Structural Design Principles: A Comprehensive Report

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PRINCIPLES OF STRUCTURAL DESIGN

Contents
Introduction......................................................................................................................................2
LO1..................................................................................................................................................3
Task 1(A):....................................................................................................................................3
Task 1(B):..................................................................................................................................13
Task 1(C):..................................................................................................................................14
Task 1(D):..................................................................................................................................17
LO 2...............................................................................................................................................19
LO3................................................................................................................................................19
Task 3.........................................................................................................................................19
LO4................................................................................................................................................20
Task 4.........................................................................................................................................20
Conclusion.....................................................................................................................................24
References......................................................................................................................................25
Figure 1: Provided...........................................................................................................................4
Figure 2: Diagram 1.........................................................................................................................6
Figure 3: Diagram 2.........................................................................................................................9
Figure 4: Provided...........................................................................................................................9
Figure 5: Diagram 3.......................................................................................................................11
Figure 6: Diagram 4.......................................................................................................................13
Figure 7: Diagram 5.......................................................................................................................16
Figure 8: Given..............................................................................................................................21
Figure 9: Diagram 6.......................................................................................................................21
Figure 10: Table.............................................................................................................................24
Figure 11: Diagram 7.....................................................................................................................26

Introduction
This particular assignment contains the detailed discussion on reaction forces that a particular
beam faces. Also, the shear forces, as well as bending movement, are also discussed. The report
has four different tasks that need to perform in which various subparts are included. Calculation
and proper math are performed in order to provide a proper illustration for calculating shear
forces, bending moments and other various things that are discussed in the overall report.
Discussion on alternative material is also added in the file that includes the use of timbre and its
disadvantages and benefits are also mentioned in the file.

LO1
Calculations for the bending movements as well as for shear forces:
Task 1(A):
a) Given:
Figure 1: Provided
Load point given as: 150KN
Span length: l = 14m
Need to calculate:
Reaction forces at A and B, BM and SF:
Calculations:
(a) Reaction forces on A and B:
Upward force Sum – Downward force sum = 0
RA + RB – 150 KN = 0
RA + RB = 150 KN  equation (i)
Moments sum = 0
RB x 14 – 150 x 7 = 0

RB = 75 KN
As seen from equation (i),
RA + RB = 150 KN
As RB = 75 KN, substitute values:
RA = 75KN
(b) Bending Moment on A, B and C points:
At point A:
BM= 0
At point B:
BM= 0
At point C:
(BM)C = RA x (L/2)
As we know: RA = (W/2)
Replace values: (BM) C = (W/2) x (L/2) = WL/4
Substitute the values:
W= 150, L=14
(BM) C = (150 x 14)/ 4 = 525 KN. m
(c) Shear forces:
At point A:
(SF)A = W / 2
W= 150 given:

(SF)A = W / 2 = 150/ 2 = 75 KN
Therefore, (SF) A =75 KN
At point B:
(SF) B = W/2 = (- W + W) / 2 = 0
Therefore, (SF) B = 0
At point C:
(SF) C = (W/2) – W = (W – 2W)/ 2 = - W/ 2
Substitute value of W: -75 KN
(SF) C = -75 KN
Diagram:
Figure 2: Diagram 1
b) Given:

Load point given as: 65KN
Span length: l = 8m
Need to calculate:
Reaction forces at A and B, BM and SF
Calculations:
(a) Reaction forces on A and B:
RA = (W x b)/ l
Where, W= 65, b=6, l=8: On substituting and calculation, we’ll get:
RA = 48.75
Therefore, on point A, reaction force is 48.75KN
For RB, use formula:
RB = (W x a)/ l
Where, W= 65, a=2, l=8: On substituting and calculation, we’ll get:
RB = 16.25
Therefore, on point B, reaction force is 16.25KN
(b) Bending Moment on A, B and C points:
At point A:
BM= 0
At point B:
BM= 0
At point C:

(BM)C = RA x a
As seen above, RA is (W x b)/ l, substitute in the equation of (BM) C as well as the values of
the W, b and l:
(BM)C = (W x b)/ l x a = (65 x 6)/8 x 2 = 97.5KN.m
(c) Shear forces:
At point A:
SF = (W x b)/ l
On substituting values of W, b and l:
SF = (65 x 6)/ 8 = 48.75KN
(SF)A = 48.75KN
At point B:
(SF)B = - [(W x a)/ l] + [(W x a)/ l] = - (W x a) + (W x a) / l = 0
(SF)B = 0
At point C:
(SF)C = [(W x b)/ l] – W
Take W common from above equation:
(SF)C = W [(b / l) - 1] = W [(b - l) / l] = -W [(l - b)/ l]
As it is already known:
l = a + b, a = l – b
Substitute a = l – b:
(SF)C = -W [(a)/ l]

Substitute values here:
(SF)C = -16.25 KN
Diagram:
Figure 3: Diagram 2
(c) Given:
Figure 4: Provided
UDL = 25KN

Span Length = 25m
Need to calculate:
Reaction forces at A and B, BM and SF
Calculations:
(a) Reaction forces on A and B:
Upward force sum – Downward force sum = 0
RA + RB – (25 x 25) = 0
RA + RB = 625 KN  equation (i)
Sum of the moments = 0
Calculate t at B:
RB x 25 = (25 x 25) (25/2)
RB = 312.5KN
As we have value of RB, on proceeding further we’ll get RA by using equation (i):
RA + RB = 625 KN
RA + 312.5 = 625
RA = 312.5 KN
(b) Bending Moment on A, B and C points:
At point A:
(BM) = 0
At point B:
BM= 0

At point C:
(BM)C = ½ x 12.5 x 32.5 = 1953.12 KN
(c) Shear force calculation
At point A:
SF = - (W x l) / 2
Substitute the values of W and l:
SF = -312.5KN
At point B:
(SF)B = (W x l) / 2 = 312.5KN
Diagram:
Figure 5: Diagram 3

(d) Given:
UDL = 20 KN/m
Load point = 150KN
Need to calculate:
Reaction forces at A and B, BM and SF
Calculations:
(a) Reaction forces on A and B:
RA + RB = 430 KN
Sum of moment force: 0
Calculate moment considering point A:
-RB x 14 + (150 x 7) + (20 x 14) (14/ 2)
3010 = 14 RB
RB = 215KN
Use equation RA + RB = 430 KN:
Replace value of RB => RA + 215 = 430 KN
RA = 215KN
(b) Shear Forces:
(SF)B =215KN
For (SF) C: