CE Board Exam Randomizer

Question 1

Tension Members

Load Combinations for Load and Resistance Factor Design (LRFD)

Combination 1: $1.4D$
Combination 2: $1.2D + 1.6L + 0.5(L_r \text{ or } S \text{ or } R)$
Combination 3: $1.2D + 1.6(L_r \text{ or } S \text{ or } R) + (L \text{ or } 0.5W)$
Combination 4: $1.2D + 1.0W + L + 0.5(L_r \text{ or } S \text{ or } R)$
Combination 5: $1.2D + 1.0E + L + 0.2S$
Combination 6: $0.9D + 1.0W$
Combination 7: $0.9D + 1.0E$

Where:
$D$ = dead load
$L$ = live load due to occupancy
$L_r$ = roof live load
$S$ = snow load
$R$ = rain or ice load
$W$ = wind load
$E$ = earthquake (seismic load)

Load Combinations for Allowable Strength Design (ASD)

Combination 1: $D$
Combination 2: $D + L$
Combination 3: $D + (L_r \text{ or } S \text{ or } R)$
Combination 4: $D + 0.75L + 0.75(L_r \text{ or } S \text{ or } R)$
Combination 5: $D \pm (0.6W \text{ or } 0.7E)$
Combination 6a: $D + 0.75L + 0.75(0.6W) + 0.75(L_r \text{ or } S \text{ or } R)$
Combination 6b: $D + 0.75L \pm 0.75(0.7E) + 0.75S$
Combinations 7 & 8: $0.6D \pm (0.6W \text{ or } 0.7E)$

Where:
$D$ = dead load
$L$ = live load due to occupancy
$L_r$ = roof live load
$S$ = snow load
$R$ = rain or ice load
$W$ = wind load
$E$ = earthquake (seismic load)

NSCP 2010/2015 — Tensile Strength

A tension member is a structural element that carries load primarily through pulling forces along its length. Unlike compression members that resist crushing, tension members resist stretching and are designed to handle axial tensile stresses. Common examples include the diagonal and vertical ties in trusses, bracing in towers, and cables in suspension bridges. Typically made of steel rods, plates, or cables, these members are essential in ensuring stability and strength by effectively transferring loads without buckling.

General strength format

$$ R_u \le \phi R_n \quad \text{(LRFD)} $$ $$ R_a \le \frac{R_n}{\Omega} \quad \text{(ASD)} $$

1) Tensile yielding in the gross section

$$P_n = F_y A_g$$ $$\phi = 0.90 \;(\text{LRFD}), \quad \Omega = 1.67 \;(\text{ASD})$$ $$P_u = 0.90F_yA_g, \qquad P_a = 0.60F_yA_g$$

2) Tensile rupture in the net section

$$P_n = F_u A_e$$ $$\phi = 0.75 \;(\text{LRFD}), \quad \Omega = 2.00 \;(\text{ASD})$$ $$P_u = 0.75F_uA_e, \qquad P_a = 0.50F_uA_e$$

Effective Net Area

The effective area of tension members shall be determined as:

$$A_e = A_n \, U$$

where: $U$ = shear lag factor
$A_n=A_g-A_h; A_n\le 0.85A_g$ for design of splice plates with holes.

$$ A_n = A_g-\sum A_h, \qquad A_h=d_h\,t\,n $$ $$ d_h=d_b+\text{standard hole allowance} $$ $$ A_e=U A_n \le 0.85A_g \quad \text{for splice plates with holes only} $$

Standard Hole Adjustment — If the nominal hole diameter is given, add:
1.6 mm (2001) or 2 mm (2010/2015) to obtain the effective/standard hole diameter.

The figure of the cross section at the left describes the gross area, while that on the right illustrates the net area. The net area subtracts the area of the holes from the gross area.

Block Shear Strength

$$R_n = 0.6\,F_u A_{nv} + U_{bs} F_u A_{nt} \;\;\le\;\; 0.6\,F_y A_{gv} + U_{bs} F_u A_{nt}$$ $$\phi = 0.75 \;(\text{LRFD}), \quad \Omega = 2.00 \;(\text{ASD})$$

Sample Area Computations

Strength of Fillet Welds

$$R_n = 0.60\,F_{exx}\,(0.707\,t\,L)$$ $$\phi = 0.75 \;(\text{LRFD}), \quad \Omega = 2.00 \;(\text{ASD})$$

Shear Lag Factors

Use $U$ to reduce the net area when the tension force is not transmitted through every element of the section.

$$ U=1.0 \quad \text{when load is transmitted directly to all elements} $$ $$ U=1-\frac{\bar{x}}{L} \quad \text{for connections to some, but not all, elements} $$

Plates with longitudinal welds only

$$ L \ge 2w: \; U=1.00 $$ $$ 1.5w \le L < 2w: \; U=0.87 $$ $$ w \le L < 1.5w: \; U=0.75 $$

Common HSS shear-lag checks

$$ \text{Round HSS: } L \ge 1.3D \Rightarrow U=1.0, \qquad D \le L < 1.3D \Rightarrow U=1-\frac{\bar{x}}{L} $$ $$ \text{Rectangular HSS: } L \ge H \Rightarrow U=1-\frac{\bar{x}}{L}, \qquad \bar{x}=\frac{B^2+2BH}{4(B+H)} $$

Question 2

Question Bank: q7

PSAD - Steel Design / Tension Members / Engr. Janclyde Espinosa (Clidez)

CE Board May 2010
A 76mmx76mmx6mm angular section is welded to a gusset plate having a thickness of 8mm. L1=125mm and L2=65mm. The angular section has a cross sectional area of 929mm2 while Fy=248MPa and Fu=400MPa.
Allowable stresses:
Allowable tensile stress of the gross area = 0.6Fy
Allowable tensile stress of the net area = 0.5Fu
Allowable shear stress of the net area = 0.3Fu

Determine the value of the tensile force P based on the gross area.

138.24kN
185.8kN
859.47kN
222.96kN

Determine the value of the tensile force P based on the net area, assuming a strength reduction coefficient of 0.85.

157.93kN
185.8kN
189.516kN
117.5kN

Determine the value of the tensile force P based on block shear in the gusset plate along the weld.

304kN
233.72kN
376.96kN
276.89kN

Solution pending in psadquestions/q7.json.