
Ra = 430 – 215 or
= 215 KN
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