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Description of the collapsible item


Theoretical background example

Check of components according to IS 800

Bolts, preloaded bolts, and welds are checked $34 USD and $44 USD according to IS 800: Indian Standard, General Construction in Steel, Code of Practice (Third Revision). Concrete in bearing is designed according to IS 456: Indian Standard, Plain and Reinforced Concrete, Code of Practice (Fourth revision).


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Sample projects

 

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Plates

The resulting equivalent stress (HMH, von Mises) and plastic strain are calculated on plates. When the design yield strength, $f_y / \gamma_{m0}$ (IS:800, Cl. 5.4.1), on the bilinear material diagram is reached, the check of the equivalent plastic strain is performed. The limit value of 5 % is suggested in Eurocode (EN 1993-1-5 App. C, Par. C8, Note 1). This value can be modified in Code setup but verification studies were made for this recommended value.

Plate element is divided into 5 layers and elastic/plastic behaviour is investigated in each of them. The program shows the worst result from all of them.

Stress may be a little bit higher than design yield strength. The reason is the slight inclination of the plastic branch of the stress-strain diagram, which is used in the analysis to improve the stability of the calculation.

Welds

Butt welds

The verification of full penetration butt welds is not carried out, as it is assumed to have the same resistance as that of profile, as long as the parent material for the butt weld is superior to that of the profile (IS 800:2007, 10.5.7.1.2).

Fillet welds

Fillet welds are checked according to IS~800, Cl.~10.5.10.1.1:

\[ f_e = \sqrt{f_a^2 + 3q^2} \le f_{wd} = \frac{f_u}{\sqrt{3} \gamma_{mw}} \]

where:

  • $f_e$ -- equivalent stress in weld
  • $f_a$ -- normal stresses, compression or tension, due to axial force or bending moment
  • $q$ -- shear stress due to shear force or tension
  • $f_{wd}$ -- design strength of a fillet weld
  • $f_u$ -- smaller of the ultimate stress of the weld or of the parent metal; the ultimate strength of the weld electrode is assumed better than of the parent metal
  • $\gamma_{mw}$ -- partial safety factor for welds -- IS~800, Table~5; editable in Code setup

Bolts

The design strength of the bolt, $V_{dsb}$, as governed shear strength is given by IS~800, Cl.~10.3.3:

\[ V_{sb} \le V_{dsb} \]

where:

  • $V_{dsb} = V_{nsb}/\gamma_{mb}$ -- design shear capacity of a bolt
  • $V_{nsb} = \frac{f_{ub}}{\sqrt{3}} A_e$ -- nominal shear capacity of a bolt
  • $f_{ub}$ -- ultimate tensile strength of a bolt;
  • $A_e$ -- area for resisting shear; $A_e = A_n$ for shear plane intercepted by the threads, $A_e = A_s$ for the case where threads do not occur in shear plane
  • $A_n$ -- net tensile stress area of the bolt
  • $A_s$ -- cross-section area at the shank
  • $\gamma_{mb} = 1.25$ -- partial safety factor for bolts -- bearing type -- IS~800, Table~5; editable in Code setup

When the grip length of bolts $l_g$ (equal to the total thickness of the connected plates) is higher than $5d$, the design shear capacity $V_{dsb}$ is reduced by a factor $\beta_{lg}$ -- IS~800, Cl.~10.3.3.2:

\[ \beta_{lg} = \frac{8}{3+l_g/d}  \]

Bearing capacity of bolts

The design bearing strength of a bolt on any plate, as governed by bearing is given by IS~800, Cl.~10.3.4:

\[ V_{sb} \le V_{dpb} \]

where:

  • $V_{dpb} = V_{npb} / \gamma_{mb}$ -- design bearing strength of a bolt
  • $V_{npb} = 2.5 k_b d t f_u$ -- nominal bearing strength of a bolt
  • $k_b = \min \left \{ \frac{e}{3d_0}, \, \frac{p}{3d_0}-0.25, \, \frac{f_{ub}}{f_u}, \, 1.0 \right \}$ -- factor for joint geometry and material strength
  • $e$ -- end distance of the fastener along bearing direction
  • $p$ -- pitch distance of the fastener along bearing direction
  • $f_{ub}$ -- ultimate tensile strength of the bolt
  • $f_u$ -- ultimate tensile strength of the plate
  • $d$ -- nominal diameter of the bolt
  • $d_0$ -- diameter of bolt hole
  • $t$ -- plate thickness
  • $\gamma_{mb} = 1.25$ -- partial safety factor for bolts -- bearing type -- IS~800, Table~5; editable in Code setup

According to IS~800, Cl.~10.3.3.3, the design shear capacity of bolts carrying shear through a packing plate with the thickness $t_{pk} \ge 6$~mm shall be decreased by a factor:

\[ \beta_{pk} = (1-0.0125 t_{pk}) \]

Each shear plane is checked separately and the worst result is shown.

Bearing capacity of bolts

The design bearing strength of a bolt on any plate, as governed by bearing is given by IS~800, Cl.~10.3.4:

\[ V_{sb} \le V_{dpb} \]








IDEA StatiCa Version 10.1 will not overwrite the 10.0 installation. You will have both on your PC. All future versions of IDEA StatiCa will behave the same way to ensure a smooth transition between versions.

From September 2019, IDEA StatiCa no longer supports installation on Windows 7.

Version 10.1 does not require a license update or any other action from the end-users, just download the latest setup.

Single licenses – IDEA StatiCa 10.1 can be downloaded on all PCs of the one user (sharing the license with other users is forbidden according to EULA – no change from previous versions)

Network licenses – IDEA StatiCa 10.1 has to be deployed on all end-machines and multiple users can access it based on purchased number of seats

IDEA StatiCa Version 10.1 will not overwrite the 10.0 installation. You will have both on your PC. All future versions of IDEA StatiCa will behave the same way to ensure a smooth transition between versions.

From September 2019, IDEA StatiCa no longer supports installation on Windows 7.

Version 10.1 does not require a license update or any other action from the end-users, just download the latest setup.

Single licenses – IDEA StatiCa 10.1 can be downloaded on all PCs of the one user (sharing the license with other users is forbidden according to EULA – no change from previous versions)

Network licenses – IDEA StatiCa 10.1 has to be deployed on all end-machines and multiple users can access it based on purchased number of seats

\[ t_s = \sqrt{2.5 w c^2 \gamma_{m0} / f_y} > t_f \]

The formula can be rewritten to determine the overlap with the assumption that \( w = 0.6 f_{ck} \):

\[ c = t_s \sqrt{\frac{f_y}{1.5 f_{ck} \gamma_{m0}}} \]

Ondrej Fridrich

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