Page 262 - Handbook of Structural Steel Connection Design and Details

Welded Joint Design and Production

Welded Joint Design and Production 247

σ 2 = 0

ε 2 = –.0003

σ 3 = 30 ksi
ε 1 = –.0003 Figure 3.28 Unit cube showing
ε 3 = +.001 applied stress from Fig. 3.27.
(Courtesy of The Lincoln
Electric Company.)
σ 1 = 0

Consider the unit cube in Fig. 3.28. It is an element of the beam
flange from point B in Fig. 3.27. The applied force due to the moment
is . There is no restraint against the longitudinal stress of 30 ksi in
3
the flange, so 0. Using Poisson’s ratio of 0.3 for steel,
1 2
Eqs. (3.1a) to (3.1c) yield the following for 30 ksi:
3
ε =+ 0 001 in/in
.
3
ε =− 0 0003 in/in
.
2
ε =− 0 0003 in/in
.
n
1
From Eqs. (3.2a) to (3.2c), it is found that
σ = 0 ksi
1
σ = 0 ksi
2
σ = 30 ksi
3
These stresses are plotted in Fig. 3.29 as a dotted circle. The larger
solid-line circle is for a stress of 70 ksi or ultimate tensile stress.

τ 1–3 = 35

27.5
σ c1

σ 3 = 70
σ 1 σ 2
Figure 3.29 A plot of the tensile
stress and shear stress from
Fig. 3.27. (Courtesy of The
Lincoln Electric Company.)

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