Influence Line of determinate beams and frames( Various types) are drawn using Brute force method with detailed calculation. Professor Dr. Tarif Uddin Ahmed, Dept. of CE, RUET asked us to solve 23 different sets of problems. I made a solution of these problems using Autocad by myself with all details that can be possibly shown. - Md. Ragib Nur Alam, CE -13, RUET. Copyright reserved.

1. Problem 1:
When 1k at point D:
𝑅 𝐵 = 1.5 𝑘
𝑅 𝐸 = −0.5𝑘
𝑉𝐴 = −1 𝑘
𝑉𝐵 = 0.5𝑘
𝑉𝐶 = 0.5 𝑘
𝑀𝐴 = −20 𝑘 − 𝑓𝑡
𝑀 𝐵 = −30 𝑘 − 𝑓𝑡
𝑀 𝐶 = −15 𝑘 − 𝑓𝑡
When 1k at point A:
𝑅 𝐵 = 1.167 𝑘
𝑅 𝐸 = −0.167 𝑘
𝑉𝐴 = −1 𝑘 (𝑤ℎ𝑒𝑛 1𝑘 𝑎𝑡 𝑠𝑙𝑖𝑔ℎ𝑡 𝑙𝑒𝑓𝑡 𝑜𝑓 𝐴)
𝑉𝐴 = 0 𝑘 (𝑤ℎ𝑒𝑛 1𝑘 𝑎𝑡 𝑠𝑙𝑖𝑔ℎ𝑡 𝑟𝑖𝑔ℎ𝑡 𝑜𝑓 𝐴)
𝑉𝐵 = 0.167 𝑘
𝑉𝐶 = 0.167 𝑘
𝑀𝐴 = 0 𝑘 − 𝑓𝑡
𝑀 𝐵 = −10 𝑘 − 𝑓𝑡
𝑀 𝐶 = −5 𝑘 − 𝑓𝑡
When 1k at point B:
𝑅 𝐵 = 1 𝑘
𝑅 𝐸 = 0 𝑘
𝑉𝐴 = 0 𝑘
𝑉𝐵 = 0 𝑘 (𝑤ℎ𝑒𝑛 1𝑘 𝑎𝑡 𝑠𝑙𝑖𝑔ℎ𝑡 𝑙𝑒𝑓𝑡 𝑜𝑓 𝐵)
𝑉𝐵 = 1 𝑘 (𝑤ℎ𝑒𝑛 1𝑘 𝑎𝑡 𝑠𝑙𝑖𝑔ℎ𝑡 𝑟𝑖𝑔ℎ𝑡 𝑜𝑓 𝐵)
𝑉𝑐 = 0 𝑘

2. 𝑀𝐴 = 0 𝑘 − 𝑓𝑡
𝑀 𝐵 = 0 𝑘 − 𝑓𝑡
𝑀 𝐶 = 0𝑘 − 𝑓𝑡
When 1k at point C:
𝑅 𝐵 = 0.5 𝑘
𝑅 𝐸 = 0.5𝑘
𝑉𝐴 = 0 𝑘
𝑉𝐵 = 0.5 𝑘
𝑉𝐶 = −0.5 𝑘 (𝑤ℎ𝑒𝑛 1𝑘 𝑎𝑡 𝑠𝑙𝑖𝑔ℎ𝑡 𝑙𝑒𝑓𝑡 𝑜𝑓 𝐶)
𝑉𝐶 = 0.5 𝑘 (𝑤ℎ𝑒𝑛 1𝑘 𝑎𝑡 𝑠𝑙𝑖𝑔ℎ𝑡 𝑟𝑖𝑔ℎ𝑡 𝑜𝑓 𝐶)
𝑀𝐴 = 0 𝑘 − 𝑓𝑡
𝑀 𝐵 = 0 𝑘 − 𝑓𝑡
𝑀 𝐶 = 15 𝑘 − 𝑓𝑡
When 1k at point E:
𝑅 𝐵 = 0 𝑘
𝑅 𝐸 = 1𝑘
𝑉𝐴 = 0 𝑘
𝑉𝐵 = 0 𝑘
𝑉𝐶 = 0 𝑘
𝑀𝐴 = 0 𝑘 − 𝑓𝑡
𝑀 𝐵 = 0 𝑘 − 𝑓𝑡
𝑀 𝐶 = 0 𝑘 − 𝑓𝑡
When 1k at point F:

3. 𝑅 𝐵 = −0.167 𝑘
𝑅 𝐸 = 1.167𝑘
𝑉𝐴 = 0 𝑘
𝑉𝐵 = −0.167 𝑘
𝑉𝐶 = −0.167 𝑘
𝑀𝐴 = 0 𝑘 − 𝑓𝑡
𝑀 𝐵 = 0 𝑘 − 𝑓𝑡
𝑀 𝐶 = −5 𝑘 − 𝑓𝑡

5. c
A B
ba
20 17 33 28
1k1k
0.4k
0.34k
0.66k
0.56k
0.4k
0.34k
1k
0.56k
20k' 11.2k'
9.52k'
13.2k'
28k'
0k
0k
0k
0k
0k
0k
0k' 0k'
0k'
0k'
0k'
0k'
IL for Va
IL for Vb
IL for Vc
IL for Ma
IL for Mb
IL for Mc
Problem 2
Draw ILD for shear and moment at a,b & c

6. 1
1
6 6 12 18
12
8
A
B C
c
b
d
Problem 4
Draw ILD for shear and moment at b,c & d
RC
RC cos 45
RC sin 45
RA
RA
Vb
Vc
Vd
Mb
Mc
Md
1.4k 1.2k 1k
0.6k
-1k
0.4k 0.2k
0.4k
1k
0.4k
0.2k
0.6k
0.4k
-6k'
8k'
3.2k' 1.6k'
3.2k'
7.2k' 3.6k'
7.2k'
Md. Ragib Nur Alam
CE 13
ragibnur.ce@gmail.com

7. 1
1
6 6 12 18
12
8
A
B C
c
b
d
RC
RC cos 45
RC sin 45
1k
when 1k at B
RA=42
30 =1.4k
RC cos 45 =-0.4k
RC=- 0.4
cos 45 =-0.566k
RC sin 45=-0.566sin 45=-0.4k
Vb=-1k
Vc=-RC sin 45=0.4
Vd=-RC cos 45 =0.4
Mb=-6k'
Mc=3.2k'
Md=-7.2k'
6
6
12
18
128
A
B
C
c
b
dRC
RC cos 45
RCsin45
RCsin45
8
A
c
RARA
Vc
1
1
18
C
d RC
RC cos 45
RC sin 45
Vd Right side of d so, Vd
in opposite direction
6
B
b
1k Vb
Left side of b
Left side of c
RCsin45
6 6 12 18
12
8
A
B C
c
b
d
1k
When 1k at b
RA=36
30 =1.2k
RC cos 45 =-0.2k
RC=- 0.2
cos 45 =-0.283k
RC sin 45=-0.283sin 45=-0.2k
Vb=-1k(left)
Vb=0(right)
Vc=-RC sin 45=0.2
Vd=-RC cos 45 =0.2
Mb=0k'
Mc=1.6k'
Md=-3.6k'
D
1
1
6 6 12 18
12
8
A
B C
c
b
d RC
RC cos 45
RC sin 45
D
1
1
6 6 12 18
12
8
A
B C
c
b
d RC
RC cos 45
RC sin 45
1k
D
When 1k at left of b
When 1k at right of b
1
1
1k
RA
RA
1k
RA
RA
1
1
RC
RC cos 45
RC sin 45
Md. Ragib Nur Alam
CE 13
ragibnur.ce@gmail.com

8. 6 6 12 18
12
8
A
B C
c
b
d
When 1k at C
RA=0k
RC cos 45 =1k
RC= 1
cos 45 =1.414k
RC sin 45=1.414sin 45=1k
Vb=0k
Vc=-RC sin 45=-1k
Vd=0k
Mb=0k'
Mc=-8k'
Md=0k'
D
1k
6 6 12 18
12
8
A
B C
c
b
d
When 1k at d
RA=18
30 =0.6k
RC cos 45 =0.4k
RC= 0.4
cos 45 =0.566k
RC sin 45=0.566sin 45=0.4k
Vb=0k
Vc=-RC sin 45=-0.4k
Vd=-RC cos 45 =-0.4k (left)
Vd=-RC cos 45 =0.6k (right)
Mb=0k'
Mc=-3.2k'
Md=7.2k'
D
1k
6 6 12 18
12
8
A
B C
c
b
d
1k
D
6 6 12 18
12
8
A
B C
c
b
d
1k
D
When 1k at left of d
When 1k at right of d
1
1
RC
RC cos 45
RC sin 45
RA
RC
RC cos 45
RC sin 45
RA
RC
RC cos 45
RC sin 45
RA
6 6 12 18
12
8
A
B C
c
b
d
1k
When 1k at D
RA=1k
RC cos 45 =0k
RC=- 0.2
cos 45 =0k
RC sin 45=-0.283sin 45=0k
Vb=0k
Vc=-RC sin 45=0
Vd=-RC cos 45 =0
Mb=0k'
Mc=0k'
Md=0k'
D
RC
RC cos 45
RC sin 45
RA
RC
RC cos 45
RC sin 45
RA
Md. Ragib Nur Alam
CE 13
ragibnur.ce@gmail.com

9. 10 8 12 10
5
8
4
C
A D Ba
b
c
1k
RBC sin 26.565
RBC cos 26.565
Rab
10
8
12
10
5
8 4
C
ADBa
b
c
1k
1
3k
4
3
1k
1
When 1k at a
Rb= 1 k
RBC sin 26 .565=0k
RBC =- 0
sin 26.565=0k
RBC cos 26.565=0k
VC=0k
VD=0
MC= 0
MD= 0
10 8 12 10
5
8
4
C
A D Ba
b
c
1k 0k
1k
0 k
When 1k at a
Rb= 3
5 k
RBC sin 26 .565=-2
5 k
RBC =- 2/5
sin 26.565=-0.894k
RBC cos 26.565=-0.8k
VC=0.8k
VD=- 2
5 k(left)
VD=3
5 k(right)
MC= -3.2k'
MD= 24
5 k
10 8 12 10
5
8
4
C
A D Ba
b
c
1k
-2
5 k
3
5 k
-0.8k
10 8 12 10
5
8
4
C
A D Ba
b
c
1k
-2
5 k
3
5 k
-0.8k
10 8 12 10
5
8
4
C
A D Ba
b
c
1k
-2
5 k
3
5 k
-0.8k
When 1k atB
Rb= 0 k
RBC sin 26 .565=-1k
RBC =- 1
sin 26.565=-2.236k
RBC cos 26.565=-2k
VC=2k
VD=0k
MC= -8k'
MD= 0k'
10 8 12 10
5
8
4
C
A D Ba
b
c
-1k
0k
-2 k
1k
Problem 5
Draw ILD for shear and moment at C & D. If the load moves from A to B.
10 8 12 10
5
8
4
C
A D Ba
b
c
1k
0.8k
2k
1
2 k
2
5 k
3
5 k
4 k'
3.2k'
8k'
24
5 k'
6k'
10 8 12 10
5
8
4
C
A D Ba
b
c
RB
RC
A D Ba
RBC=RB=RC
Ra
Ra
Rb
When 1k at A
Rb= 30
20= 3
2 k
RBC sin 26 .565=1
2 k
RBC = 1/2
sin 26.565=1.118k
RBC cos 26.565=1k
VC=-1
VD=RBC sin 26 .565=1
2 k
MC= 4k'
MD= -6k'
10 8 12 10
5
8
4
C
A D Ba
b
c
1k 1
2 k
3
2
0.667k
Md. Ragib Nur Alam
CE 13
ragibnur.ce@gmail.com

10. 10 15 4 6
1
2 3
A
B C D
Problem 06
Find shear and moment at 1,2 &3 and reaction at
C if load moves from B to D
When 1k at B
RC= 12
37= 0.324 k
RA=1-RC=0.675 k
V1=RA cos 53= 0.405 k
V2= -0.324 k
V3= 0k
M1= 10 * 0.406 = 4.06 k'
M2= 15 * 0.324 = 4.86 k'
M3= 0 k'
Shear force is sum of all the transverse forces
(perpendicular to the member or beam)
37
53°
53°
16
6 6
RA cos 53=0.406k
10
10 15 4 6
1
2 3
A
B C D
1k
When 1k at 2-2
RC= 22
37= 0.595 k
RA=1-RC=0.405 k
V1=RA cos 53= 0.244 k
V2= -0.595 k (When 1k at left of 2-2)
V2= 0.405 k (When 1k at right of 2-2)
V3= 0k
M1= 10 * 0.244 = 2.44 k'
M2= 15 * 0.595 = 8.925 k'
M3= 0 k'
RA=0.405K
RC=0.595K
37
53°
53°
16
6 6
RA cos 53=0.244k
10
10 15 4 6
1
2 3
A
B C D
1k
RA= 0 K
RC=1K
37
53°
53°
16
6 6
RA cos 53=0 k
10
10 15 4 6
1
2 3
A
B C D
1k
RA=0.405K
RC=0.595K
37
53°
53°
16
6 6
RA cos 53=0.244k
10
10 15 4 6
1
2 3
A
B C D
1k
RA=0.405K
RC=0.595K
37
53°
53°
16
6 6
RA cos 53=0.244k
10
When 1k at left of 2-2
When 1k at right of 2-2
When 1k at C
RC= 1 k
RA=1-RC=0 k
V1=RA cos 53= 0 k
V2= 0 k
V3= 0k
M1= 0 k'
M2= 0 k'
M3= 0 k'
10 15 4 6
1
2 3
A
B C D
1k
RC=1.108K
37
53°
53°
16
6 6
10
When 1k at 3-3
RC= 41
37= 1.108 k
RA=1-RC= 0.108 k
V1=RA cos 53= -0.065 k
V2= -0.108k
V3= 0 k (When 1k at left of 3-3)
V3= 1 k (When 1k at right of 3-3)
M1= -10 * 0.065 =- 0.65k'
M2= 15 * 1.108 -19*1 = -2.38 k'
M3= 0 k'
RA= 0.108 K
RA cos 53= 0.065 k
10 15 4 6
1
2 3
A
B C D
1k
RC=1.108K
37
53°
53°
16
6 6
10
RA= 0.108 K
RA cos 53= 0.065 k
10 15 4 6
1
2 3
A
B C D
1k
RC=1.108K
37
53°
53°
16
6 6
10
RA= 0.108 K
6 6 10 15 4 6
1
2 3
A
B C D
0.324 k
0.595 k
1 k
1.108 k
1.27 k
0.405k
0.244 k
0.065 k 0.162 k
0.324k
0.595k
0.405k
0.108k 0.27k
1k 1k
4.06k'
2.44k'
0.65k' 1.62k'
4.86k'
8.925 k'
2.38 k' 5.95k'
6 k'
IL for RC
IL for V1
IL for V2
IL for V3
IL for M1
IL for M2
IL for M3
1k
RC=0.324K
RA=0.675 K
RA cos 53= 0.065 k
10 15 4 6
1
2 3
A
B C D
1k
RC=1.27K
37
53°
53°
16
6 6
10
When 1k at D
RC= 47
37= 1.27 k
RA=1-RC= 0.27 k
V1=RA cos 53= -0.162 k
V2= -0.27k
V3= 1 k
M1= -10 * 0.162 = -1.62 k'
M2= 15 * 1.27 - 25*1 = -5.95 k'
M3= -6 k'
RA= 0.27 K
RA cos 53= 0.162 k
Md. Ragib Nur Alam
CE 13
ragibnur.ce@gmail.com

11. Problem 7
10 30
8
2
1
2
A
B
D
C
6 4 10 30
8
2
1
2
A
B
D
C
6 4
1k
When 1k at A
Rc=1
3 k Rb=4
3 k
V1= 1
3 k
M1= 0k
V2= - 1
3 k
M2= -2 k
1k
1
3 k
10 30
8
2
1
2
A
B
D
C
6 4
1k
When 1k at B
Rc=0 k Rb=1 k
V1= 0 k
M1= 0k
V2= 0 k
M2= 0 k
10 30
8
2
1
2
A
B
D
C
6 4
1k
When 1k at 1-1
Rc=1 k Rb=0 k
V1= 1 k(left)
V1=0 k (right)
M1= 0k
V2= 1 k
M2= 6 k
Left clockwise moment +ve
up shear +ve
10 30
8
2
1
2
A
B
D
C
6 4
1k
10
1
8
30
2
D
A
2
C
46
B
1k
6 k'
When 1k at right of 1-1
When 1k at left of 1-1
10 30
8
1
2
A
D
2
B
C
1k
6
When 1k at D
Rc=4
3 k Rb=- 1
3 k
V1= - 1
3 k
M1=-10k
V2= 4
3 k
M2= 8 k'
4
1 k
1/3 k
- 1/3 k
- 2 k'
1 k
0 k'
- 10 k'
8 k
- 1/3 k
4/3 k
Md. Ragib Nur Alam
CE 13
ragibnur.ce@gmail.com

12. 8 20 28
10 20 16
6 2
g f
A B
C D
Problem 08
Draw ILD for shear and moment for f & g for the unit load moving
from A to B.
8 20 28
10 20 16
6 2
g f
A B
C D
8 20 28
A B
1k
a b c
E F
a b c
a b c
10 20 16
6 2
g fC D
a b c
8 20 28
A B
1k
a= 1.4 k b=0 c=-0.4 k
E F
Take moment about F
so, unknown reduced
to 1
sum of moments at
⅀MF = 0
⇒-1*28+a*20=0
⇒a=28
20=1.4 k
Then take moment
about E'
a and b passes
through E'
⅀ME'=0
⇒-1*8-c*20=0
⇒c=-0.4 k
⅀Fy = 0
⇒a+c+by-1=0
⇒1.4 - 0.4+by-1=0
⇒by=0
⅀Fx = 0
⇒bx=0
E' F'
E'
10 20 16
6 2
g fC D
0.957k 0.043k
E' F'
a= 1.4 k b=0 c=-0.4 k
Take moment about C
⅀MC = 0
⇒1.4*10-0.4*30-RD*46=0
⇒RD= 2
46=0.0434 k
⅀Fy = 0
⇒1.4-0.4-0.043-RC=0
⇒RC=0.957k
.
Vf=-0.0434k
Vg=-0.0434-0.4=-0.4434k
Mf= 0.0434*14=0.6076k'
Mg=0.0434*30+0.4*14=6.902k'
When 1k at A
-0.043 k
0.608 k'
6.902 k'
-0.217 k
-0.217 k
3.308 k'
6.51 k'
-0.652 k
0.348 k
9.128 k'
8.128 k'
-1.261 k
1.139 k
17.654 k'
4.23 k'
8 20 28
A B
-0.443 k
IL fror Vf
IL fror Vg
IL fror Mf
IL fror Mg

13. 8 20 28
A B
1k
a= 1 k b=0 c=-0 k
E F
Take moment about F.
so, unknown reduced
to 1. sum of moments
at
⅀MF = 0
⇒-1*20+a*20=0
⇒a=1k
Then take moment
about E' . since, a and
b passes through E'
⅀ME'=0
⇒c =0
⅀Fy = 0
⇒by=0
⅀Fx = 0
⇒bx=0
E'
10 20 16
6 2
g fC D
0.783k 0.217k
E' F'
a= 1k b=0 c=0k
Take moment about C
⅀MC = 0
⇒1*10-RD*46=0
⇒RD=10
46=0.217 k
⅀Fy = 0
⇒1-0.217-RC=0
⇒RC=0.783k
Vf=-0.217k
Vg=-0.217k
Mf=
0.217*14=3.038k'
Mg=0.217*30=6.51k'
When 1k at E
8 20 28
A B
1k
a=0k b=0 c=1k
E F
Take moment about F.
so, unknown reduced to
1. sum of moments at F
⅀MF = 0
⇒a=0k
Then take moment
about E' a and b passes
through E'
⅀ME'=0
⇒-1*20+c*20=0
⇒c =1k
⅀Fy = 0
⇒by=0
⅀Fx = 0
⇒bx=0
E'
10 20 16
6 2
g fC D
0.348k 0.652k
E' F'
a=0 k b=0 c=1k
Take moment about C
⅀MC = 0
⇒
1.4*10-0.4*30-RD*46=0
⇒RD=30
46=0.652 k
⅀Fy = 0
⇒1-0.652-RC=0
⇒RC=0.348 k
Vf=-0.652k
Vg=1-0.652=0.348k
Mf= 0.348*14=9.128k'
Mg=0.652*30-1*14=5.56k'
When 1k at F
8 20 28
A
B
1k
a= -1.4k b=0 c=2.4 k
E F
Take moment about F
so, unknown reduced
to 1.sum of moments
at F
⅀MF = 0
⇒a*20+1*28=0
⇒a=-1.4k
Then take moment
about E' . a and b
passes through E'
⅀ME'=0
⇒1*48-c*20=0
⇒c =2.4k
⅀Fy = 0
⇒1.4+1-2.4-by=0
⇒by=0
⅀Fx = 0
⇒bx=0
E'
10 20 16
6 2
g f D
-0.261k 1.261k
E' F'
a= -1.4 k b=0 c=2.4 k
Take moment about C
⅀MC = 0
⇒-1.4*10+2.4*30-RD*46=0
⇒RD=58
46=1.261 k
⅀Fy = 0
⇒-1.4+2.4-1.261-RC=0
⇒RC=-0.261k
Vf=-1.261k
Vg=-1.261+2.4=1.139k
Mf= 1.261*14=17.654k'
Mg=1.261*30-2.4*14=4.23k'
When 1k at B
C

14. 20 8 8 10 15
6
8 10 15
20
36.87°
A B
C
2 ED 1
RBC
RBC
20 8 8 10 15
6
1k
RAy
A B
C
D E21
⅀MB = 0
⇒RA=1k
⅀Fy=0
⇒RBC sin 36.87°=0k
⇒RBC= 0
sin 36.87°=0k
⇒RBC cos 36.87°=0k
V1=0k
V2=0k
M1= 0k'
M2=0k'
20
1k
RA=1k
A
B RBC cos 36.87°
36.87°
RBC sin 36.87°=0k
RBC=0k
When 1 k at A
REy
REx
hinge
RAx
20 8 8 10 15
6
1k
RAy
A B
C
D E21
⅀MB = 0
⇒RA=0k
⅀Fy=0
⇒RBC sin 36.87°=1k
⇒RBC= 1
sin 36.87°=1.667k
⇒RBC cos 36.87°=1.333k
ME=33*1+1.333*6=41k'
V1=-1 k
V2=-1 k
M1= -41+25= -16k'
M2= -25+15= -26k'
RBC=1.667k
When 1 k at B
REy
REx
hinge
RAx
8 10 15
36.87°
ED 21
ME=25 k'
ME
ME
6
RBC cos 36.87°=1.333k
RBC sin 36.87°=1k
20
1k
RA=0k
A
B
RBC cos 36.87°=1.333k
RBC sin 36.87°=1k
0 k
-1 k
-1 k -1 k-1 k
0 k
IL for V1
IL for V2
IL for M1
IL for M2
-26 k'
10 k'
1 k
1 .667k
1.667k
REx=1.333k
-16 k'
IL for RA
IL for RBC
Problem 09
Draw ILD for RA, RBC, shear & moment at 1-1 & 2-2

15. 20 8 8 10 15
6
1k
RAy
A B
C
D E21
⅀MB = 0
⇒RA=0k
⅀Fy=0
⇒RBC sin 36.87°=0k
⇒RBC= 0
sin 36.87°=0k
⇒RBC cos 36.87°=0k
ME=-25*1=-25k'
V1=-1 k(Just left of section 1-1)
V1=0k (Just right of section 1-1)
V2=-1 k
M1= -25+25=0k'
M2= -25+15= -10k'
When 1 k at D
REy
REx
hinge
RAx
8 10 15
36.87°
E
REy=1k
D 21
ME=25 k'
ME
6
s 36.87°
sin 36.87°
1k
8 10 15
36.87°
ED 21
ME=25 k'
6
1k
8 10 15
36.87°
ED 21
ME=25 k'
6
1k
Just left of section 1-1
Just right of section 1-1
20 8 8 10 15
6
1k
RAy
A B
C
D E21
⅀MB = 0
⇒RA=0k
⅀Fy=0
⇒RBC sin 36.87°=0k
⇒RBC= 0
sin 36.87°=0k
⇒RBC cos 36.87°=0k
ME=-15*1=-15k'
V1=0 k
V2=-1 k(Just left of section 2-2)
V2=0k (Just right of section 2-2)
M1= -15-10+25=0k'
M2= -15+15= 0k'
When 1 k at 2-2
REy
REx
hinge
RAx
8 10 15
E
REy=1k
D 21
ME=15 k'ME
6
1k
8 10 15
ED 21
ME=15 k'
6
1k
8 10 15
ED 21
ME=15 k'
6
1k
Just left of section 2-2
Just right of section 2-2
20 8 8 10 15
6
1k
RAy
A B
C
D E21
⅀MB = 0
⇒RA=0k
⅀Fy=0
⇒RBC sin 36.87°=0k
⇒RBC= 0
sin 36.87°=0k
⇒RBC cos 36.87°=0k
ME=0k'
V1=0 k
V2=0 k
M1= -25+25=0k'
M2= -15+15= 0k'
When 1 k at E
REy
REx
hinge
RAx
8 10 15
E
REy=1k
D 21
ME=0 k'
ME
6
1k
1.667k
REx=0k
s 36.87°
sin 36.87°
1.667k
REx=0k
s 36.87°
sin 36.87°
1.667k
REx=0k
36.87°
s 36.87°
sin 36.87°
1.667k
REx=0k
36.87°
s 36.87°
sin 36.87°
1.667k
REx=0k
36.87°
s 36.87°
sin 36.87°
1.667k
REx=0k
36.87°
s 36.87°
sin 36.87°
1.667k
REx=0k
Created By -
Md.Ragib Nur Alam(CE- 13)
ragibnur.ce@gmail.com
REy=1k

16. Problem 10
Draw ILD for RBE, shear & moment at F & G
A B F C D
G
H
5 5 10 3 3
E
8 A B F C D
G
H
5 5 10 3 3
E
8
RHx
RHy
A B F C D
G
H
5 5 10 3 3
8
RBE=1.238 k
1k
⅀MH = 0
⇒RBE*21-26*1=o
⇒RBE=1.238 k
⅀Fy=0
⇒RH= -0.238k
⇒RH cos 53.13°= -0.143k
VF=1.238-1=0.238k
VG=0.143k
MF= -0.238*16= -3.808k'
MG= -0.143*5= -0.715k'
When 1 k at A
x=53.13°
21
⅀MH = 0
⇒RBE*21-21*1=o
⇒RBE=0 k
⅀Fy=0
⇒RH = 0k
⇒RH cos 53.13°= 0k
VF=0k
VG=0k
MF= 0k'
MG=0k'
When 1 k at B
RH=-0.238k
RH cos x= -0.143k
A B F C D
G
H
5 5 10 3 3
8
RBE=1k
1k
x=53.13°
21
RH=0k
RH cos x=0k
A B F C D
G
H
5 5 10 3 3
8
RBE=0.768 k
1k
⅀MH = 0
⇒RBE*21-16*1=o
⇒RBE=0.762 k
⅀Fy=0
⇒RH= 0.238 k
⇒RH cos 53.13°= 0.143k
VF= -0.238k(1k at left of F)
VF= 0.768k(1k at right of F)
VG= -0.143k
MF= 3.808k'
MG= 0.715k'
When 1 k at F
x=53.13°
21
RH= 0.238k
RH cos x= 0.143k
16
5
16
5
5
A B F C D
G
H
5 5 10 3 3
8
RBE=0.768 k
1k
x=53.13°
RH= 0.238k
RH cos x= 0.143k
5
A B F C D
G
H
5 5 10 3 3
8
RBE=0.768 k
1k
x=53.13°
RH= 0.238k
RH cos x= 0.143k
5
When 1k at left of F When 1k at right of F
⅀MH = 0
⇒RBE*21-6*1=o
⇒RBE=0.286 k
⅀Fy=0
⇒RH = 0.714k
⇒RH cos 53.13°= 0.429k
VF=0.286k
VG=-0.429k
MF= 0.714*16-10=1.424k'
MG=0.429*5=2.145k'
When 1 k at C
A B F C D
G
H
5 5 10 3 3
8
RBE=0.286k
1k
x=53.13°
21
RH=0.714k
RH cos x=0.429k
5
16
⅀MH = 0
⇒RBE*21-3*1=o
⇒RBE=0.143 k
⅀Fy=0
⇒RH = 0.857k
⇒RH cos 53.13°= 0.514k
VF=0.143k
VG=-0.514k
MF= 0.143*5=0.715k'
MG=0.514*5=2.571k'
When 1 k at D
A B F C D
G
H
5 5 10 3 3
8
RBE=0.143k
1k
x=53.13°
21
RH=0.857k
RH cos x=0.514k
5
16
0.143 k
0.238 k
-0.715 k'
-3.808 k'
1.238 k 1k
-0.143 k
-0.238 k
0.715 k'
0.762k
-0.429 k
1.424 k'
0.286 k
0.143 k
0.143 k
-0.514 k
2.571 k'
2.145 k'
0.286 k
0.715 k'
3.808 k'
0.768 k IL for RBE
IL for VF
IL for VG
IL for MF
IL for MG
Created By -
Md.Ragib Nur Alam(CE- 13)
ragibnur.ce@gmail.com

17. Problem :11
When 1k at point A:
𝐴 𝑦 = 1 𝑘
𝑅 𝐵𝐷 = 0 𝑘
𝑉𝑎 = 0𝑘
𝑉𝑏 = 0𝑘
𝑀𝐴 = 0 𝑘 − 𝑓𝑡
𝑀 𝑎 = 0 𝑘 − 𝑓𝑡
𝑀 𝑏 = 0 𝑘 − 𝑓𝑡
When 1k at point a:
𝐴 𝑦 = 1 𝑘
𝑅 𝐵𝐷 = 0 𝑘
𝑉𝑎 = 0𝑘 (𝑤ℎ𝑒𝑛 1𝑘 𝑎𝑡 𝑙𝑒𝑓𝑡)
𝑉𝑎 = 1𝑘 (𝑤ℎ𝑒𝑛 1𝑘 𝑎𝑡 𝑟𝑖𝑔ℎ𝑡)
𝑉𝑏 = 0𝑘
∑ 𝑀𝐴 = 0
⟹ 𝑀𝐴 + 6 = 0
⟹ 𝑀𝐴 = −6 𝑘 − 𝑓𝑡
∑ 𝑀 𝑎 = 0
⟹ 𝑀𝐴 + 𝑀 𝑎 + 6 = 0
⟹ 𝑀 𝑎 − 6 + 6 = 0
⟹ 𝑀 𝑎 = 0 𝑘 − 𝑓𝑡
𝑀 𝑏 = 0 𝑘 − 𝑓𝑡
When 1k at point B:
𝐴 𝑦 = 1 𝑘
𝑅 𝐵𝐷 = 0 𝑘
𝑉𝑎 = 1𝑘
𝑉𝑏 = 0𝑘
∑ 𝑀𝐴 = 0
⟹ 𝑀𝐴 + 12 = 0
⟹ 𝑀𝐴 = −12 𝑘 − 𝑓𝑡
∑ 𝑀 𝑎 = 0
⟹ 𝑀 𝑎 + 𝑀𝐴 + 9 = 0
⟹ 𝑀 𝑎 − 12 + 6 = 0
⟹ 𝑀 𝑎 = −6 𝑘 − 𝑓𝑡
𝑀 𝑏 = 0 𝑘 − 𝑓𝑡
When 1k at point C:
𝑅 𝐵𝐷 sin 36.87 = 1.4 𝑘
𝑅 𝐵𝐷 =
1.4
sin 36.87
= 2.33
𝑅 𝐸 = −0.4
𝐴 𝑦 = 1.4 𝑘
𝑉𝑎 = 1.4 𝑘
𝑉𝑏 = 0.4 𝑘
∑ 𝑀𝐴 = 0
⟹ 𝑀𝐴 + 1.4 ∗ 12 = 0
⟹ 𝑀𝐴 = −16.8 𝑘 − 𝑓𝑡
∑ 𝑀 𝑎 = 0
⟹ 𝑀 𝑎+ 𝑀𝐴 + 6 ∗ 1.4 = 0
⟹ 𝑀 𝑎 − 16.8 + 8.4 = 0
⟹ 𝑀 𝑎 = −8.4 𝑘 − 𝑓𝑡
∑ 𝑀 𝑏 = 0
⟹ 𝑀 𝑏 + 0.4 ∗ 6 = 0
⟹ 𝑀 𝑏 = −2.4 𝑘 − 𝑓𝑡
When 1k at point D:
𝑅 𝐵𝐷 sin 36.87 = 1𝑘
𝑅 𝐵𝐷 =
1
sin 36.87
= 1.667 𝑘
𝑅 𝐸 = 0𝑘
𝐴 𝑦 = 1 𝑘
𝑉𝑎 = 1 𝑘
𝑉𝑏 = 0 𝑘
∑ 𝑀𝐴 = 0
⟹ 𝑀𝐴 + 12 ∗ 1 = 0
⟹ 𝑀𝐴 = −12 𝑘 − 𝑓𝑡
∑ 𝑀 𝑎 = 0
⟹ 𝑀 𝑎+ 𝑀𝐴 + 6 ∗ 1 = 0
⟹ 𝑀 𝑎 − 12 + 6 = 0
⟹ 𝑀 𝑎 = −6𝑘 − 𝑓𝑡
𝑀 𝑏 = 0 𝑘 − 𝑓𝑡
When 1k at point b:
𝑅 𝐵𝐷 sin 36.87 = 0.6 𝑘
𝑅 𝐵𝐷 =
0.6
sin 36.87
= 1𝑘
𝑅 𝐸 = 0.4𝑘
𝐴 𝑦 = 0.6 𝑘

18. 𝑉𝑎 = 0.6 𝑘
𝑉𝑏 = −0.4 𝑘(𝑗𝑢𝑠𝑡 𝑎𝑡 𝑙𝑒𝑓𝑡 𝑜𝑓 𝑏)
𝑉𝑏 = 0.6 𝑘(𝑗𝑢𝑠𝑡 𝑎𝑡 𝑟𝑖𝑔ℎ𝑡 𝑜𝑓 𝑏)
∑ 𝑀𝐴 = 0
⟹ 𝑀𝐴 + 0.6 ∗ 12 = 0
⟹ 𝑀𝐴 = −7.2 𝑘 − 𝑓𝑡
∑ 𝑀 𝑎 = 0
⟹ 𝑀 𝑎+ 𝑀𝐴 + 6 ∗ 0.6 = 0
⟹ 𝑀 𝑎 − 7.2 + 3.6 = 0
⟹ 𝑀 𝑎 = −3.6 𝑘 − 𝑓𝑡
∑ 𝑀 𝑏 = 0
⟹ 𝑀 𝑏 − 0.4 ∗ 6 = 0
⟹ 𝑀 𝑏 = 2.4 𝑘 − 𝑓𝑡
When 1k at point E:
𝑅 𝐵𝐷 = 𝑅 𝐸 = 𝐴 𝑦 = 0𝑘
𝑀𝐴 = 𝑀 𝑎 = 𝑀 𝑏 = 0
𝑉𝑎 = 𝑉𝑏 = 0 𝑘
IL for 𝑅 𝐵𝐷
IL for 𝑣 𝑎
IL for 𝑣 𝑏
IL for 𝑀𝐴
IL for 𝑀 𝑏
IL for 𝑀 𝑎

19. E
b
a
C D
F
A B
Problem 12
Draw ILD for RBD, MF, Va, Vb, Ma, Mb
RBD
RBDa
A
9 9 6 6 9
4.5
E
b
a
C D
F
A B
9 9 6 6 9
4.5
EC
D
F MF
RFy6 6 9
4.5
b
9 9
B
1k
⅀MB = 0
⇒RA=1 k
⅀Fy=0
⇒RBD sin 36.87°= 0k
⇒RBD= 0
sin 36.87° = 0k
⇒RBD cos 36.87°= 0k
Va=0 k
Vb=0k
Ma= 0k'
Mb= 0k'
When 1 k at A
RBD sin x
RBD
RBD cos x
RBD
RBD cos x
RBD sin x
a
A
9 9
B
RBD sin x=0k
RBD=0k
RBD cos x=0k
1k
36.87°
x=36.87°
RA=1k
E
b
a
C D
F
A B
9 9 6 6 9
4.5
1k
⅀MB= 0
⇒RA*18-9*1=0
⇒RA=0.5 k
⅀Fy=0
⇒RBD sin 36.87°=0.5k
⇒RBD= 0.5
sin 36.87° = 0.833k
⇒RBD cos 36.87°= 0.667k
MF=0.5*15+0.667*4.5=10.5 k'
Va=-0.5 k(When 1k at left of a)
Va=0.5k(When 1k at right of a)
Vb=-0.5k
Ma= 0.5*9=4.5k'
Mb= -3k'
When 1 k at a
a
A
9 9
B
RBD sin x=0.5k
RBD=0.833k
RBD cos x=0.667k
1k
36.87°
x=36.87°
RA=0.5k
0.833 k
EC
D
F MF=10.5k'
b
RBD=0.833k
RBD cos x=0.667k
RBD sin x=0.5k
36.87°
RFy=0.5k6 6 9
4.5
a
A
9 9
B
RBD sin x=0.5k
RBD=0.833k
RBD cos x=0.667k
1k x=36.87°
RA=0.5k
a
A
9 9
B
RBD sin x=0.5k
RBD=0.833k
RBD cos x=0.667k
1k x=36.87°
RA=0.5k
When 1k at left of a When 1k at right of a
-0.5 k
4.5 k'
-3 k'
E
b
a
C D
F
A B
9 9 6 6 9
4.5
1k
⅀MB= 0
⇒RBD sin 36.87*18-9*1=0
⇒RBD sin 36.87=1 k
⇒RBD= 1
sin 36.87° =1.667k
⇒RBD cos 36.87°=1.333k
⅀Fy=0
⇒RA=0k
MF=1*15+1.333*4.5=21 k'
Va=0 k
Vb=-1k
Ma= 0k'
Mb= -6k'
When 1 k at B
a
A
9 9
B
RBD sin x=1k
RBD=1.667k
RBD cos x=1.333k
1k
36.87°
x=36.87°
RA=0k
EC
D
F MF=21k'
b
RBD=1.667k
RBD cos x=1.333k
RBD sin x=1k
36.87°
RFy=1k6 6 9
4.5
1.667 k
-1 k
-6 k'
E
b
a
C D
F
A B
9 9 6 6 9
4.5
1k RBD sin 36.87=0
⇒RBD= 0
sin 36.87° =0k
⇒RBD cos 36.87°=0k
⅀Fy=0
⇒RF=0k
MF=1*21=21 k'
Va=0 k
Vb=-1k
Ma= 0k'
Mb= -21+9=-12k'
36.87°
EC
D
F
MF=21k'b
RBD=0k
RBD cos x=0 k
RBD sin x=0k
36.87°
RFy=1k
6 6 9
4.5
1k
RFx=1.333k
RFx=0k
RFx=0.667k
RFx
-1 k
-12 k'
When 1 k at C
E
b
a
C D
F
A B
9 9 6 6 9
4.5
1k RBD sin 36.87=0
⇒RBD= 0
sin 36.87° =0k
⇒RBD cos 36.87°=0k
⅀Fy=0
⇒RF=0k
MF=1*15=15 k'
Va=0 k
Vb=-1k
Ma= 0k'
Mb= -6k'36.87°
EC
D
F MF=15k'
b
RBD=0k
RBD cos x=0 k
RBD sin x=0k
36.87°
RFy=1k
6 6 9
4.5
1k
RFx=0k
When 1 k at D
-6 k'
E
b
a
C D
F
A B
9 9 6 6 9
4.5
1k
RBD sin 36.87=0
⇒RBD= 0
sin 36.87° =0k
⇒RBD cos 36.87°=0k
⅀Fy=0
⇒RF=0k
MF=1*9=9 k'
Va=0 k
Vb=-1k(When 1k at left of b)
Vb=0k(When 1k at right of b)
Ma= 0k'
Mb= 0k'
36.87°
EC
D
F MF=9k'
b
RBD=0k
RBD cos x=0 k
RBD sin x=0k
36.87°
RFy=1k
6 6 9
4.5
1k
RFx=0k
When 1 k at b
EC
D
F MF=9k'
b
RBD=0k
RBD cos x=0 k
RBD sin x=0k
36.87°
RFy=1k
6 6 9
4.5
1k
RFx=0k
EC
D
F MF=9k'
b
RBD=0k
RBD cos x=0 k
RBD sin x=0k
36.87°
RFy=1k6 6 9
4.5
1k
RFx=0k
When 1k at left of b
When 1k at right of b
E
b
a
C D
F
A B
9 9 6 6 9
4.5
1k
RBD sin 36.87=0
⇒RBD= 0
sin 36.87° =0k
⇒RBD cos 36.87°=0k
⅀Fy=0
⇒RF=0k
MF=0 k'
Va=0 k
Vb=0k
Ma= 0k'
Mb= 0k'
36.87°
EC
D
F MF=0k'
b
RBD=0k
RBD cos x=0 k
RBD sin x=0k
36.87°
RFy=1k
6 6 9
4.5
1k
RFx=0k
When 1 k at E
-0.5 k
0.5 k
21 k'
10.5 k' 15 k'
ILD for RBD
ILD for MF
ILD for Va
ILD for Vb
ILD for Ma
ILD for Mb
9k'
Created By -
Md.Ragib Nur Alam(CE- 13)
ragibnur.ce@gmail.com

20. 9 9 6 6 9
4.5
4.5
E
C
D
A
B a
b
1k
6 6 9
4.5
E
C
D bRC sin x
RC sin x
RC cos x
9 9
4.5
C
A
B a
RE
MF=10.5k'
What is going on at the
link???
As at left side of link there is a fixed support so the link transmits the
load from right to left.
The load transmitted from right to left should be considered minor.
RC sin x=0k
9 9
4.5
C
A
B a
MA=0k'
1k
RAy = 1k
RAy
RAx
Taking, RC sin 36.87°
as minor
RC sin 36.87°= 0k
RAy=1 k
MA=0k'
Va=0 k
Vb=0k
Ma= 0k'
Mb= 0k'
36.87°
RC sin x=0k
RC cos x
9 9
4.5
C
A
B a
MA=9k'
1k
RAy = 1k
Taking, RC sin 36.87° as minor
RC sin 36.87°= 0k
RAy=1 k
MA=9k'
Va=0 k(When 1k at left of a)
Va=1k(When 1k at right of a)
Vb=0k
Ma= 0k'
Mb= 0k'
RC sin x=0k
9 9
4.5
C
A
B a
MA=9k'
1k
RAy = 1k
RC sin x=0k
9 9
4.5
C
A
B a
MA=9k'
1k
RAy = 1k
When 1k at left of a
When 1k at right of a
9 9 6 6 9
4.5
4.5
E
C
D
A
B a
b
1k
RC cos x
9 9
4.5
C
A
B a
MA=18k'
1k
RAy = 1k
RAx
Taking, RC sin 36.87° as minor
RC sin 36.87°= 0k
RAy=1 k
MA=18k'
Va=1 k
Vb=0k
Ma= -9k'
Mb= 0k'
RC sin x=0k
9 9 6 6 9
4.5
4.5
E
C
D
A
B a
b
1k
9 9
4.5
C
A
B a
MA=12.852k'
RAy = 0.714 k
⅀ME= 0
⇒RC sin 36.87*21-15*1=0
⇒RC sin 36.87=0.714 k=RAy
⇒RC cos 36.87°=0=RAx(Roller)
⅀Fy=0
⇒RE=0.286 k
MA=12.852k'
Va=0.714 k
Vb= -0.286k
Ma= -6.426k'
Mb= 2.574 k'
RC sin x=0.714k
6 6 9
4.5
E
C
D b
RC sin x=0.714 k
36.87°
1k
9 9 6 6 9
4.5
4.5
C
D
A
B a
b
1k
9 9
4.5
C
A
B a
MA=-3.861k'
RAy = 0.429 k
⅀ME= 0
⇒RC sin 36.87*21-9*1=0
⇒RC sin 36.87=0.429 k=RA
⇒RC cos 36.87°=0k=RAx
⅀Fy=0
⇒RE=0.571 k
MA=0.429*18=7.722k'
Va=0.429 k
Vb= -0.571k(When 1k at left of b)
Vb=0.429 k(When 1k at right of b)
Ma=0.429*9= -3.861k'
Mb= 0.571*9=5.139 k'
RC sin x=0.429k
6 6 9
4.5
E
C
D b
RC sin x=0.429 k
RE=0.571k
36.87°
1k
Horizontal Component is
not possible cause there
is no hinge support
When 1 k at B
When 1 k at a
When 1 k at C
When 1 k at D
When 1 k at b
When 1k at left of b
When 1k at right of b
6 6 9
4.5
E
C
D b
RC sin x=0.429 k
RE=0.571k
36.87°
1k
6 6 9
4.5
E
C
D b
RC sin x=0.429 k
RE=0.571k
36.87°
1k
Horizontal force need not to be counted as right side of link/ hinge is
roller. let us assume the internal hinge as link :)
9 9 6 6 9
4.5
4.5
E
C
D
A
B a
b
1k
9 9
4.5
C
A
B a
MA=0k'
RAy = 0 k
⅀ME= 0
⇒RC sin 36.87*21-0*1=0
⇒RC sin 36.87=0 k=RAy
⇒RC cos 36.87°=0=RAx(Roller)
⅀Fy=0
⇒RE=1 k
MA=0k'
Va=0 k
Vb=0k
Ma=0k'
Mb=0k'
RC sin x=0k
6 6 9
4.5
E
C
D b
RC sin x=0 k
RE=1k
36.87°
1k
Horizontal Component is not
possible cause there is no hinge
support
When 1 k at D
Problem 13
Draw ILD for MA, Va, Vb, Ma, Mb
9 9 6 6 9
4.5
4.5
E
C
D
A
B a
b
9k'
1k
18k'
1k
9k'
12.852k'
0.714k
-0.286k
-6.426k'
2.574k'
7.722k'
0.429k
-0.571k
-3.861k'
5.139k'
ILD for MA
ILD for Va
ILD for Vb
ILD for Ma
ILD for Mb
0.429k
9 9 6 6 9
4.5
4.5
E
C
D
A
B a
b
1k
RE=0.286k

21. 4 2 8 4
4
2
2
2
4
A
aC
D E F
G
4 2 8 4
4
2
2
2
4
A
aC
D E F
G RG cos x
RG sin x
RA
4 2 8 4
4
2
2
2
4
A
aC
D b
c
E F
G RG cos x=0k
RG sin x=Ry=-0.286k
RA=1.286k
1k
⅀MG= 0
⇒RA*14-18*1=0
⇒RA=18
14 =1.286 k
⅀Fx=0
⇒RG cos 45°=0k
⅀Fy=0
⇒RA+RG sin 45°=1k
⇒RG sin 45°= -0.286k
Va= 0 k
Vb= 0.286k
Vc= -Ry cos
45°=-(-0.202)=0.202k
Ma= 0k'
Mb=-0.286*12=-3.432k'
Mc= (-0.202*3)= -0.606k'
When 1 k at D
x=45°
b
c
b
c
Ry cos x= -0.202k
x
6
4 2 8 4
4
2
2
2
4
A
aC
D b
c
E F
G RG cos x=0k
RG sin x=Ry=0.143k
RA=0.857k
1k
⅀MG= 0
⇒RA*14-12*1=0
⇒RA=12
14 =0.857 k
⅀Fx=0
⇒RG cos 45°=0k
⅀Fy=0
⇒RA+RG sin 45°=1k
⇒RG sin 45°= 0.143k
Va= 0 k
Vb= -0.143k(When 1k at left of b)
Vb=0.857k(When 1k at left of b)
Vc= -Ry cos 45°=-(0.101)= -0.101k
Ma= 0k'
Mb=0.101*12=1.212k'
Mc= (0.101*3)= 0.303k'
When 1 k at b
x=45°Ry cos x= 0.101k
x
6
When 1k at left of b
4 2 8 4
4
2
2
2
4
A
aC
D b
c
E F
G RG cos x=0k
RG sin x=Ry=0.143k
RA=0.857k
1k
x=45°Ry cos x= 0.101k
x
6
4 2 8 4
4
2
2
2
4
A
aC
D b
c
E F
G RG cos x=0k
RG sin x=Ry=0.143k
RA=0.857k
1k
x=45°Ry cos x= 0.101k
x
6
When 1k at right of b
4 2 8 4
4
2
2
2
4
A
aC
D b
c
E F
G RG cos x=0k
RG sin x=Ry=0.714k
RA=0.286k
1k
⅀MG= 0
⇒RA*14-4*1=0
⇒RA=4
14 =0.286 k
⅀Fx=0
⇒RG cos 45°=0k
⅀Fy=0
⇒RA+RG sin 45°=1k
⇒RG sin 45°= 0.714k
Va= 0 k
Vb=0.286k
Vc= -Ry cos 45°=-0.505k
Ma= 0k'
Mb=0.286*2=0.572k'
Mc= (0.505*3)= 1.515k'
When 1 k at E
x=45°Ry cos x= 0.505k
x
6
-3.432k'
1.212k'
4
0.857k
2 8
0.286k
4
4
2
1.515k'
0k
-0.707k
2
2
0k'
4
2.121k'
A
aC
D b
c
E F
G RG cos x= 1k
RG sin x=Ry=1k
RA=0k
1k
⅀MG= 0
⇒Ry*14-14*1=0
⇒Ry=1 k
⅀Fx=0
⇒RG cos 45°=0k
⅀Fy=0
⇒RA+RG sin 45°=1k
⇒RA= 0k
Va= 0 k
Vb=0k
Vc= -Ry cos 45°=-0.707k
Ma= 0k'
Mb=0k'
Mc= (0.707*3)= 2.121k'
When 1 k at F
x=45°Ry cos x= 0.707k
x
6
Problem 14:
Draw ILD for Va,Vb,Vc,Ma,Mb,Mc if the unit load moves from D to F.
0.286k
0.202k
-0.606k'
-0.143k
-0.101k
0.303k'
0.572k'
-0.505k
IL for Va
IL for Vb
IL for Vc
IL for Ma
IL for Mb
IL for Mc
The distance of A from b is not given in assignment if
assumed infinitesimal influence line would be different.
Created By -
Md.Ragib Nur Alam(CE- 13)
ragibnur.ce@gmail.com

22. a b l c
RA RC RD
A C D EB F G
RA=1k RC RD
A C D EB F G
1k
a b l c
RA=1k
A
B
RB=0 RC RD
C D EF G
b l c
B
RB
a
1k
⅀MB= 0
⇒RA*a-a*1=0
⇒RA=a
a =1 k
⅀Fy=0
⇒RA+RB=1k
⇒RB=0k
for right of link
RC=RD=0k
VF= 0 k
VG= 0 k
MF= 0k'
MG= 0k'
RA RC RD
A C D EB F G
1k
a b l c
RA=1k
A B
RB=1k
RC=(1+ b
l)k RD= -b
l k
C D EF G
b l c
B
RB=1k
a
1k
When 1 k at A
When 1 k at B
B
RC=1k RD= 0 k
C D EF G
b l c
RB=0k
RB is minor
for right of link
⅀MD= 0
⇒l*1-RC*l=0
⇒RC=1k
⅀Fy=0
⇒RC+RD=1
⇒RD= 0k
VF= 0 k
VG= 0 k
MF= 0 k'
MG= 0 k'
1k
⅀MA= 0
⇒RB*a-a*1=0
⇒RB=a
a =1 k
⅀Fy=0
⇒RA+RB=1k
⇒RA=0k
for right of link
⅀MD= 0
⇒(l+b)*1-RC*l=0
⇒RC=l+b
l =(1+ b
l)k
⅀Fy=0
⇒RC+RD=RB
⇒RD= -b
l k
VF= -1 k
VG= b
l k
MF= -1*b
2 = -b
2k'
MG=- b
l * l
2= b
2k'
RA RC RD
A C D EB F G
1k
a b l c
B
RC=1
2 k RD= 1
2 k
C D EF G
b l c
RB=0k
RB is minor
for right of link
⅀MD= 0
⇒ l
2 *1-RC*l=0
⇒RC=1
2 k
⅀Fy=0
⇒RC+RD=1
⇒RD= 1
2k
VF= 0 k
VG= - 1
2 k(When 1k at left of G)
VG= 1
2 k(When 1k at right of G)
MF= 0 k'
MG= 1
2 * l
2 =l
4k'
When 1 k at G
1k
RA RC RD
A C D EB F G
1k
a b l c
B
RC= (1+
b
2l)k RD= -b
2l k
C D EF G
b l c
When 1 k at F
Rb is minor
for right of link
⅀MD= 0
⇒(l+b
2)*1-RC*l=0
⇒RC==(1+ b
2l)k
⅀Fy=0
⇒RC+RD=1
⇒RD= -b
2l k
VF= -1 k(When 1k at left of F)
VF= 0 k(When 1k at right of F)
VG= b
2l k
MF=0k'
MG=- b
2l * l
2= b
4k'
1k
B
RC= (1+
b
2l)k RD= -b
2l k
C D EF G
b l c
1k
B
RC= (1+
b
2l)k RD= -b
2l k
C D EF G
b l c
RB=0k
1k
When 1k at left of F
When 1k at right of F
B
RC=1
2 k RD= 1
2 k
C D EF G
b l c
RB=0k 1k
B
RC=1
2 k RD= 1
2 k
C D EF G
b l c
RB=0k 1k
When 1k at left of G
When 1k at right of G
RA RC RD
A C D EB F G
1k
a b l c
B
RC=0k RD= 1k
C D EF G
b l c
RB=0k RB is minor
for right of link
⅀MC= 0
⇒l*1-RD*l=0
⇒RD=1k
⅀Fy=0
⇒RC+RD=1
⇒RC= 0k
VF= 0 k
VG= 0 k
MF= 0 k'
MG= 0 k'
When 1 k at D
1k
RA RC RD
A C D EB F G
1k
a b l c
B
RC=- c
l k RD= (1+ c
l)k
C D EF G
b l c
RB=0k
RB is minor
for right of link
⅀MC= 0
⇒(l+c)*1-RD*l=0
⇒RD=(1+ c
l) k
⅀Fy=0
⇒RC+RD=1
⇒RC=-c
l k
VF= 0 k
VG= (1+ c
l) k
MF= 0 k'
MG= (1+ c
l)* l
2=(l
2+ c
2)k'
When 1 k at E
1k
1k
1k
(1+ b
l)k
- b
lk
-1k
b
lk
- b
2k'
b
2k'
(1+ b
2l)k
- b
2lk
-1k
b
2lk
b
4k'
1k
1
2 k
1
2 k
(1+ c
l)k
- c
lk
(1+ c
l)k
(l
2+ c
2)k'
- 1
2 k
1
2 k
RA
A B
RB RC RD
C D EF G
b l c
B
RB
a
RB=0k
RB=0k
RA RC RD
A C D EB F G
1k
a b l c
When 1 k at C
Created By -
Md.Ragib Nur Alam(CE- 13)
ragibnur.ce@gmail.com

23. For a particular position of load, shear force at any point in panel will have
same value. Therefore, instead of influence line for shear force at a point,
influence line for shear force in a panel is drawn.
BA C D E F G
6 panels @ 10' = 60'
1k
When 1k at Mid span of BC
RM=3
4 k
RN=1
4 k
ME =1
4 *20=1
5 k'
VBC= 1
4 k
RM=3
4
RN=1
4 k
BA
0.5k
3
4 k
VBC
100.5k 0.5k
1k
BA C D E F G
6 panels @ 10' = 60'
1k
When 1k at C
RM=2
3 k
RN=1
3 k
ME =1
3 *20=20
3 k'
VBC= 2
3 k
RM=2
3 k
RN=1
3
C D E F G
1k
10 10 10 10
VBC
BA C D E F G
6 panels @ 10' = 60'
1k
When 1k at D
RM=0.5 k
RN=0.5 k
ME =0.5 *20=10 k'
VBC= 0.5 k
RM=0.5 k RN=0.5
C D E F G
1k
RN=0.5
10 10 10 10
VBC
BA C D E F G
6 panels @ 10' = 60'
1k
When 1k at E
RM=1
3 k
RN=2
3 k
ME =2
3 *20=40
3 k'
VBC= 1
3 kRM=1
3 k RN=2
3 k
C D E F G
1k
RN=2
3 k
10 10 10 10
VBC
BA C D E F G
6 panels @ 10' = 60'
1k
When 1k at F
RM=1
6 k
RN=5
6 k
ME =-1*10+ 5
6 *20=20
3 k'
VBC=1
6 kRM=1
6 k RN=5
6 k
C D E F G
1k
RN=5
6 k
10 10 10 10
VBC
BA C D E F G
6 panels @ 10' = 60'
1k
When 1k at G
RM=0 k
RN=1 k
ME =0 k'
VBC= 0 k
RM=0 k RN=1 k
C D E F G
1k
RN=1 k
10 10 10 10
VBC
BA C D E F G
6 panels @ 10' = 60'
0.5k
RM=3
4 k
RN=1
4 k
0.5k
BA C D E F G
6 panels @ 10' = 60'
Problem 16
FInd shear in panel BC and moment at E
-1/6 k
0 k
1/4 k
2/3 k
1/2 k
1/3 k
1/6 k
0 k
0 k'
10
3 k'
1
5 k'
20
3 k'
10 k'
40
3 k'
20
3 k'
0 k'
BA C D E F G
6 panels @ 10' = 60'
1k
When 1k at 'A'
RM=1k
RN=0k
ME =0k'
VBC= 0k
RM=1
BA
1k
1k
VBC
BA C D E F G
6 panels @ 10' = 60'
1k
When 1k at B'
RM=5
6 k
RN=1
6 k
ME =1
6 *20=10
3 k'
VBC= -1
6 k
RM=5
6 RN=1
6
BA
1k
5
6 k
VBC
10
Md. Ragib Nur Alam
CE 13
ragibnur.ce@gmail.com
RN=0
RN=1
3

24. Problem 17
Construct ILD for moment of Girder at E?
10 5 5 10 5
A B C D E F G
10 5 5 10 5
A B C D E F G
When 1k at 'A'
RM=1k
RN=0k
ME =0k'1k
0k
RM RN
1k
10 5 5 10 5
A B C D E F G
When 1k at B
RM=2
3 k
RN=1
3 k
ME =10
3 k'
1k
0k
RM RN
1k
10 5 5 10 5
A B C D E F G
When 1k at 'C'
RA=-0.5k
RB=1.5k
RM=0.5k
RN=0.5k
ME =5k'
RM RN
1k
A B C
1k
0.5k
1.5k 10 5 5 10 5
A B C D
E F G
0.5k 0.5k
RM RN0.5k
1.5k
10 5 5 10 5
A B C D E F G
When 1k at 'D'
RF=-0.5k
RE=1.5k
RM=0.5k
RN=0.5k
ME =0.5*10+0.5*10=10k'
RM RN
1k
10 5 5 10 5
A B C D
E F G
0.5k
0.5k
RM RN
D E F G
1k 0.5
1.5k
1.5k
0.5
10 5 5 10 5
A B C D E F G
RM RN
1k
10 5 5 10 5
A B C D E F G
RM RN
1k
10 5 5 10 5
A B C D E F G
RM RN
1k
When 1k at 'E'
RM=1
3 k
RN=2
3 k
ME =2
3*10=20
3 k'
When 1k at F'
RM=0 k
RN=1k
ME =1*10-1*10=0k'
When 1k at G'
RE=-0.5
RF=1.5k
RM=-0.167 k
RN=1.167k
ME =1.167*10-1.5*30=-3.333k'
10 5 5 10 5
D E F
G0.167
1.167
RM RN
1k
1.5k
0.5k
1.5k
0.5k
5k'
10k'
0k'
10
3 k'
20
3 k'
0k'
0k'
ILD for moment at E
3.333k'
Find moment for 1k at A,C,D,F,G
Md. Ragib Nur Alam
CE 13
ragibnur.ce@gmail.com
Girder with conventional end supported stringers , panel points are key points where influence
line might change slope or have discontinuities.
For unusual stringer arrangement like this, Discontinuities in ILD may occur at key points
other than panel points Like "C-D"
Md. Ragib Nur Alam
CE 13
ragibnur.ce@gmail.com

25. Problem 18
Find Shear and moment a,b,c,d, If load moves from A to D
4 4 8 4 5
6
2
A a B b C D
c
E
d
F
4 4 8 4 5
6
2
A
a B b C D
c
E
d
F
1k When 1k at A
RF=1k
RE=0k
Va= -1k
Vb=0k
Vc=0k
Vd= 1* 4
sqrt(4^2 +3^2)=0.8k
Ma=-4 k'
Mb=0k'
Mc=0k'
Md=0.8*sqrt(4^2 +3^2) =4k'
1k
normal
force/axial
force
H
V
X
X
Vsinx
H cos x
Hsin x + V cos x
Shear force= V sin x - H cos x
4 4 8 4 5
6
2
A
a B b C D
c
E
d
F
1k When 1k at a
RF=0.8k
RE=0.2k
Va= -1k(left)
Va= 0k(right)
Vb=-0.2k
Vc=0k
Vd= 0.8* 4
sqrt(4^2 +3^2)=0.64k
Ma=0 k'
Mb=0.8k'
Mc=0k'
Md=0.64*sqrt(4^2 +3^2)
=3.2k'
0.8k
4 4 8 4 5
6
2
A
a B b C D
c
E
F
0.8k
4 4 8 4 5
6
2
A
a B b C D
c
E
d
F
1k
0.8k
When 1k at just left of a When 1k at just right of a
1k
0.2k
d
0.2k 0.2k
4 4 8 4 5
6
2
A
a B b C D
c
E
d
F
1k
When 1k at B
RF=0.6k
RE=0.4k
Va= 0k
Vb=-0.4k
Vc=0k
Vd= 0.6* 4
sqrt(4^2 +3^2)=0.48k
Ma=0 k'
Mb=1.6k'
Mc=0k'
Md=0.48*sqrt(4^2 +3^2)=2.4k'0.6k
0.4k
4 4 8 4 5
6
2
A
a B b C D
c
E
d
F
1k When 1k at b
RF=0.2k
RE=0.8k
Va= 0k
Vb=-0.8k (left)
Vb=0.2k (right)
Vc=0k
Vd= 0.2* 4
sqrt(4^2 +3^2)=0.16k
Ma=0 k'
Mb=3.2k'
Mc=0k'
Md=0.16*sqrt(4^2 +3^2)
=0.8k'
0.2k
0.8k
4 4 8 4 5
6
2
A
a B b C D
c
E
F
0.2k
4 4 8 4 5
6
2
A
a B b C D
c
E
d
F
1k
0.2k
When 1k at just left of b
When 1k at just right of b
1k
d
0.8k
0.8k
4 4 8 4 5
6
2
A
a B b C D
c
E
d
F
1k
0k
1k
1k
When 1k at C
RF=0k
RE=1k
Va= 0k
Vb=0k
Vc=0k
Vd= 0k
Ma=0 k'
Mb=0k'
Mc=0k'
Md=0k'
4 4 8 4 5
6
2
A
a B b C D
c
E
d
F
1k When 1k at D
RF=-0.25k
RE=1.25k
Va= 0k
Vb=-0.25k
Vc=0k
Vd= -0.25* 4
sqrt(4^2 +3^2)=-0.2k
Ma=0 k'
Mb=-4k'
Mc=0k'
Md=-0.2*sqrt(4^2 +3^2)
=-1k'
-1k
Va
Vb
Vc
Vd
Ma
Mb
Mc
Md
-0.4k
-0.2k
-0.8k
0.2k
0.25k
0k
0.8k 0.64k
0.48k
0.16k
0.2k
-4k'
0.8k'
1.6k'
3.2k'
4k'
-1k'
0.8k'
2.4k'
3.2k'
4k'
2
1.25k
0.25k
Md. Ragib Nur Alam
CE 13
ragibnur.ce@gmail.com

26. 103
Problem 19
3 3
4
A
When 1k at A
RF= 1.6 k RB=-0.6k
Va= 0.6k
Vb= -1.6k
Vc= 0k
Ma=-0.6*10=-6k'
Mb=1.6*3=4.8k'
Mc= 0k
3
F
6
4
4
C
D
aA c
Eb
B
F
0.6k
10 6
c
4
D
F
B
a
b
c
1k
E
C
A
1k
1.6k
a
B
C
b
4
3
4
103 6
b
D
4
4
10 6
When 1k at a
RF= 1 k RB= 0k
Va=0 k (left)
Va= 1k (right)
Vb= -1k
Vc= 0k
Ma= 0k'
Mb=1*3=3k'
Mc= 0k'
E
0k
C
1k
F
1k
3 3
10
D
6
A a
D
c
B
4
1k
E
0k
A C
B
F
1k
3
4
3
b
a c
E
1k
0k
Shear = sum of transeverse forces either left or right of section
When 1k at just left of a
When 1k at just Right of a
3 3 10 6
4
4
D
A a c
C
B
b E
F
1k
0k
Total transverse
force = 0
Total transverse
force = 1kWhen 1k at B
RF= 0 k RB= 1k
Va=0 k
Vb= 0 k
Vc= 0k
Ma= 0k'
Mb=0 k'
Mc= 0k'
1k
3 3 10 6
4
4
D
A a c
C
B
b E
F
1k
0k
When 1k at c
RF= 0 k RB= 1k
Va=0 k
Vb= 0 k
Vc= 0k(left)
Vc= 1k(Right)
Ma= 0k'
Mb=0 k'
Mc= 0k'
1k
3 3 10 6
4
4
D
A a c
C
B
b E
F
1k
When 1k at C
RF= 6
10 =0.6 k RB= 16
10 =1.6 k
Va=-1.6+1=-0.6 k
Vb= 0.6 k
Vc= 1k
Ma= -1*16+1.6*10=0k'
Mb=-0.6*3=-1.8 k'
Mc= -6k'
1.6k
3 3 10 6
4
4
D
A a c
C
B
b E
F
1k
0k
3 3 10 6
4
4
D
A a c
C
B
b E
F
1k
0k
When 1k at just left of c
When 1k at just Right of c
Total transverse
force = 0
Total transverse
force = 1k
1k
1k
0.6k
Va
Vb
Vc
Ma
Mb
Mc
0.6k
1.6k
1k
6k'
4.8k'
-6k'
1k
0.6k
1k
1.8k'
3k'
0.6 k
Md. Ragib Nur Alam
CE 13
ragibnur.ce@gmail.com

27. Problem 20
6 10 3 3 6
4
4
A
B C D
F
G
E
a
b
Va
Vb
Ma
Mb
RFG
6 10 3 3 6
4
4
A
B C D
F
G
E
a
b
When 1k at A
Va= 1.6 -1 = 0.6k
Vb= -0.6k
Ma=0k'
Mb= 1.8k'
RFG=-0.6 k
1k
6 10 3 3 6
4
4
A
B C D
F
G
E
a
b
When 1k at B
Va= 0 k
Vb= 0k
Ma=0k'
Mb= 0k'
RFG=0k
1k
6 10 3 3 6
4
4
A
B C D
F
G
E
a
b
When 1k at a
Va= -1 k(Left of a)
Va= 0 k(Right of a)
Vb= 1k
Ma= 0k'
Mb= 3k'
RFG= 1k
1k
6 10 3 3 6
4
4
A
B C D
F
G
E
a
b
1k
0k
1k
6 10 3 3 6
4
4
A
B C D
F
G
E
a
b
1k
6 10 3 3 6
4
4
A
B C D
F
G
E
a
b
When 1k at C
Rb=-0.6
Va= -0.6 k
Vb=1.6 k
Ma=-6k'
Mb= 4.8k'
RFG=1.6k
1k
6 10 3 3 6
4
4
A
B C
D
F
G
Eb
When 1k at D
Rb= -1.2
Va= -1.2 k
Vb= 2.2k
Ma=-12k'
Mb= 6.6 k'
RFG=22
10 = 2.2k
1k
1.2k
0.6k
-0.6k
0k'
1.8k'
-0.6 k
-1k
1k
4.8k'
When 1k at left of a
When 1k at right of a
a
2.2k
-0.6k
1k
3k'
1.6k
- 6k'
1.6k
-1.2k
2.2k
- 12k'
6.6k'
2.2k
Md. Ragib Nur Alam
CE 13
ragibnur.ce@gmail.com