Design a welded connection for an MC 9 × 23.9 of A572 Grade 50 steel connected to a 3 8 -inch-thick gusset plate. The gusset plate is A36 steel. Show your results on a sketch, complete with dimensions. a . Use LRFD. b . Use ASD.

(a)

The design of a welded connection using Load and Resistance Factor Design (LRFD).

Given:

 Figure-(1)

Concept Used:

Write the expression for the maximum size of the weld.

wmax=t−116 in     ...... (I)

Here, the maximum size of the weld is wmax, the thickness of the member is t.

Write the expression for the minimum size of the weld.

wmin=14 in

Write the expression for the required length of the weld in longitudinal direction.

Ll=Lreq−Lt2     ...... (II)

Here, the length of the weld in the longitudinal direction is Ll, the required length of the weld is Lreq, the length of the weld in the transverse direction is Lt.

Write the expression for the required length of the weld.

Lreq=PuϕRn     ...... (III)

Here, the factored load is Pu, the load reduction factor is ϕ, the design strength of the connection is Rn.

Write the expression for the factored load.

Pu=1.2D+1.6L     ...... (IV)

Here, the dead load is D, the live load is L.

Write the nominal weld strength equation per unit length.

wn=ϕRnn     ...... (V)

Here, the nominal weld strength is wn, the nominal strength of the weld per unit length is Rnn.

Write the expression for the nominal strength of the weld per unit length.

Rnn=0.707w(0.6FEXX)     ...... (VI)

Here, the weld size is w, the strength of the weld material is FEXX.

Write the expression for the shear strength of the base metal in yielding.

Puy=1.0(0.6Fyt)     ...... (VII)

Here, the shear strength of the base metal in yielding is Puy, the strength of the weld in yielding is Fy,

Write the expression for the shear strength of the base metal in rupture.

Pur=0.75(0.6Fut)     ...... (VIII)

Here, the shear strength of the base metal in rupture is Pur, the strength of the weld in rupture is Fu.

Write the expression for the longitudinal length of the weld.

Ll=Pl2wl     ...... (IX)

Here, the capacity of weld in longitudinal direction is Pl and the size of the weld in longitudinal direction is wl.

Write the expression for the capacity of the longitudinal weld.

Pl=Pu−Ltwt     ...... (X)

Here, the load capacity of the weld is wt.

Write the expression for the load capacity of the weld in longitudinal direction.

wl=0.85ϕRn     ....... (XI)

Here, the load capacity of the weld in longitudinal direction.

Write the expression for the load capacity of the weld in transverse direction.

wt=1.5ϕRn     ...... (XII)

Here, the load capacity of the weld in transverse direction.

Check the strength of the weld in block shear.

Write the expression for the block shear.

Rreq=0.6FyAnv+UbsFuAnt     ...... (XIII)

Here, the required strength of the weld is Rreq, the gross area along the longitudinal direction is Agv, the net area along the transverse direction is Ant, the load reduction factor is Ubs.

Write the expression for the gross area of the weld.

Agv=Anv=2tLl     ...... (XIV)

Write the expression for the net area of the weld.

Ant=tLt     ...... (XV)

Calculation:

Calculate the maximum size of the weld.

Substitute 38 in for t in the Equation (I).

wmax=38 in−116 in=516 in

Calculate the minimum size of the weld.

wmin=316 in

Hence, adopt weld of thickness 316 in of E70XX material.

Calculate the factored load.

Substitute 48 kips for D and 120 kips for L in the Equation (IV).

Pu=(1.2×48 kips)+(1.6×120 kips)=249.6 kips

Calculate the nominal strength of the weld per unit length.

Substitute 14 in for w, 70 ksi for FEXX in the Equation (VI).

Rnn=0.707×14×(0.6×70 ksi)=7.424 kips/in

Calculate the nominal weld strength equation per unit length.

Substitute 0.75 for ϕ, 7.424 kips/in for Rnn in the Equation (V).

wn=0.75×7.424 kips/in=5.568 kips/in

Calculate the shear strength of the base metal in yielding.

Substitute 36 ksi for Fy, 38 in for t in the Equation (VII).

Puy=1.0×(0.6×36 ksi×38 in)=8.1 kips/in

Calculate the shear strength of the base metal in rupture.

Substitute 50 ksi for Fy, 38 in for t in the Equation (VIII).

Pur=0.75×(0.6×50 ksi×38 in)=8.438 kips/in

The strength of the base metal is the minimum of the shear strength in yielding and rupture.

Thus, the strength of the base metal is 8

(b)

The design of a welded connection using Allowable Strength Design (ASD).