THREE-HINGED FRAMES

• Statically determinate systems which have three hinges are called as «three-
hinged frames».
• In Figure 1, one of the hinges is on the system and the other two of them are on
supports.
• In Figure 2, if the supports are fixed supports, 3 hinges can be on the system
elsewehre from the supports.

40 kN 60 kN/m
B C
E G

α
Figure 1
D 100 kN
A
F
20 kN/m

F
2,0
G m
E α
100 kN 100 kN 3,0
m
C D H I
Figure 2
5,0
m

A B
MA=1840 kNm 1,2 m 5,8m 5,8m 1,2
14,0 m MB=1840 kNm
Example 1:
40 kN 60 kN/m
B C
E G

m
4

4
α 100 kN
RAx D
A

m
2
RAy F RFx
3m 10 m 5m
RFy

Estimate support reactions and draw internal force diagrams of the given system.
Solution
There are 4 support reaction on this system, thus these reactions are obtained
by 3 equilibrium equations and 1 hinge condition. System is statically
determinate.

Memeber lengths and their trigonometric identities:

L AB  3 2  4 2  5 m cos   3 / 5  0,60 Sin  4 / 5  0,80

 40 kN 60 kN/m
B C LAB  5 m
E G cos   3 / 5  0,60
Sin  4 / 5  0,80

m
4

4
α 100 kN
D
RAx=370,90 kN A

m
2
RAy=384,90 kN F RFx=270,90 kN
3m 10 m 5m
RFy=555,10 kN

Support reactions
For ABG

M G 0 R Ay  13,0  R Ax  4,0  40  13,0  60  10  5,0  0

13R Ay  4R Ax  3520  0
For whole system M F 0
R Ay  18,0  R Ax  2,0  40  18,0  60  15  7,50  100  2,0  0
18R Ay  2R Ax  7670  0

R Ax  370,90 kN R Ay  384,90 kN
 40 kN 60 kN/m
B C
E G

m
4

4
α 100 kN
D
RAx=370,90 kN A

m
2
RAy=384,90 kN F RFx=270,90 kN
3m 10 m 5m
RFy=555,10 kN
For GCF M G  0 Hinge condition

 R Fy  5,0  R Fx  6,0  100  4,0  60  5,0  2,5  0

 5R Fy  6R Fx  1150  0
Whole system M A 0

 R Fy  18,0  R Fx  2,0  40  0,0  100  0,0  60  15,0  10,5  0

 18R Fy  2R Fx  9450  0

R Fx  270,90 kN R Fy  555,10 kN

Check the results:

P x  370,90  100  270,90  0

P y  384,90  40  900  555,10  0

 40 kN 60 kN/m
B C LAB  5 m
E G cos   3 / 5  0,60
Sin  4 / 5  0,80

4m

4m
α α 100 kN
D

2m
RAx=370,90 kN A
RAy=384,90 kN F RFx=270,90 kN
α 3m 10 m 5m
RFy=555,10 kN
Internal Forces
N AB  N BA  370,90  cos   384,90  sin 
N AB  N BA  370,90  0,6  384,90  0,80  530,46 kN
VAB  VBA  370,90  sin   384,90  cos 
VAB  VBA  370,90  0,8  384,90  0,60  65,78 kN
M AB  0 M BA  384,90  3,0  370,90  4,0  328,90 kNm
N EB  N BE  0 VEB  VBE  40 kN M EB  0
M BE  40  3,0  120 kNm

N BG  N GB  N GC  N CG  370 ,90 kN
VBG  384,90  40  344,90 kN
M BG  384,90  3,0  370,90  4,0  40  3  448,90 kNm
VGB  VGC  555,10  60  5,0  255,10 kN
 40 kN 60 kN/m
B C LAB  5 m
E G cos   3 / 5  0,60
Sin  4 / 5  0,80

m
4

4
α α 100 kN
D
RAx=370,90 kN A

m
2
RAy=384,90 kN F RFx=270,90 kN
α 3m 10 m 5m
RFy=555,10 kN
VBG  344,90 kN MBG  448,90 kNm
VGB  255,10 kN

Since the sign of the shear forces at B and G ends of BG frame changes,
shear force is zero at any location and moment is maximum at that point.

x 0  VBG / q  344,90 / 60  5,75 m

M enb.  M BG  VBG
2
/(2q)  448,90  344,9 2 /(2  60)  542,40 kNm
 40 kN 60 kN/m
B C
E G

m
4

4
α α 100 kN
D
RAx=370,90 kN A

m
2
RAy=384,90 kN F RFx=270,90 kN
α 3m 10 m 5m
RFy=555,10 kN

VCG  555,1 kN
M CG  M CD  270,90  6,0  100  4,0  2025,40 kNm
N CD  N DC  N DF  N FD  555,10 kN
VCD  VDC  270,90  100  370,90 kN
M DC  M DF  270,90  2,0  541,80 kNm
VDF  VFD  270,90 kN M FD  0
 344,90
40 E B + G C 370,90

555,10
5,75 m +

255,10
370,90
A D 270,90
(b) V (kN) +
F
270,90

2025,40
448,90

542,40
E G 2025,40
120B + C

(c) M (kNm)
A D 541,80

F
370,90

E B G 555,10 C

370,90
370,90

(d) N (kN)
A D

F
555,10
Example 2:
G
Estimate support reactions and

3,6 m
30 kN/m 100 kN draw internal force diagrams of the
given system.
C

2,4 m
α
RAx B RBx
A Solution
RAy 5m 3m 2 m RBy L AG  L BG  5 2  6 2  7,81 m
sin   6 / 7,81  0,7682
cos   5 / 7,81  0,6402

Support Reactions
Resultant force on AG member

Q AG  30  7,81  234,30 kN

For whole system

M B 0 R Ay  (234,30  3,0  100  2,0) / 10  50,29 kN

M A 0 R By  (234,30  3,0  100  8,0) / 10  150,29 kN

Check:  Py  50,29  100  150,29  0
 G
LAG  7,81 m

3,6 m
30 kN/m 100 kN
sin   6 / 7,81  0,7682
cos   5 / 7,81  0,6402
C

2,4 m
QAG  234,30 kN
α
RAx B RBx
A
RAy 5m 3m 2 m RBy

Horizontal support reactions are obtained by writing the hinge

condition on AG and BG systems, separately.

For AG system M G 0
R Ax  (50,29  5,0  234,30  3,0) / 6  159,06 kN

For GB system M G 0
R Bx  (150,29  5,0  100  3,0) / 6  75,24 kN

Check P x  159,06  234,30  75,24  0

 G
Internal Forces

α 100 kNα LAG  7,81 m

234,3 kN C sin   6 / 7,81  0,7682
cos   5 / 7,81  0,6402
α B α
A 159,06 kN
α
50,29 kN 150,29 kN

Resultant forces of the external loads, which are parallel and perpendicular
to inclined members, are estimated to obtain the axial and shear forces of the
members
N AG  38,63  101,83  140,46 kN
VAG  32,20  122,19  89,99 kN M AG  0
N GA  38,63  101,83  150  9,54 kN
VGA  32,20  122,19  180  90,01 kN M GA  0
N BC  N CB  115,45  48,17  163,62 kN
VBC  VCB  96,22  57,80  38,42 kN M BC  0
M CB  M CG  150,29  2,0  75,24  2,4  120,00 kNm M GC  0
N CG  N GC  115,45  48,17  76,82  86,80 kN
VCG  VGC  96,22  57,80  64,02  25,60 kN
 Mx Nx
Vx Since the sign of the shear forces at A and G
ends of AG frame changes , shear force is zero
at any location and moment is maximum at
α that point.
q y

VAG
α
MAG
A
NAG x

Vx  VAG  q  sin   ( y / sin )  0

Vx  VAG  q  y  0
y 0  VAG / q Height of the zero shear point from the support A
x 0  y 0 / tg
M x  M AG  VAG  ( y / sin )  q  sin   ( y / sin )( y / 2 sin )
M x  M AG  VAG  y / sin   q  y 2 / 2 sin 
y  y 0  VAG / q M enb.  M AG  VAG 2
/(2q  sin )
y 0  89,99 / 30  3,0 m x 0  3,0 / 1,2  2,50 m
M enb.  0  89,99 2 /( 2  30  0,7682)  175,70 kNm
 G

3,6 m
100 kN G
30 kN/m

C +

2,4 m
α C
RAx B RBx
A +
RAy 5m 3m 2 m RBy A (c) V (kN) B

G
G

α 100 kN α + C
234,3 kN +
C
A B
(d) M (kNm)
α B α
A 159,06 kN 75,24 kN
G
α
50,29 kN 150,29 kN

C
+
B
A
(e) N (kN)
Example 3:

Estimate support reactions and draw internal force diagrams of the given system.
G

3,0m
20 kN/m
100 kN
C α2 R Bx
1,5m

1,0m
R Ax A α1
R By
1,125 1,875 4,0 2,0
R Ay 6,0m
3,0m

Solution:
Horizontal distance between A and C is 1,125 m.
Length of BG member:
L BG  32  6 2  6,7082 m
 G
Q=67,082 kN

3,0m
20 kN/m
100 kN
C α2 R Bx

1,5m

1,0m
B
A α1
LBG  6,7082 m
R Ax R By
1,125 1,875 4,0 2,0
R Ay 3,0m 6,0m

Resultant force on BG member

Q  0,5  20  6,7082  67,082 kN
For AG system:  M G  0
R A y .3,0  R A x  4,0  100  2,50  0 3R A y  4R A x  250
For whole system:  M B  0
R A y .9,0  R A x  1,0  100  0,5  67,082  2,0  0 9R A y  R A x  84,164
R Ax  60,531 kN R A y  2,626 kN
 G
Q=67,082 kN

3,0m
20 kN/m
100 kN
C α2 R Bx
1,5m

1,0m
B
A α1
LBG  6,7082 m
R Ax R By
1,125 1,875 4,0 2,0
R Ay 3,0m 6,0m

For GB system: M G 0
 R B y .6,0  R Bx  3,0  67,082  4,0  0  2R B y  R Bx  89,443
For whole system:  M A  0
 R B y .9,0  R Bx  1,0  100  1,5  67,082  7,0  0
 9R B y  R Bx  619,574

R Bx  39,469 kN R B y  64,456 kN
 G
Q=67,082 kN
L BG  6,7082 m

3,0m
20 kN/m
100 kN
C α2 R Bx  39,469 kN

1,5m

1,0m
B
A α1
R By  64,456 kN
R A x  60,531 kN
1,125 1,875 4,0 2,0
3,0m 6,0m
R A y  2,626 kN

Check the results:

Fro whole system:
P  R
x Ax  R Bx  100  60,531  39,469  100  0
P  Ry Ay  R Bx  67,082  2,626  64,456  67,082  0
The estimated support reactions are correct
Member lenghts and their trigonometric identities:
L BG  3 2  6 2  6,7082 m L AG  4 2  3 2  5,0 m
cos 1  3 / 5  0,60 sin 1  4 / 5  0,80
cos  2  6 / 6,7082  0,8944 sin  2  3 / 6,7082  0,4472
 67,082 kN
Internal forces:
G L BG  6,7082 m
L AG  5,0 m
α2
cos 1  0,60
α1 C sin 1  0,80
B α2 α2
100 kN
cos  2  0,8944
α1
A α1 sin  2  0,4472
A
60,531 kN α2

α1
2,626 kN

AC and CG members
NAC  NCA  2,101  36,319  34,218 kN
VAC  VCA  48,425  1,575  50,00 kN M AC  0
M CA  M CG  60,531 1,5  2,626  1,125  92,75 kNm
NCG  NGC  2,101  36,319  60,00  25,782 kN
VCG  VGC  48,425  1,575  80,00  30,00 kN MGC  0
BG member
NBG  28,825  35,301  64,126 kN
VBG  57,650  17,650  40,00 kN M BG  0
NGB  28,825  35,301  30  34,126 kN
VGB  57,650  17,650  60  20,00 kN MGB  0
 G
q α VBG  40,00 kN
2 q(x)=3,333x
M BG  0
VGB  20,00 kN
α2 Mx MGB  0
x
Vx Nx
Şekil 4.71c

Since the sign of the shear forces at B and G ends of BG frame changes,
shear force is zero at any location and moment is maximum at that point.
The function of triangular distributed load: q(x)  (x / 6)q  (20 / 6)x  3,333x
Vx  20  0,5  q(x). cos  2 .(x / cos  2 )
Vx  20  1,667x 2  0 x 0  20 / 1,667  3,464 m
M x  20  x / cos  2  0,5  q(x)  cos  2  (x / cos  2 )  x /(3  cos  2 )
veya
M x  20  x / cos  2  0,5  q(x)  (x / cos  2 )  x / 3
M x  (20  x  0,555x 3 ) / cos  2
Substitute 3,464 m instead of x
M enb.  (20  3,464  0,555  3,464 3 ) / 0,8944  51,64 kNm
 G G
+ 2op

3,0m
100 kN 20 kN/m
α2

1,5m
R Bx

1,0m
3,464 m
C
+ C R Ax B
B A α1 R By
RA
1,125 1,875 4,0 2,0
A y
3,0m 6,0m
V (kN)

C + 3op +
B

A M (kNm)

C 2op
B
+
A
N (kN)
Example 4:
40 kN/m
100 kN 30 kNm

E I
F G

2m
H
50 kN
C D
4m 1m

RAx A M α B
A
2m 3m 8m 3m 2m
RAy RBy
Şekil 4.75a Taşıyıcı sistem
Estimate support reactions and draw internal force diagrams of the
given system.

Solution:
Member lenghts and their trigonometric identities:
L BH  6 2  3 2  6,71 m
cos   3 / 6,71  0,447 sin   6 / 6,71  0,894
 40 kN/m
100 kN 30 kNm

E I
F G

2m
H
50 kN
C D
1m

4m RAx A M α B
A
2m 3m 8m 3m 2m
RAy RBy

40 kN/m
30 kNm
Gx G
I
H
Gy 8m 3m 2m

100 kN 40 kN/m Gy
Gx
F G
2m

E
50 kN
B
C D
1m
RBy α
4m

RAx A Structural layout

MA
2m 3m
RAy
 40 kN/m
30 kNm
Gx=0 G
I
H
Gy=215,455 kN 8m 3m 2m

100 kN 40 kN/m 215,455 kN

Gx=0
2m
E F G Support Reactions:
50 kN
B
C D For GHB system
1m
α
P 0 G 0
4m

x x

M  0
RBy=304,545 kN
RAx=0 A B
MA=-676,37 kNm
2m 3m
G y  [40  13,0  (6,5  2,0)  30] / 11  215,455 kN
M
RAy=485,455 kN
G 0
R By  (40 13,0  6,5  30] /11  304,545 kN
Check the results: P y  215,455  520,000  304,545  0

For AFG system

P 0 x
R Ax  0

P 0 y R Ay  215,455  40  3,0  100  50  485,455 kN

M  0 A

M A  215,455  3,0  40  3,0  1,5  100  2,0  50  1,0  676,37 kNm

 40 kN/m
30 kNm
Gx=0 G
I
H Internal Forces
Gy=215,455 kN 8m 3m 2m
40 kN/m
AC member
100 kN 215,455 kN
N AC  N CA  485,455 kN
Gx=0
F G VAC  VCA  0
2m
E
50 kN
B
C D
1m M CA  M CA  676,37 kNm
α
4m

RBy=304,545 kN
RAx=0 A
CF member
N CF  N FC  100  120  215,455  435,455 kN,
MA=-676,37 kNm
2m 3m
RAy=485,455 kN MCF  MFC  100  2,0  120 1,5  215,455  3,0
MCF  M FC  626,37 kNm VCF  VFC  0
CD member
N CD  N DC  0 VCD  VDC  50 kN
M CD  50  1,0  50,00 kNm M DC  0
EF member
N EF  N FE  0 VEF  VFE  100 kN
M FE  100  2,0  200,00 kNm M EF  0
FG member
N FG  N GF  0 VFG  215,455  120  335,455 kN
VGF  215,455 kN
M FG  215,455  3,0  120  1,5  826,37 kNm M GF  0
 40 kN/m
30 kNm
Gx=0 G
I
H
Gy=215,455 kN 8m 3m 2m

100 kN 40 kN/m 215,455 kN

Gx=0
2m E F G
50 kN
B
C D
1m
α
4m

RBy=304,545 kN
RAx=0 A
MA=-676,37 kNm
2m 3m
RAy=485,455 kN

GH member
N GH  N HG  0
VGH  VGF  215,455 kN
M GH  M GF  0
VHG  215,455  320  104,445 kN
M HG  215,455  8,0  320  4,0  443,64 kNm

Distance of zero shear force location to G: x 0  VGH / q

x 0  215,455 / 40  5,39 m
Maximum value of moment: Menb.  MGH  VGH
2
/(2q)
M enb.  215,455 2 /( 2  40)  580,26 kNm
 40 kN/m
30 kNm
Gx=0 G
I
H
Gy=215,455 kN 8m 3m 2m

100 kN 40 kN/m 215,455 kN

Gx=0 cos   0,447
F G
sin   0,894
2m
E
50 kN
B
C D
1m
α
4m

RBy=304,545 kN
RAx=0 A
MA=-676,37 kNm
2m 3m
RAy=485,455 kN
,
HI member
N HI  N IH  0 VHI  40  5,0  200,00 kN VIH  0
M HI  40  5,0 2 / 2  30  470,00 kNm M IH  30 kNm
HB member
N HB  N BH  304,545  sin   304,545  0,894  272,263 kN
VHB  VBH  304,545  cos   304,545  0,447  136,132 kN
M HB  304,545  3,0  913,64 kNm M BH  0
 335,455

215,455

200,00
+
+ 40 kN/m 30 kNm
E F 100 kN

100
I

100
G 10H
I

2m
4,4 E F

50
+ D 5,39 m G H
50 kN
C 45 C D
1m

4m
A B
V (kN) RAx A M α B
A
2m 3m 8m 3m 2m
RAy RBy
200 826,37

470,00
580,26
626,37 I

30,00
E F H +
G +

443,64
626,37
50
676,37 C D
,0
0
+
A M (kNm) B
676,37

E 435,455 H I
F G
435,455 D
C 485,455

A N (kN)
B
485,455
Example 5:
20 kN/m

3,0m 2,0m
E G
100 kN 100 kN

C D H I

5,0m
R Ax R Bx
A B
1,2m 5,8m 5,8m 1,2m
MA MB
14,0 m
R Ay R By

Estimate support reactions and draw internal force diagrams of the given
system.
 20 kN/m

F Solution:

3,0m 2,0m
Number of support
E G
reactions:
100 kN 100 kN nr  3  3  6
C D H I Number of hinge
conditions:
nm 111 3

5,0m
n  6  (3  3)  0
R Ax
A B
MA 1,2m 5,8m 5,8m 1,2m MB
R Ay 14,0 m R By

Support reactions:
R A x  R Bx
Since the system is symmtrical R A y  R By
MA  MB
 20 kN/m

3,0m 2,0m
E G
100 kN 100 kN

C D H I

5,0m
R Ax
A B R Bx
MA 1,2m 5,8m 5,8m 1,2m MB
R Ay 14,0 m R By
ACE system: M E 0
 R Ax  8,0  M A  100  1,2  0  8R Ax  M A  120
ACEF system  MF  0
 R Ax  10,0  R A y  7,0  M A  100  5,8  20  7,0  3,5  0
 10R Ax  7 R A y  M A  1070
ACEFG system  MG  0
 R Ax  8,0  R A y  14,0  M A  100  12,8  20  14,0  7  0
 8R Ax  14R A y  M A  3240
R A x  245 kN R A y  240 kN M A  1840 kNm
 20 kN/m

2,0m
E G
100 kN

3,0m
100 kN

C D H I

5,0m
R A x  245 kN
A B R Bx  245 kN
1,2m 5,8m 5,8m 1,2m
MB=1840 kNm
MA=1840 kNm 14,0 m
R A y  240 kN R B y  240 kN

R Bx  R A x  245 kN R B y  R A y  240 kN
M B  M A  1840 kNm
Check the results:
For whole system:  Py  0
P  240  240  100  100  20  14  0
y

For whole system:  Px  0 

Px  245  245  0
For whole system: M B 0
240  14,0  1840  100  12,8  280  7,0  100  1,2  1840  0
 20 kN/m
20 kN/m
F
F

3,0m 2,0m
Ex
E G Gx
E G
100 kN 100 kN Ey Gy

C D H I Ey Gy
Ex Gx

5,0m
E G
100 kN 100 kN
R Ax
A B R Bx
1,2m 5,8m 5,8m 1,2m
MB
MA 14,0
R Ay m R By
R Ax R Bx
A B
MB
MA
R Ay R By

For EFG subsystem:  M G  0

E y  14,0  20  14,0  7,0  0 E y  1960 / 14  140 kN
F section of EF member of EFG subsystem  M F  0
E y  7,0  E x  2,00  20  7,0  3,5  0
7E y  2E x  490 E x  (140  7  490) / 2  245 kN
Using symmetry rule G x  E x  245 kN G y  E y  140 kN
 20 kN/m

Ex=245 E G Gx=245

Ey=140 Gy=140

140 kN 140 kN
245 kN 245 kN
E G
100 kN 100 kN

R Ax R Bx
A B

MA MB
R Ay R By

Reactions of support E of EFG system are applied to the E point

of ACE system:
P x 0 R A x  245 kN P y 0 R A y  140  100  240 kN

M A 0 M A  245  8,0  100  1,2  1840 kNm

Obtaining the reactions of support B using symmetry condition
R Bx  R A x  245 kN R B y  R A y  240 kN M B  M A  1840 kNm
Internal forces 20 kN/m

2,0m
E G
100 kN

3,0m
100 kN

C D H I

5,0m
R A x  245 kN
A B R Bx  245 kN
1,2m 5,8m 5,8m 1,2m
MB=1840 kNm
MA=1840 kNm 14,0 m
R A y  240 kN R B y  240 kN

L EF  L FG  7 2  2 2  7,28 m sin   2 / 7,28  0,2747

cos   7 / 7,28  0,9615
Since the system is symmetrical, internal forces of ACEF system are
determined and BIGF system will have the same magnitude and sign of
moment and axial force with the ACEF system, but BIGF system will
have same magnitude and opposite sign of shear force with ACEF system.
AC member
N AC  N CA  240,000 kN VAC  VCA  245,000 kN
M AC  1840,00 kNm
M CA  1840,00  245,00  5,0  615,00 kNm
 20 kN/m

2,0m
E G
100 kN

3,0m
100 kN

C D H I

5,0m
R A x  245 kN
A B R Bx  245 kN
1,2m 5,8m 5,8m 1,2m
MB=1840 kNm
MA=1840 kNm 14,0 m
R A y  240 kN R B y  240 kN

CD member
N CD  N DC  0 VCD  VDC  50,000 kN
M CD  100  1,20  120,00 kNm M CD  0
CE member
N CE  N EC  240  100  140,000 kN
VCE  VEC  245,000 kN M EC  0
M CE  1840  245  6,0  100  1,2  735,00 kNm
 20 kN/m

2,0m
E G
α
100 kN sin   0,2747

3,0m
100 kN
cos   0,9615
C D H I

5,0m
α α
A B R Bx  245 kN
R A x  245 kN 1,2m 5,8m 5,8m 1,2m
MB=1840 kNm
MA=1840 kNm 14,0 m
R A y  240 kN R B y  240 kN
EF member
N EF  (240  100)  sin   245  cos 
N EF  140  0,2747  245  0,9615  274,026 kN
VEF  (240  100)  cos   245  sin 
VEF  140  0,9615  245  0,2747  67,302 kN M EF  0
N FE  (240  100  20  7,0)  sin   245  cos 
N FE  245  0,9615  235,57 kN
VFE  (240  100  20  7,0)  cos   245  sin 
VFE  245  0,2747  67,302 kN M FE  0
 20 kN/m

2,0m
E G
100 kN

3,0m
100 kN

C D H I

5,0m
R A x  245 kN
A B R Bx  245 kN
1,2m 5,8m 5,8m 1,2m
MB=1840 kNm
MA=1840 kNm 14,0 m
R A y  240 kN R B y  240 kN

The distance of zero shear force location to point E:

Vx  67,392  20x  x / cos   0
x 0  67,302 /(20  0,9615)  3,50 m
Maximum moment on EF member
M enb.  67,302  (7,28 / 2)  20  3,50  1,75  122,49 kNm
 +
+ 3,5m
245
245
100 100
+ 100
100

+
V (kN)

245 245

140
+ 140
+

120 + + 120
735 735 140 140
615 615 240 240

+
+
M (kNm) N (kN)

1840 1840 240 240

 50 kN
Example 6:
D E
Estimate support reactions and

2m
C
α draw internal force diagrams of the
given system.

3m
10 kN/m
RBx=0
B

2m
RBy=77,50 kN
Solution:
A
A´
RAx=-50 kN
10 m 3m
L CD  100  4  10,198 m
RAy=-27,5kN sin   2 / 10,198  0,196
cos   10 / 10,198  0,981
Support reactions
M D 0 at point D of BD frame due to hinge condition
R BX  5  0 R BX  0
For the system Px  0  R Ax  10  5,0  50 kN
Moment respect to point A´ where RAx and RBy are intersected
M A´ 0 R Ay  (10  5,0  2,5  50  3,0) / 10  27,50 kN
For the system  M A  0
R By  (10  5,0  2,5  50  13,0) / 10  77,50 kN
 50 kN
D E

2m
C
α

3m
10 kN/m
RBx=0
B

2m
A RBy=77,50 kN
α A´
RAx=-50 kN
10 m 3m
α
RAy=-27,5kN

Check the results: P y  27,5  77,5  50  0

Internal forces
AC frame
N AC  N CA  27,50 kN VAC  (50)  50 kN M AC  0
VCA  (50)  10  5,0  0
M CA  (50)  5,0  10  5,0  2,5  125 kNm
CD frame
N CD  N DC  (50) cos   (27,50) sin   10  5,0  cos 
N CD  N DC  27,50  0,196  5,39 kN
VCD  VDC  (50) sin   (27,50) cos   10  5,0  sin 
VCD  VDC  27,50  0,981  26,98 kN
M CD  M CA  125 kNm
 50 kN
D E

2m
C
α

3m
10 kN/m
RBx=0
B

2m
A RBy=77,50 kN
α A´
RAx=-50 kN
10 m 3m
α
RAy=-27,5kN

DE frame
N DE  N ED  0 VDE  VED  50 kN M ED  0

BD frame
N BD  N DB  77,50 kN VBD  VDB  0 M BD  M DB  0
 50 kN

50
D

50
D +

2m
C E
α
C

3m
10 kN/m RBx=0
B
B

2m
A RBy=77,50 kN +
α A´
RAx=-50 kN A
10 m 3m 50
α V (kN)
RAy=-27,5kN

150
D E
77,5 D E
27,5
+ C 125 +
C
+
B
+
B
77,5
A
A
27,5 N (kN) M (kNm)
Reference:
 Prof. İbrahim EKİZ, YAPI STATİĞİ 1, BİRSEN YAYINEVİ

Acknowledgement:
Thank you to Prof. İbrahim EKİZ for sharing this presentation
with me.