BME Department of Mechanics, Materials and Structures Special loadbearing structures 2018/2019 fall

T2. VIERENDEEL STRUCTURES AND FRAMES 1/11

T2. VIERENDEEL STRUCTURES

NOTE: The Picture Window House can be designed using a Vierendeel structure, but now we consider a

simpler problem to discuss the calculation algorithm of Vierendeel structures.

What’s the difference between a truss and a Vierendeel structure?

Truss Vierendeel structure

statically determinate statically indeterminate

can be calculated using statics there are too many unknowns in the equilibrium

equations we use an approximate method

all joints are pin joints, only normal and

shear force can be transmitted

fixed joints, capable of transferring bending

moments

the internal forces are either compressive or

tensile normal forces, no bending occurs

bending occurs in the members; normal force,

shear force and moment as internal forces.

Exercise 1. Vierendeel truss

Calculate the internal forces of the given Vierendeel structure!

Approximate method:

Hinges (pin-joints) are assumed at the mid-point of the rods (where 𝑀 ≈ 0), utilize symmetry and assume the

same stiffness for each members. These assumptions lead to a good approximation of the real behavior.

Step 1: Calculation of the reaction forces

The global statical model is a simply supported beam.

∑𝑀𝐴 = 0

550 ⋅ (2,5 + 5,0 + 7,5 + 10,0 + 12,5) +

+ 275 ⋅ 15,0 − 𝐵𝑦 ⋅ 15 = 0

→ 𝑩𝒚 = 𝟏𝟔𝟓𝟎 𝒌𝑵

∑𝐹𝑉 = 0

𝑨𝒚 = 𝐵𝑦 = 𝟏𝟔𝟓𝟎 𝒌𝑵

BME Department of Mechanics, Materials and Structures Special loadbearing structures 2018/2019 fall

T2. VIERENDEEL STRUCTURES AND FRAMES 2/11

Step 2: Equilibrium of the sections, calculation of the internal forces

Now we calculate the equilibrium equations of the sections of the structure to determine the internal forces.

Since we separate the parts of the structure at pins, 4 unknowns would occur in the ith section: 𝐻𝑖1, 𝐻𝑖2, 𝑉𝑖1, 𝑉𝑖2.

We assume, that the stiffness is constant for each beam, therefore 𝑉𝑖1 = 𝑉𝑖2(= 𝑉𝑖) and 𝐻𝑖1 = 𝐻𝑖2(= 𝐻𝑖). For

Section 1 we use the correct notation, but for the further sections we simply use 𝑉𝑖 for the vertical and 𝐻𝑖 for

the horizontal forces.

Section 1

We assume: 𝑉11 = 𝑉12, 𝐻11 = 𝐻12.

Equilibrium equations:

ƩFV=0 V1=(1650-275)/2=687,5 kN

ƩMA’=0 H1=[(1650-275)·1,25]/3=573 kN

Section 2 Equilibrium equations:

ƩFV=0 V2=(2·687,5-550)/2=412,5 kN

ƩMB’=0 H2=[2·687,5·2,5-550·1,25+573·3]/3=1490 kN

Section 3 Equilibrium equations:

ƩFV=0 V3=(2·412,5-550)/2=137,5 kN

ƩMC’=0 H3=[2·412,5·2,5-550·1,25+1490·3]/3=1948 kN

Free body diagram:

BME Department of Mechanics, Materials and Structures Special loadbearing structures 2018/2019 fall

T2. VIERENDEEL STRUCTURES AND FRAMES 3/11

Step 3: Internal forces and moments

We can draw the normal force, shear and moment diagrams from the free body diagram.

normal force [kN]

(internal forces paralell

to the axis)

shear force [kN]

(internal forces

perpendicular

to the axis)

bending moment [kNm]

(we draw it on the tensile side

of the axis)

moment equilibrium of joints

BME Department of Mechanics, Materials and Structures Special loadbearing structures 2018/2019 fall

T2. VIERENDEEL STRUCTURES AND FRAMES 4/11

Step 4: Determine the maximal internal normal forces and moments

For designing the cross-section, we need the maximal internal normal forces and moments.

We have to identify the heavily loaded members. Method: mark the elements under either maximal normal

force or bending moment. In this problem, we can identify 4 members (meaning 4 different combinations of

N and M).

column:

1. Nmax= -962,5 kN; M = 859,4 kNm

2. N = -275 kN; Mmax = 1375 kNm

beam (chord):

3. Nmax= ± 1948 kN; M = 171,9 kNm

4. N = ± 573 kN; Mmax= 859,4 kNm

Step 5: Evaluation of the required cross section

We use the Dunkerley-formula:

0,1+Rd

Edyy

Rd

Ed

M

Mk

N

N (𝑘𝑦𝑦 ≈ 1 → interaction factor)

utilization of the cross section

𝑁𝐸𝑑 , 𝑀𝐸𝑑 are the normal force and moment from the effects (from Step 4)

𝑁𝑅𝑑 , 𝑀𝐸𝑑 are the resistances

The resistance of a cross section:

NRd ≈ χy,z ∙ A ∙ fyd (χy,z = 0,7 → buckling reduction factor)

MRd ≈ χLT ∙ Wpl ∙ fyd (χLT = 0,8 → lateral torsional buckling factor)

Try to find a proper cross-section! → cross section catalogue

We choose S235 material quality → fyd= 235 N/mm2

Let’s try cross-section HEB 700 (A=306,4 cm2, Wpl,y=8327 cm3) for the 2nd combination of Step 4!

NEd=-275 kN, MEd=1375 kNm

193,02351083278,0

101375

2351004,3067,0

102753

63

=

+

OK!

HEB 700 is good for the other combinations as well!

(utilization for all combinations: 0,74; 0,93; 0,49; 0,66)

The structure is heavy compared to

a truss and the construction of the

joints is complicated.

BME Department of Mechanics, Materials and Structures Special loadbearing structures 2018/2019 fall

T2. VIERENDEEL STRUCTURES AND FRAMES 5/11

T2. FRAMES

Most of our structures are under compression/tension and subjected to bending at the same time. Such

structures are: slabs, frames, beams and columns. Various loads increase the complexity of the calculation,

therefore we need an approximate method.

Portal-method

Frames are statically indeterminate with multiple degrees of indeterminacy. We simplify the calculation by

separating the vertical and horizontal loads.

Vertical loads:

Horizontal loads:

Approximate model: hinges are assumed at the middle of the columns. We utilize symmetry and assume that

the members have the same stiffness. The method results in a good approximation of the internal forces.

Vertical loads

Horizontal loads

The approximate method can be applied, if

- the spans are approximately equal

- the loads are constant and uniformly distributed

- the stiffness of the columns and the beams are around the same

BME Department of Mechanics, Materials and Structures Special loadbearing structures 2018/2019 fall

T2. VIERENDEEL STRUCTURES AND FRAMES 6/11

Exercise 2. Frames

Calculate the internal forces of one frame of a monolithic reinforced concrete (RC) framed office building

with the portal-method! Draw the N,V,M diagrams and sketch the deformation of the structure!

Additional information:

- frames are in every 6 m

- stiffness of the columns and beams are approximately the same

- the wind load can be approximated by a distributed load on one side* (due to the stiffness of

the slabs)

* otherwise the upwind and downwind sides must be handled separately

Steps:

1. free-body diagram, taking apart the frame

2. vertical and horizontal loads separately

3. specifying the bending moments

4. specifying the shear and normal forces

5. superposition of the N,V,M diagrams

BME Department of Mechanics, Materials and Structures Special loadbearing structures 2018/2019 fall

T2. VIERENDEEL STRUCTURES AND FRAMES 7/11

VERTICAL LOADS

Equilibrium of the stories:

BME Department of Mechanics, Materials and Structures Special loadbearing structures 2018/2019 fall

T2. VIERENDEEL STRUCTURES AND FRAMES 8/11

VERTICAL LOADS

Internal forces and moments

loads

support forces

deformation

BME Department of Mechanics, Materials and Structures Special loadbearing structures 2018/2019 fall

T2. VIERENDEEL STRUCTURES AND FRAMES 9/11

HORIZONTAL LOADS

Equilibrium of the stories:

𝑤 =1,2 ⋅ 6,0 ⋅ 6,0

2= 21,6𝑘𝑁

BME Department of Mechanics, Materials and Structures Special loadbearing structures 2018/2019 fall

T2. VIERENDEEL STRUCTURES AND FRAMES 10/11

HORIZONTAL LOADS

Internal forces and moments

loads

support forces

deformation

BME Department of Mechanics, Materials and Structures Special loadbearing structures 2018/2019 fall

T2. VIERENDEEL STRUCTURES AND FRAMES 11/11

HORIZONTAL AND VERTICAL LOADS

Internal forces and moments of the original structure (the rule of the superposition has been used)

loads

support forces