This problem answers step by step how to obtain stress values in plane frame systems subjected to bending.

1. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Block A.
Stress in frame systems.
Step by step problem.

2. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
 Study the following structural system and the corresponding cross-sections of its bars (z-axis is perpendicular to the plane of the screen).
Then, answer:
1. Moment of inertia (cm4) and first moment of area of half cross-section (cm3) (both related to the z-axis) of both members, CD and DE.
2. Maximum normal stress, positive, in segment DE (kN/cm2). State section and y-coordinate where it takes place.
3. Maximum normal stress, negative, in segment DE (kN/cm2). State section and y-coordinate where it takes place.
4. Maximum shear stress in segment CD (kN/cm2), stating section and y-coordinate where it takes place.
5. Normal stress in section B, in the superior part of the cross-section (kN/cm2)
6. Equivalent stress (Von Mises) in section A, in the top web-flange joint, (kN/cm2)
7. Yield stress in section B (Von Mises) (kN/cm2)
(+)
(+)
Postitive sign convention:
y
z
G
y
z G
Cross-section for
members CD y DE
Cross-section for
member AD
1 cm
1 cm
25 cm
25 cm
1 cm
1 cm
25 cm
25 cm
1 cm1 cm
C
A
B
D E
2,5 m1 m
1,5 m
1,5 m
50 kN
10 kN/m
20 kN/m
50 kNm
50 kN
10 kN/m

3. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
 Study the following structural system and the corresponding cross-sections of its bars (z-axis is perpendicular to the plane of the screen).
Then, answer:
1. Moment of inertia (cm4) and first moment of area of half cross-section (cm3) (both related to the z-axis) of both members, CD and DE.
First we find the “y” coordinate of the centroid (as we already know that it has to be located on the symmetry axis of the cross-section.


    
  
  
1
· 25 1 24,5 24 1 12
18,38
25 1 24 1
i n
i i
G
i i
A d
y cm
A
y
G
Cross-section for
members CD y DE
1 cm
1 cm
25 cm
25 cm


      2
1
25 1 24 1 49
i n
i
i
A A cm

4. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
 Study the following structural system and the corresponding cross-sections of its bars (z-axis is perpendicular to the plane of the screen).
Then, answer:
1. Moment of inertia (cm4) and first moment of area of half cross-section (cm3) (both related to the z-axis) of both members, CD and DE.
First we find the “y” coordinate of the centroid (as we already know that it has to be located on the symmetry axis of the
cross-section.


    
  
  
1
· 25 1 24,5 24 1 12
18,38
25 1 24 1
i n
i i
G
i i
A d
y cm
A
18,38 cm
y
z
G
Cross-section for
members CD y DE
1 cm
1 cm
25 cm
25 cm


      2
1
25 1 24 1 49
i n
i
i
A A cm

5. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
 Study the following structural system and the corresponding cross-sections of its bars (z-axis is perpendicular to the plane of the screen).
Then, answer:
1. Moment of inertia (cm4) and first moment of area of half cross-section (cm3) (both related to the z-axis) of both members, CD and DE.
First we find the “y” coordinate of the centroid (as we already know that it has to be located on the symmetry axis of the cross-section.
Once we know where the centroid is located, we can obtain the moment of inertia using Steiner’s Theorem:
And the first moment of area.


    
  
  
1
· 25 1 24,5 24 1 12
18,38
25 1 24 1
i n
i i
G
i i
A d
y cm
A
18,38 cm
y
z
G
Cross-section for
members CD y DE
1 cm
1 cm
25 cm
25 cm 


 
          
3 3
22 2 4
1
25 1 1 24
( · ) 25 1 6,12 24 1 6,38 3067,35
12 12
i n
z i i i
i
I I A d cm
1
sec
32
1
5,62
· 25 1 6,12 5,62 1 168,8
2
i n
z i i
i
S A d cm


       


      2
1
25 1 24 1 49
i n
i
i
A A cm

6. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
 Study the following structural system and the corresponding cross-sections of its bars (z-axis is perpendicular to the plane of the screen).
Then, answer:
1. Moment of inertia (cm4) and first moment of area of half cross-section (cm3) (both related to the z-axis) of both members, CD and DE.
First we find the “y” coordinate of the centroid (as we already know that it has to be located on the symmetry axis of the cross-section.
Once we know where the centroid is located, we can obtain the moment of inertia using Steiner’s Theorem:
And the first moment of area.


    
  
  
1
· 25 1 24,5 24 1 12
18,38
25 1 24 1
i n
i i
G
i i
A d
y cm
A
18,38 cm
y
z
G
Cross-section for
members CD y DE
1 cm
1 cm
25 cm
25 cm 


 
          
3 3
22 2 4
1
25 1 1 24
( · ) 25 1 6,12 24 1 6,38 3067,35
12 12
i n
z i i i
i
I I A d cm
1
sec
32
1
5,62
· 25 1 6,12 5,62 1 168,8
2
i n
z i i
i
S A d cm


       


      2
1
25 1 24 1 49
i n
i
i
A A cm
Remember that if the centroid has been
correctly obtained, the first moment of area
will be the same no matter if you choose the
superior or the inferior half of the cross-
section. You can use it as a way to check if
you did it right.

7. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
 Study the following structural system and the corresponding cross-sections of its bars (z-axis is perpendicular to the plane of the screen).
Then, answer:
1. Moment of inertia (cm4) and first moment of area of half cross-section (cm3) (both related to the z-axis) of both members, CD and DE.
First we find the “y” coordinate of the centroid (as we already know that it has to be located on the symmetry axis of the cross-section.
Once we know where the centroid is located, we can obtain the moment of inertia using Steiner’s Theorem:
And the first moment of area.
Same procedure for the second cross-section:


    
  
  
1
· 25 1 24,5 24 1 12
18,38
25 1 24 1
i n
i i
G
i i
A d
y cm
A
18,38 cm
y
z
G
Cross-section for
members CD y DE
1 cm
1 cm
25 cm
25 cm 


 
          
3 3
22 2 4
1
25 1 1 24
( · ) 25 1 6,12 24 1 6,38 3067,35
12 12
i n
z i i i
i
I I A d cm
1
sec
32
1
5,62
· 25 1 6,12 5,62 1 168,8
2
i n
z i i
i
S A d cm


       


      2
1
25 1 24 1 49
i n
i
i
A A cm


    
4 4
2 4
1
25 23
( · ) 9232
12 12
i n
z i i i
i
I I A d cm
 


       
1
sec
32
1
· 25 12 11,5 0,5 11,5 2 432,25
i n
z i i
i
S A d cm


    2 2 2
1
25 23 96
i n
i
i
A A cm
y
z G
1 cm
1 cm
25 cm
25 cm
1 cm1 cm

8. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
2. Maximum normal stress, positive, in segment DE (kN/cm2). State section and y-coordinate where it takes place.
As stress values depend on the values of the internal forces, first thing to do is to draw the axial force, shear force and bending moment diagrams.
-50 kN
+50 kN
-60 kN
-50 kN
10 kN
-100 kN
-115 kN
50 kNm
12,5 kNm5 kNm
7,5 kNm
157,5 kNm
318,75 kNm
Axial force Shear force Bending moment

9. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
2. Maximum normal stress, positive, in segment DE (kN/cm2). State section and y-coordinate where it takes place.
As stress values depend on the values of the internal forces, first thing to do is to draw the axial force, shear force and bending moment diagrams.
Once the values have been obtained, we can obtain the maximum normal stress, positive, in segment DE.
-50 kN
+50 kN
-60 kN
-50 kN
10 kN
-100 kN
-115 kN
50 kNm
12,5 kNm5 kNm
7,5 kNm
157,5 kNm
318,75 kNm
Axial force Shear force Bending moment
 
  ,
max
DE
z
N M
y
A I
As studied during classes, normal stress depends on the value of the axial force, the area of the cross-section, the
value of the bending moment, the moment of inertia and the distance from the centroid where the normal stress is
obtained.
Thus each combination of these factors must be studied to find the greatest value of the normal stress, positive, in
this segment.

10. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
2. Maximum normal stress, positive, in segment DE (kN/cm2). State section and y-coordinate where it takes place (cont.)
Let’s find the possible combinations that provide positive values of normal stress in segment DE:
-50 kN
+50 kN
-60 kN
50 kNm
12,5 kNm5 kNm
7,5 kNm
157,5 kNm
318,75 kNm
Axial
force
Bending moment
y
z
G
1 cm
25 cm
18,38 cm
6,62 cm
 
  ,
max
DE
z
N M
y
A I
2
49Área cm
4
3067,35zI cm

11. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
2. Maximum normal stress, positive, in segment DE (kN/cm2). State section and y-coordinate where it takes place (cont.)
Let’s find the possible combinations that provide positive values of normal stress in segment DE:
-50 kN
+50 kN
-60 kN
50 kNm
12,5 kNm5 kNm
7,5 kNm
157,5 kNm
318,75 kNm
Axial
force
Bending moment
y
z
G
1 cm
25 cm
6,62 cm
 
  ,
max
DE
z
N M
y
A I
2
49Área cm
4
3067,35zI cm
  
 

    ,
22 4
12,5 100
50
6,62 3,72
49 3067,35
DE
D
cm
kNm
kN m kNcm
cmcm cm

12. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
2. Maximum normal stress, positive, in segment DE (kN/cm2). State section and y-coordinate where it takes place (cont.)
Let’s find the possible combinations that provide positive values of normal stress in segment DE:
-50 kN
+50 kN
-60 kN
50 kNm
12,5 kNm5 kNm
7,5 kNm
157,5 kNm
318,75 kNm
Axial
force
Bending moment
y
z
G
1 cm
25 cm
18,38 cm
 
  ,
max
DE
z
N M
y
A I
2
49Área cm
4
3067,35zI cm
  
 

    ,
22 4
12,5 100
50
6,62 3,72
49 3067,35
DE
D
cm
kNm
kN m kNcm
cmcm cm
  
 

    ,
22 4
50 100
50
18,38 30,99
49 3067,35
DE
E
cm
kNm
kN m kNcm
cmcm cm

13. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
2. Maximum normal stress, positive, in segment DE (kN/cm2). State section and y-coordinate where it takes place (cont.)
Let’s find the possible combinations that provide positive values of normal stress in segment DE:
-50 kN
+50 kN
-60 kN
50 kNm
12,5 kNm5 kNm
7,5 kNm
157,5 kNm
318,75 kNm
Axial
force
Bending moment
y
z
G
1 cm
25 cm
18,38 cm
 
  ,
max
DE
z
N M
y
A I
2
49Área cm
4
3067,35zI cm
  
 

    ,
22 4
12,5 100
50
6,62 3,72
49 3067,35
DE
D
cm
kNm
kN m kNcm
cmcm cm
  
 

    ,
22 4
50 100
50
18,38 30,99
49 3067,35
DE
E
cm
kNm
kN m kNcm
cmcm cm
Thus, the maximum tensile stress in segment DE
happens on section E, in the inferior part of the cross
section (y=-18,38cm) and the value is 30.99 kN/cm2

14. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
3. Maximum normal stress, positive, in segment DE (kN/cm2). State section and y-coordinate where it takes place (cont.)
Let’s find the possible combinations that provide negative values of normal stress in segment DE:
  
 

     ,
22 4
12,5 100
50
18,38 6,47
49 3067,35
DE
D
cm
kNm
kN m kNcm
cmcm cm
-50 kN
+50 kN
-60 kN
50 kNm
12,5 kNm5 kNm
7,5 kNm
157,5 kNm
318,75 kNm
Axial
force
Bending moment
 
  ,
max
DE
z
N M
y
A I
2
49Área cm
4
3067,35zI cm
y
z
G
1 cm
25 cm
18,38 cm

15. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
3. Maximum normal stress, positive, in segment DE (kN/cm2). State section and y-coordinate where it takes place (cont.)
Let’s find the possible combinations that provide negative values of normal stress in segment DE:
  
 

     ,
22 4
12,5 100
50
18,38 6,47
49 3067,35
DE
D
cm
kNm
kN m kNcm
cmcm cm
-50 kN
+50 kN
-60 kN
50 kNm
12,5 kNm5 kNm
7,5 kNm
157,5 kNm
318,75 kNm
Axial
force
Bending moment
 
  ,
max
DE
z
N M
y
A I
2
49Área cm
4
3067,35zI cm
y
z
G
1 cm
25 cm
  
 

     ,
22 4
50 100
50
6,62 9,77
49 3067,35
DE
E
cm
kNm
kN m kNcm
cmcm cm
6,62 cm

16. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
3. Maximum normal stress, positive, in segment DE (kN/cm2). State section and y-coordinate where it takes place (cont.)
Let’s find the possible combinations that provide negative values of normal stress in segment DE:
  
 

     ,
22 4
12,5 100
50
18,38 6,47
49 3067,35
DE
D
cm
kNm
kN m kNcm
cmcm cm
-50 kN
+50 kN
-60 kN
50 kNm
12,5 kNm5 kNm
7,5 kNm
157,5 kNm
318,75 kNm
Axial
force
Bending moment
 
  ,
max
DE
z
N M
y
A I
2
49Área cm
4
3067,35zI cm
y
z
G
1 cm
25 cm
  
 

     ,
22 4
50 100
50
6,62 9,77
49 3067,35
DE
E
cm
kNm
kN m kNcm
cmcm cm
6,62 cm
Thus, the maximum compressive stress in segment
DE happens on section E, in the superior part of the
cross section (y=+6,62cm) and the value is -9,77
kN/cm2

17. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
y
z
G
1 cm
25 cm
Problem 1
4. Maximum shear stress in segment CD (kN/cm2), stating section and y-coordinate where it takes place.
4
3067,35zI cmShear stress
1
sec
32
280,76zS cm
Shear stress depends on the value of the shear, on the first moment of area and the moment of inertia of
the cross section, and on the thickness of the cross section at the point where the shear stress is being
obtained.
max
fibra
zCD
z
V S
I b




1 cm18,38 cm
6,62 cm
-50 kN
10 kN
-100 kN
-115 kN

18. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
y
z
G
1 cm
25 cm
Problem 1
4. Maximum shear stress in segment CD (kN/cm2), stating section and y-coordinate where it takes place.
4
3067,35zI cmShear stress
1
sec
32
280,76zS cm
Shear stress depends on the value of the shear, on the first moment of area and the moment of inertia of
the cross section, and on the thickness of the cross section at the point where the shear stress is being
obtained.
max
fibra
zCD
z
V S
I b




1 cm18,38 cm
6,62 cm
-50 kN
10 kN
-100 kN
-115 kN
In a rectangular cross section, the distribution of shear stress is parabollic and the máximum value happens at the centroid.
Thus the first moment of área that has to be used is the máximum (the one that corresponds to half cross-section).


 

3
2max 4
10 168,8
0,55
3067,35 1
CD kN cm kN
cmcm cm

19. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
y
z
G
1 cm
25 cm
Problem 1
4. Maximum shear stress in segment CD (kN/cm2), stating section and y-coordinate where it takes place.
4
3067,35zI cmShear stress
1
sec
32
280,76zS cm
Shear stress depends on the value of the shear, on the first moment of area and the moment of inertia of
the cross section, and on the thickness of the cross section at the point where the shear stress is being
obtained.
max
fibra
zCD
z
V S
I b




1 cm18,38 cm
6,62 cm
-50 kN
10 kN
-100 kN
-115 kN
In a rectangular cross section, the distribution of shear stress is parabollic and the máximum value happens at the centroid.
Thus the first moment of área that has to be used is the máximum (the one that corresponds to half cross-section).
The maximum shear stress in segment CD appears in section
D, at the y=0 coordinate and the value is 0,55 kN/cm2

 

3
2max 4
10 168,8
0,55
3067,35 1
CD kN cm kN
cmcm cm

20. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
5. Normal stress in section B, in the superior part of the cross-section (kN/cm2)
To answer this question, take into account that segment AD is made with a different cross-section:
-50 kN
+50 kN
-60 kN
50 kNm
12,5 kNm5 kNm
7,5 kNm
157,5 kNm
318,75 kNm
Axial
force
Bending moment
   sup
B
z
N M
y
A I
2
96Área cm
4
9232zI cm
25 cm

21. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
5. Normal stress in section B, in the superior part of the cross-section (kN/cm2)
To answer this question, take into account that segment AD is made with a different cross-section:
-50 kN
+50 kN
-60 kN
50 kNm
12,5 kNm5 kNm
7,5 kNm
157,5 kNm
318,75 kNm
Axial
force
Bending moment
   sup
B
z
N M
y
A I
2
96Área cm
4
9232zI cm
25 cm
 
 

      2sup 2 4
157,5 100
60
12,5 20,70
96 9232
B
cm
kNm
kN m kNcm
cmcm cm

22. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
6. Equivalent stress (following Von Mises’ criterion) in section A, in the top web-flange joint, (kN/cm2)
The equivalent stress in a determinate y-coordinate provides the value of the combined effect of normal and shear stress in a specific point of the
cross-section:
-50 kN
+50 kN
-60 kN
50 kNm
12,5 kNm5 kNm
7,5 kNm
157,5 kNm
318,75 kNm
Axial
force
Bending moment
2
96Área cm
4
9232zI cm
25 cm
Shear force
     
 
22
3ALA ALMA ALA ALMA ALA ALMA
co
Before finding the equivalent stress, we must obtain the values of the normal and
shear stress in that specific y-coordinate of the cross-section.
A A A
-50 kN
10 kN
-100 kN
-115 kN
Web-flange
sup.
eq

23. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
6. Equivalent stress (following Von Mises’ criterion) in section A, in the top web-flange joint, (kN/cm2)
The equivalent stress in a determinate y-coordinate provides the value of the combined effect of normal and shear stress in a specific point of the
cross-section:
-50 kN
+50 kN
-60 kN
50 kNm
12,5 kNm5 kNm
7,5 kNm
157,5 kNm
318,75 kNm
Axial
force
Bending moment
2
96Área cm
4
9232zI cm
25 cm
Shear force
     
 
22
3ALA ALMA ALA ALMA ALA ALMA
co
Before finding the equivalent stress, we must obtain the values of the normal and
shear stress in that specific y-coordinate of the cross-section.
A A A
-50 kN
10 kN
-100 kN
-115 kN
Web-flange
sup.
eq
1. Normal stress in the superior web-flange joint of the cross-section in section A of the structure
  
 

      2sup 2 4
318,75 100
60
11,5 39,08
96 9232
A
ALA ALMA
cm
kNm
kN m kNcm
cmcm cmWeb-flange
sup.

24. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
6. Equivalent stress (following Von Mises’ criterion) in section A, in the top web-flange joint, (kN/cm2)
The equivalent stress in a determinate y-coordinate provides the value of the combined effect of normal and shear stress in a specific point of the
cross-section:
-50 kN
+50 kN
-60 kN
50 kNm
12,5 kNm5 kNm
7,5 kNm
157,5 kNm
318,75 kNm
Axial
force
Bending moment
2
96Área cm
4
9232zI cm
25 cm
Shear force
     
 
22
3ALA ALMA ALA ALMA ALA ALMA
co
Before finding the equivalent stress, we must obtain the values of the normal and
shear stress in that specific y-coordinate of the cross-section.
A A A
-50 kN
10 kN
-100 kN
-115 kN
Web-flange
sup.
eq
1. Normal stress in the superior web-flange joint of the cross-section in section A of the structure
  
 

      2sup 2 4
318,75 100
60
11,5 39,08
96 9232
A
ALA ALMA
cm
kNm
kN m kNcm
cmcm cmWeb-flange
sup.
11,5 cm

25. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
6. Equivalent stress (following Von Mises’ criterion) in section A, in the top web-flange joint, (kN/cm2)
The equivalent stress in a determinate y-coordinate provides the value of the combined effect of normal and shear stress in a specific point of the
cross-section:
-50 kN
+50 kN
-60 kN
50 kNm
12,5 kNm5 kNm
7,5 kNm
157,5 kNm
318,75 kNm
Axial
force
Bending moment
2
96Área cm
4
9232zI cm
25 cm
Shear force
     
 
22
3ALA ALMA ALA ALMA ALA ALMA
co
Before finding the equivalent stress, we must obtain the values of the normal and
shear stress in that specific y-coordinate of the cross-section.
A A A
-50 kN
10 kN
-100 kN
-115 kN
Web-flange
sup.
eq
1. Normal stress in the superior web-flange joint of the cross-section in section A of the structure
11,5 cm
2. Shear stress in the superior web-flange joint of the cross-section in section A
The first moment of area to be used is the one that
corresponds only to the superior flange of the cross-
section, because that’s the y-coordinate where the
shear stress has to be found:
y
z G
12 cm
25 cm
1 cm

     3
1
· 25 12 1 300
i n
ALA
z i i
i
S A d cm

26. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
6. Equivalent stress (following Von Mises’ criterion) in section A, in the top web-flange joint, (kN/cm2)
The equivalent stress in a determinate y-coordinate provides the value of the combined effect of normal and shear stress in a specific point of the
cross-section:
-50 kN
+50 kN
-60 kN
50 kNm
12,5 kNm5 kNm
7,5 kNm
157,5 kNm
318,75 kNm
Axial
force
Bending moment
2
96Área cm
4
9232zI cm
25 cm
Shear force
     
 
22
3ALA ALMA ALA ALMA ALA ALMA
co
Before finding the equivalent stress, we must obtain the values of the normal and
shear stress in that specific y-coordinate of the cross-section.
A A A
-50 kN
10 kN
-100 kN
-115 kN
Web-flange
sup.
eq
1. Normal stress in the superior web-flange joint of the cross-section in section A of the structure
11,5 cm
2. Shear stress in the superior web-flange joint of the cross-section in section A
  
 

3
24
115 300
1,86
9232 2
ALA ALMA
A
kN cm kN
cmcm cm
y
z G
12 cm
25 cm
1 cm

     3
1
· 25 12 1 300
i n
ALA
z i i
i
S A d cm
The thickness of the cross section at the point where the flange
meets the web is 2 cm (1 cm for each of the webs). Thus, the
final result of the shear stress is:

27. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
6. Equivalent stress (following Von Mises’ criterion) in section A, in the top web-flange joint, (kN/cm2)
The equivalent stress in a determinate y-coordinate provides the value of the combined effect of normal and shear stress in a specific point of the
cross-section:
-50 kN
+50 kN
-60 kN
50 kNm
12,5 kNm5 kNm
7,5 kNm
157,5 kNm
318,75 kNm
Axial
force
Bending moment
2
96Área cm
4
9232zI cm
25 cm
Shear force
     
 
22
3ALA ALMA ALA ALMA ALA ALMA
co
Before finding the equivalent stress, we must obtain the values of the normal and
shear stress in that specific y-coordinate of the cross-section.
A A A
-50 kN
10 kN
-100 kN
-115 kN
Web-flange
sup.
eq
1. Normal stress in the superior web-flange joint of the cross-section in section A of the structure
11,5 cm
2. Shear stress in the superior web-flange joint of the cross-section in section A
The last step left is combining both values of stress using Von Mises’ criterion:
 
   2 2
2, 39,08 3 1,86 39,21ALA ALMA
co A
kN
cm
Web-flange
sup.
eq

28. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
7. Yield stress (using Von Mises’ criterion) in section B (kN/cm2)
The equivalent stress has a physical meaning because it is obtained on a specific point of the cross-section. On the contrary, the yield stress will
combine the maximum value of both, normal and shear stresses. During the lectures we have already studied that the maximum of both stresses
can’t coincide on the same point of the cross-section.
-50 kN
+50 kN
-60 kN
50 kNm
12,5 kNm5 kNm
7,5 kNm
157,5 kNm
318,75 kNm
Axial
force
Bending moment
2
96Área cm
4
9232zI cm
25 cm
Shear force
    
2 2
, max, max,3env B B B
1. Maximum normal stress in point B (as an absolute value)
B B B
1
sec
32
432,25zS cm
   max,B
z
N M
y
A I
-50 kN
10 kN
-100 kN
-115 kN
yield

29. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
7. Yield stress (using Von Mises’ criterion) in section B (kN/cm2)
The equivalent stress has a physical meaning because it is obtained on a specific point of the cross-section. On the contrary, the yield stress will
combine the maximum value of both, normal and shear stresses. During the lectures we have already studied that the maximum of both stresses
can’t coincide on the same point of the cross-section.
-50 kN
+50 kN
-60 kN
50 kNm
12,5 kNm5 kNm
7,5 kNm
157,5 kNm
318,75 kNm
Axial
force
Bending moment
2
96Área cm
4
9232zI cm
25 cm
Shear force
    
2 2
, max, max,3env B B B
1. Maximum normal stress in point B (as an absolute value)
B B B
1
sec
32
432,25zS cm
   max,B
z
N M
y
A I
 


      2sup 2 4
157,5 100
60
12,5 20,70
96 9232
B
cm
kNm
kN m kNcm
cmcm cm
-50 kN
10 kN
-100 kN
-115 kN
yield

30. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
7. Yield stress (using Von Mises’ criterion) in section B (kN/cm2)
The equivalent stress has a physical meaning because it is obtained on a specific point of the cross-section. On the contrary, the yield stress will
combine the maximum value of both, normal and shear stresses. During the lectures we have already studied that the maximum of both stresses
can’t coincide on the same point of the cross-section.
-50 kN
+50 kN
-60 kN
50 kNm
12,5 kNm5 kNm
7,5 kNm
157,5 kNm
318,75 kNm
Axial
force
Bending moment
2
96Área cm
4
9232zI cm
25 cm
Shear force
    
2 2
, max, max,3env B B B
1. Maximum normal stress in point B (as an absolute value)
B B B
1
sec
32
432,25zS cm
   max,B
z
N M
y
A I
 


      2sup 2 4
157,5 100
60
12,5 20,70
96 9232
B
cm
kNm
kN m kNcm
cmcm cm
-50 kN
10 kN
-100 kN
-115 kN
 
 

      2inf 2 4
157,5 100
60
12,5 21,95
96 9232
B
cm
kNm
kN m kNcm
cmcm cm
yield

31. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
7. Yield stress (using Von Mises’ criterion) in section B (kN/cm2)
The equivalent stress has a physical meaning because it is obtained on a specific point of the cross-section. On the contrary, the yield stress will
combine the maximum value of both, normal and shear stresses. During the lectures we have already studied that the maximum of both stresses
can’t coincide on the same point of the cross-section.
-50 kN
+50 kN
-60 kN
50 kNm
12,5 kNm5 kNm
7,5 kNm
157,5 kNm
318,75 kNm
Axial
force
Bending moment
2
96Área cm
4
9232zI cm
25 cm
Shear force
    
2 2
, max, max,3env B B B
1. Maximum normal stress in point B (as an absolute value)
2. Maximum shear stress in point B
B B B
1
sec
32
432,25zS cm
-50 kN
10 kN
-100 kN
-115 kN


 

3
24
100 432,25
2,34
9232 2
kN cm kN
cmcm cm
yield

32. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Problem 1
7. Yield stress (using Von Mises’ criterion) in section B (kN/cm2)
The equivalent stress has a physical meaning because it is obtained on a specific point of the cross-section. On the contrary, the yield stress will
combine the maximum value of both, normal and shear stresses. During the lectures we have already studied that the maximum of both stresses
can’t coincide on the same point of the cross-section.
-50 kN
+50 kN
-60 kN
50 kNm
12,5 kNm5 kNm
7,5 kNm
157,5 kNm
318,75 kNm
Axial
force
Bending moment
2
96Área cm
4
9232zI cm
25 cm
Shear force
    
2 2
, max, max,3env B B B
1. Maximum normal stress in point B (as an absolute value)
2. Maximum shear stress in point B
B B B
1
sec
32
432,25zS cm
-50 kN
10 kN
-100 kN
-115 kN


 

3
24
100 432,25
2,34
9232 2
kN cm kN
cmcm cm
     
2 2
2, 21,95 3 2,34 22,32env B
kN
cm
yield
Yield
,B

33. Structural Systems – Academic Year 2017/18 Instructor: Maribel Castilla Heredia @maribelcastilla
Block A.
Stress in frame systems.
Step by step problem.