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 statically determinate can be calculated using statics all joints are pin joints, only normal and shear force can be transmitted the internal forces are either compressive or tensile normal forces, no bending occurs Vierendeel structure statically indeterminate there are too many unknowns in the equilibrium equations we use an approximate method fixed joints, capable of transferring bending moments 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. M A = 0 550 (2,5 + 5,0 + 7,5 + 10,0 + 12,5) + + 275 15,0 B y 15 = 0 B y = 1650 kn F V = 0 A y = B y = 1650 kn
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: H i1, H i2, V i1, V i2. We assume, that the stiffness is constant for each beam, therefore V i1 = V i2 (= V i ) and H i1 = H i2 (= H i ). For Section 1 we use the correct notation, but for the further sections we simply use V i for the vertical and H i for the horizontal forces. Section 1 We assume: V 11 = V 12, H 11 = H 12. 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:
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
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: N Ed M Ed + k yy 1,0 (k yy 1 interaction factor) N M Rd Rd utilization of the cross section N Ed, M Ed are the normal force and moment from the effects (from Step 4) N Rd, M Ed are the resistances The resistance of a cross section: NRd χy,z A fyd MRd χlt Wpl fyd (χy,z = 0,7 buckling reduction factor) (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/mm 2 Let s try cross-section HEB 700 (A=306,4 cm 2, Wpl,y=8327 cm 3 ) for the 2 nd combination of Step 4! NEd=-275 kn, MEd=1375 knm 3 275 10 1375 10 + 0,7 306,4 100 235 0,8 8327 10 6 3 = 0,93 1 235 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.
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
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
T2. VIERENDEEL STRUCTURES AND FRAMES 7/11 VERTICAL LOADS Equilibrium of the stories:
T2. VIERENDEEL STRUCTURES AND FRAMES 8/11 VERTICAL LOADS Internal forces and moments loads support forces deformation
T2. VIERENDEEL STRUCTURES AND FRAMES 9/11 HORIZONTAL LOADS Equilibrium of the stories: w = 1,2 6,0 6,0 2 = 21,6kN
T2. VIERENDEEL STRUCTURES AND FRAMES 10/11 HORIZONTAL LOADS Internal forces and moments loads support forces deformation
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