structural analysis

1. CEE-311
Structural Analysis-I
(3.0 credit)
Lecture:14
Bijit Kumar Banik
Assistant Professor, CEE, SUST
Room No.: 115 (“C” building)
bijit_sustbd@yahoo.com
Department of Civil and Environmental Engineering

2. Moment due to moving concentrated loads (S&V; pp-119)
M = (Mʹ1 + Mʹ2) – (M1 + M2)
Initial moment = M1 + M2
M2M1
W2W1
Mʹ2Mʹ1
Wʹ2Wʹ1
IL of MC
Final moment = Mʹ1 + Mʹ2
= (Mʹ2 – M2 ) - (M1 - Mʹ1)
= I - D
I = Increase in moment at the section due to the movement of
those loads on the right of the section
C
D = Decrease in moment at the section due to the movement of
those loads on the left of the section

3. Derive the criteria to investigate position of maximum moment
M = I - D
I = Increase in moment at the section due to the movement of
those loads on the right of the section
D = Decrease in moment at the section due to the movement of
those loads on the left of the section
For a unit movement to the left
I = W2*(i/b)
W2W1
IL of MCC
i
a b
A
B
O
[(i/b) = slope of IL BO]
D = W1*(i/a) [(i/a) = slope of IL AO]

4. Derive the criteria to investigate position of maximum moment
M = W2*(i/b) - W1*(i/a)
W2W1
IL of MCC
i
a b
A
B
O
BM at ‘C’ cannot change from an increasing to a decreasing
function unless a load passes from right of C to the left.
In order to pass through a maximum value
The change in moment must equal to zero
M = 0
W1*(i/a) = W2*(i/b)
W1/a = W2/b

5. Derive the criteria to investigate position of maximum moment
W2W1
IL of MCC
i
a b
A
B
O
W1/a = (W1+W2)/(a+b)
W1/a = W/L
L
Where, W = total load on span
W1 = Loads on left of the section
L = Length of span
This is the required criteria

6. Example
Find maximum Moment at ‘C’
1 2 3 4 5 6 7 8 9 10 11 12 13 14
5ʹ 8ʹ 6ʹ 6ʹ 6ʹ 8ʹ 4ʹ 9ʹ 4ʹ 4ʹ 12ʹ 4ʹ 4ʹ
10k 10k 36k 36k 36k 36k
10k 10k 20k 20k 20k
20k 20k 20k
A B
20ʹ 40ʹ
C
MC
(20*40)/60=40/3

7. Example
Solution:
1 2 3 4 5 6 7 8 9 10 11 12 13 14
5ʹ 8ʹ 6ʹ 6ʹ 6ʹ 8ʹ 4ʹ 9ʹ 4ʹ 4ʹ 12ʹ 4ʹ 4ʹ
10k 10k 36k 36k 36k 36k
10k 10k 20k 20k 20k
20k 20k 20k
A B
20ʹ 40ʹ
C
Position-1: wheel 3 at ‘C’
W = No.1 to No.9 = 204 k
W/L = 204/60 = 3.4 k/ft
Wheel 3 right to C: W1/a = 20/20 = 1 k/ft <W/L
Wheel 3 left to C: W1/a = 56/20 = 2.8 k/ft <W/L
Not ok

8. Example
Solution:
1 2 3 4 5 6 7 8 9 10 11 12 13 14
5ʹ 8ʹ 6ʹ 6ʹ 6ʹ 8ʹ 4ʹ 9ʹ 4ʹ 4ʹ 12ʹ 4ʹ 4ʹ
10k 10k 36k 36k 36k 36k
10k 10k 20k 20k 20k
20k 20k 20k
A B
20ʹ 40ʹ
C
Position-2: wheel 4 at ‘C’
W = No.1 to No.10 = 224 k
W/L = 224/60 = 3.73 k/ft
Wheel 4 right to C: W1/a = 56/20 = 2.8 k/ft <W/L
Wheel 4 left to C: W1/a = 92/20 = 4.8 k/ft >W/L
ok

9. Example
Solution:
1 2 3 4 5 6 7 8 9 10 11 12 13 14
5ʹ 8ʹ 6ʹ 6ʹ 6ʹ 8ʹ 4ʹ 9ʹ 4ʹ 4ʹ 12ʹ 4ʹ 4ʹ
10k 10k 36k 36k 36k 36k
10k 10k 20k 20k 20k
20k 20k 20k
A B
20ʹ 40ʹ
C
Position-3: wheel 5 at ‘C’
W = No.3 to No.11 = 224 k
W/L = 224/60 = 3.73 k/ft
Wheel 5 right to C: W1/a = 72/20 = 3.6 k/ft <W/L
Wheel 5 left to C: W1/a = 108/20 = 5.4 k/ft>W/L
Also ok
So, both position 2 and 3 satisfied the condition

10. Example
1 2 3 4 5 6 7 8 9 10 11 12 13 14
5ʹ 8ʹ 6ʹ 6ʹ 6ʹ 8ʹ 4ʹ 9ʹ 4ʹ 4ʹ 12ʹ 4ʹ 4ʹ
10k 10k 36k 36k 36k 36k 10k 10k 20k 20k20k 20k 20k20k
A B
20ʹ 40ʹ
C
MC
(20*40)/60=40/3
Wheel 4 at C: MC= 1793 ft-k
Wheel 5 at C: MC= 1756 ft-k
So, wheel 4 at ‘C’ will produce Maximum moment and it
is 1793 k-ft
Please calculate by yourself !