Appunti corso Advanced structural design

Optimal Design for Orientation K

We can develop our analysis by introducing the parametric approach to trigonometric functions. By using the relationship between cos, sin, and tg of angles and putting k = tg (teta), we get:

By substituting this parametric change of variables in the previous equation, we obtain 2 new equations in parametric variables. If we reintroduce or solve the 1 derivative in terms of the new variable k, we get the terms on the right with respect to the equal sign. Developing further, we get:

Introducing the condition:

From this and the last system equations of the previous page, we get:

At the end, we get:

Minimum Steel Reinforcement

Now we can try to develop some calculations by considering a real value as a function of the typical properties of our plate (e.g., the value of z). On the left, we have the simplified formulas to get, under particular conditions, the yielding moment as a function of the distance. To minimize the quantity of the area of reinforcement (usually for geometric conditions of the reinforcement, we have...

different distance dx wrt dy):

Introducing:

Developing the calculations:

POSITIVE-MOMENT FIELDS:

What happen if we start with positive- moment field?

(POSITIVE MOMENT FIELDS: mean that yielding resisting moment > 0 and in the same way Muy)

By substituting:

Only for positive moment fields, the WOOD ARMER DESIGN EQUATIONS are:

Here the yielding condition in terms of bending in x direction = acting bending moment in x direction + absolute value of the torque moment. ( In the same way in y direction).

NEGATIVE- AND MIXED-MOMENT FIELDS:

We have problems in this case (Johansen criterion not working).

WOOD-ARMER DESIGN EQUATIONS:

Here are summarized the design equations for bottom reinforcement (and the top one) by taking into account that the sum of acting bending moments are positive.

Is necessary to have the total distribution of bending moment on the plate to apply this criteria. Another problem in RC slabs is that if we have the compression is not necessary to put reinforcement. But when we time of

The concrete, during the lifetime of the plate, if we don't take into account the behavior at infinite steel in compression, we have a change of n.a. position and a decreasing of zx or zy. Also, this increases the deformability of our plate. If we also put reinforcement in compression, the reinforcement is not subject to creep and we don't have a change in n.a. of the C-S of our plate. So we must consider the dimension of reinforcement in terms of resistance, but there is a possibility to introduce reinforcement for serviceability limit state.

LEZIONE 07/10/22

KINKING OF REINFORCING STEEL BARS

We consider the behavior of the bar which crosses the yield line. Johansen criterion doesn't take into account kinking. Instead, by considering the wood modification of the Johansen yield criteria, we have this example:

The angle between the yield line and the vertical direction is alpha. The hypothesis is that the bar crosses the yield line and bends completely. The bend opening is exaggerated with respect to the diameter phi. By

Introducing those hypo for wood we obtain the kinking.

INFLUENCE OF BAR KINKING:

The difference between wood method and real one is 41%. The real contribute of kinking is 18% wrt the Johansen's one.

JOHANSEN'S YIELD SURFACE:

Here we can see 3D with axis Mx, My, Mxy we can represent in terms of surface the yield surface: Resistant bending moments in x,y directions. For isotropic slabs all resistant plastic moments are: Reinforcement must be yielded to ensure a good behavior.

LIMIT ANALYSIS OF RC PLATES:

BASIC HYPOTHESIS:

Here we have a material with symmetrical behavior. Ideal mean that we don't have elastic behavior (we have immediately plastic stress which is then constant for infinite). If we apply the different type of elastic behavior 1,2 (plastic limit is the same), perfectly rigid plastic behavior on the structure (supported beam on extremities and fixed point on the remaining part), we have the behavior illustrated in figure. In rigid plastic terms, we start from zero and we get

The curve on the right. When we consider 1,2 we arrive to aλc. same value of ADMISSIBLE STRESS AND STRAIN FIELDS:

We want to introduce concepts on limit analysis theorems. We have:

With statically admissible stress field we can define a statically admissible load multiplier. Also, for kinematic:

LIMIT ANALYSIS THEOREMS:

Under the previous conditions we can introduce:

In this way we define 2 limits (upper and lower limit). Under the previous conditions, we can also introduce UNIQUENESS THEOREM:

OPTIMAL SOLUTION:

Here we see in terms of curves the definition of the previous concepts. We have the kinematic or the upper bound λ-solution and the lower solution and λc the exact solution

EXAMPLE OF APPLICATIONS OF THE BEAM:

At center we see the plastic flow function of theta. The result is:

When we consider this situation we have problems in elastic terms of the unknown of a statically determinate structure with 1 unknown. To solve this problem in terms of elastic terms could be consider infinite

value of the unknown. We can also consider the unknown value =0. By doing this we have our structure statically determinate. Here we have the same shear at the end and by increasing the multiplier lambda we can develop the plastic moment at the mid span. The max acting moment at the mid span if C= 0 is FL/4 (simple supported beam). The multiplier is the value that transform the max acting moment in plastic moment (then we will have collapse of our structure). Instead if we solve our structure by compute the unknown in the fixed point we can develop this type of solution: (we can also develop a plastic bending moment in this case) By introducing the solution of unknown restrained at the end we get:

For the kinematic approach we have to consider the work due to plastic. In this case we have some kinematic conditions in which we introduce the 1 plastic hinge under the applied load and a 2 one for ex. at a distance l/4. The plastic work is given by:

By enforcing the equivalence between the 2 works we

get:Considering another kinematic solution, for ex. a plastic hinge in a fixed point and a one under a load we get:

PLASTIC HINGES IN RC BEAMS:

We have a real problem (for ex meana RC slab in plastic region). We must be able to define the plastic rotation. This mean that the single point in which we put the rigid plastic point in the hinge is the point of maximum curvature:

Note that the curvature is a quantity defined for HOMOGENEOUS MATERIALS !st After the 1 elastic branch we have a decreasing of bending moment and then a 2 branch until a change of shape. From branch 1’-2 we see that our bar has low reinforcement ratio, cause we change curvature very fast.

GENERALIZED PLASTIC HINGES IN RC PLATES:

To apply plastic behavior also to our plates we have to introduce some definitions forex the generalized plastic hinges in RC plates. We have that:

We have rotations concentrated in the yield line.

LEZIONE 17/10/22

MEMBRANE ACTIONS IN RC PLATES:

At ULS at failure we have also some additional mechanism

strength is the ratio of stress in compression and the maximum stress.

LIMIT ANALYSIS OF RC STRUCTURES:

Here we can compare the rigid plastic model for limit analysis used for concrete and steel. We can see the fundamental hypo that concrete don’t behave in tension, don’t have strength in tension, we have only rigid plastic behavior in tension.

For steel is different cause steel show a symmetric behavior in compression and tension but to have the symmetric behavior is necessary to avoid INSTABILITY IN COMPRESSION. To do this this mean that we must have a correct quantity of STIRRUPS.

EFFECTIVENESS FACTOR FOR BENDING FAILURE:

We have a dependency of effectiveness factor from the mechanical properties of concrete and steel: Putting this in excel we get the minimum value of effectiveness factor (EF). We remember that the max value of resisting bending moment depend on EF.

Also the geometric ration between area of concrete and area of steel must ensure the complete yielding of

thereinforcement (low percentage to ensure a ductile behavior):

BENDING STRENGTH: RIGID PLASTIC MODEL:

For a lot of application the computation of EF is not necessary cause we started by computing bendingmoment in rigid plastic model. Also we have defined the stress block on right:

In this picture y (length of slab) =1. So dx * 1 = area of the CS.

Enforcing the hypo that STRAIN CONCRETE DEFORMATION IS UNLIMITED (also for steel)

By computation we define the dimension in percentage of the n.a. from the top multiplying dx:

We get the resisting bending moment function of the steel yielding strength in design, by introducing the EF:stOmega sx = mechanical percentage of the reinforcement. This is the 1 formula depending on rigid plasticmodel.

EXPERIMENTAL VALIDATION

If we compare in terms of a dimensional bending resisting moment we get the formula in the middle of thegraph (function of a quadratic function of mechanical ratio of reinforcement):

The points represent the experimental results.

BENDING

STRENGTH: NON LINEAR MODEL:

We can also try to develop the same analysis of bending strength in NON LINEAR MODEL. In non linear, we don't need EF.model. In this case, we define a maximum strain for concrete and also introduce some conditions for a maximum strain for steel. We start with PABABOLA RECTANGULAR APPROACH, so we have a stress block which is a parabola:

With the same equilibrium conditions, we can write in terms of resultants in tension and compression to get the position of the neutral axis:

By developing the equilibrium equation at rotation, we get the MAXIMUM RESISTANT BENDING MOMENT.

NON LINEAR VS LIMIT ANALYSIS IN BENDING: