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Welded Joint Design and Production

                    246   Chapter Three

                    Four conditions are required for ductility:

                    1. There must be a shear stress ( ) component resulting from the
                       given load condition.
                    2. This shear stress must exceed its critical value by a reasonable
                       amount. The more it exceeds this value, the greater will be the
                       resulting ductility.
                    3. The plastic shear strain resulting from this shear stress must act
                       in the direction which will relieve the particular stress which can
                       cause cracking.
                    4. There must be sufficient unrestrained length of the member to
                       permit “necking down.”

                      If conditions (1) and (2) are not met, there will be no apparent duc-
                    tility and no yielding. The stress will simply build up to the ultimate
                    tensile strength with little or no plastic energy absorbed. This condi-
                    tion is called brittle failure.
                      Figure 3.27 shows two regions in question. Point A is at the weld
                    joining the beam flange to the face of the column flange. Here there is
                    restraint against strain (movement) across the width of the beam
                    flange (ε ) as well as through the thickness of the beam flange (ε ).
                            1                                                      2
                    Point B is along the length of the beam flange away from the connect-
                    ing weld. There is no restraint across the width of the flange or
                    through its thickness.
                      In most strength of materials texts, the following equations can be
                    found:
                                          ε =  1  ( σ − µσ −  µσ )
                                           3       3    2    1                  (3.1a)
                                               E
                                           ε =  1  − ( µσ  + σ  − µσ  )         (3.1b)
                                            2        3   2    1
                                               E
                                           ε =  1  − (  µσ − µσ +  σ )          (3.1c)
                                           1         3    2   1
                                               E
                    It can be shown that

                                             E[ µε +  µε + ( 1 −  µ ε ) ]
                                        σ =      3    2        1                (3.2a)
                                          1
                                                 ( 1 + )(  −  µ)
                                                    µ 1 2
                                             E[ µε + ( 1 − µ ε +  µε ]
                                                         )
                                        σ =      3        2    1                (3.2b)
                                                        −
                                          2
                                                     )
                                                 ( 1 +  µ 1 2 µ)
                                                      (
                                                    )
                                             E[( 1 − µ ε + µε +  µε ]
                                        σ =          3    2    1                (3.2c)
                                                        −
                                          3
                                                      (
                                                 ( 1 +  µ 1 2 µ)
                                                     )
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