This document discusses the calculation of shear forces at different points on a beam due to moving concentrated loads. It provides three conditions for calculating the change in shear force as loads move along the beam. As an example, it calculates the maximum shear forces at points C and D on a beam as 14 loads move across it from right to left. It is found that the maximum shear forces are 75.992 kN at point C and 111.94 kN at point D.

1. CEE-311
Structural Analysis-I
(3.0 credit)
Lecture:12 & 13
Bijit Kumar Banik
Assistant Professor, CEE, SUST
Room No.: 115 (“C” building)
bijit_sustbd@yahoo.com
Department of Civil and Environmental Engineering

2. Shear due to moving concentrated loads (S&V; pp-117)
Condition-1: (No wheels comes on the span during movement)
1
1.
P
L
dP
V −=∆
∑
Where, V = change in shear due to moving from one wheel at
the section to the following wheel at section
P1 = the wheel which was over the section just before
movement & is moved off during movement
d1 = the distance between P1 and the following wheel
∑P = the summation of all loads which are on the span
& stay during movement

3. Shear due to moving concentrated loads (S&V; pp-112)
Condition-2: (Wheels comes on the span during movement but
no wheel moved off the span)
Where, Pin = load which comes on during movement
ein = distance which Pin moves on the span
1
1 ..
P
L
eP
L
dP
V inin
−+=∆
∑

4. Shear due to moving concentrated loads (S&V; pp-112)
Condition-3: (wheel moved off the span)
Where, Pout = load which moved off the span
eout = distance which Pout moves on the span before
moved off
If ∆V,
Positive → continue movement
Negative → stop movement
1
1 ...
P
L
eP
L
eP
L
dP
V outoutinin
−++=∆
∑

5. Shear due to moving concentrated loads (S&V; pp-112)
Find maximum shear at ‘C’ when wheels move
right to left.
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
0.67
0.33
VC
+
-

6. Shear due to moving concentrated loads (S&V; pp-112)
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 1 at ‘C’
1ʹ

7. Shear due to moving concentrated loads (S&V; pp-112)
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 2 at ‘C’
2ʹ
P = 174 k [No.1 to no.7]
Condition-2
ContinuekV 83.410
60
2*10
60
5*174
+=−+=∆

8. Shear due to moving concentrated loads (S&V; pp-112)
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 3 at ‘C’
1ʹ
P = 184 k [No.1 to no.8]
Condition-2
ContinuekV 87.1410
60
1*20
60
8*184
+=−+=∆

9. Shear due to moving concentrated loads (S&V; pp-112)
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-4: wheel 4 at ‘C’
3ʹ
P = 204 k [No.1 to no.8]
Condition-2
StopkV 6.1436
60
3*20
60
6*204
−=−+=∆

10. Shear due to moving concentrated loads (S&V; pp-112)
So, wheel 3 at C will produce maximum shear
= 75.992 k
0.67
0.33
VC
+
-
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
( ) [ ] [ ]10*710*12
20
33.0
20*110*1010*1436*2236*2836*3436*40
40
67.0
max
+−++++++=CV

11. Shear due to moving concentrated loads (S&V; pp-112)
Find maximum shear at ‘D’ when wheels move
right to left.
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
10ʹ 50ʹ
D
0.83
0.17
VD
+
-

12. Shear due to moving concentrated loads (S&V; pp-112)
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
10ʹ 50ʹ
D
Position-1: wheel 1 at ‘D’
7ʹ

13. Shear due to moving concentrated loads (S&V; pp-112)
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
10ʹ 50ʹ
D
Position-2: wheel 2 at ‘D’
3ʹ
P = 184 k [No.1 to no.8]
Condition-2
ContinuekV 33.610
60
3*20
60
5*184
+=−+=∆

14. Shear due to moving concentrated loads (S&V; pp-112)
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
10ʹ 50ʹ
D
Position-3: wheel 3 at ‘D’
3ʹ
P = 194 k [No.2 to no.9]
Condition-3
ContinuekV 03.2010
60
5*10
60
3*20
60
7*20
60
8*194
+=−+++=∆
Pin
Pin Pout

15. Shear due to moving concentrated loads (S&V; pp-112)
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
10ʹ 50ʹ
D
Position-4: wheel 4 at ‘D’
9ʹ
P = 224 k [No.3 to no.11]
Condition-3
StopkV 27.1336
60
2*10
60
6*224
−=−+=∆
Pout

16. Shear due to moving concentrated loads (S&V; pp-112)
So, wheel 3 at D will produce maximum shear
= 111.94 k
( ) [ ]
10*2*
10
17.0
20*320*720*1110*2010*2436*3236*3836*4436*50
50
83.0
max
−
++++++++=DV
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
10ʹ 50ʹ
D
0.83
0.17
VD
+
-