Solved Problems

A built-up plate girder has a web depth $h = 60 \text{ in}$ and web thickness $t_w = 0.375 \text{ in}$. Transverse stiffeners are spaced at $a = 60 \text{ in}$. The steel has $F_y = 50 \text{ ksi}$. The web shear coefficient $C_v$ has been calculated as $0.40$. Calculate the nominal shear strength ($V_n$) assuming Tension Field Action is permitted.

A structural design firm is bidding on a highway overpass project utilizing long-span steel plate girders.

Scenario: To win the bid, they need to minimize the total weight (and cost) of the steel. They initially specify a relatively thick web that requires no stiffeners.

Solution: A value-engineering review shows that while a thick web saves labor (no stiffeners to weld), it uses an immense amount of steel over a 150-foot span.

The engineers redesign the girder using a very thin, "slender" web. By doing so, they drastically reduce the weight of the steel. However, the slender web will buckle under shear forces long before yielding. To compensate, they design transverse intermediate stiffeners. These stiffeners allow the web to develop post-buckling Tension Field Action, essentially acting like the diagonal tension members in a truss. The added labor cost of welding the stiffeners is far outweighed by the massive savings in raw steel weight.

During the erection of a deep plate girder on a bridge pier, an issue occurs.

Scenario: The girder has a slender web designed for tension field action. When the girder is lowered onto its bearing pad at the support pier, the immense localized reaction force causes the web to begin crippling (buckling locally).

Solution: Because plate girders often have highly slender webs ($h/t_w$ is very large), they are extremely vulnerable to concentrated forces, especially at the reactions (supports).

The design must include bearing stiffeners at all support locations and points of highly concentrated loads. These stiffeners are full-depth plates welded to both sides of the web and fitted tightly against the flanges. They act essentially as short columns, transferring the massive concentrated reaction force safely from the bottom flange up into the full depth of the web, preventing web crippling.

Verify if the web of a built-up plate girder is classified as "slender" for shear. The girder has a web depth $h = 72 \text{ in}$ and web thickness $t_w = 0.50 \text{ in}$. The steel yield strength is $F_y = 50 \text{ ksi}$. Transverse stiffeners are spaced at $a = 108 \text{ in}$.

Calculate the design flexural strength ($\phi_b M_n$) of a built-up plate girder based on the Tension Flange Yielding limit state. The section is a singly symmetric I-shape. $F_y = 50 \text{ ksi}$. The section modulus with respect to the tension flange is $S_{xt} = 1200 \text{ in}^3$. The web slenderness $h_c/t_w = 160$, and the limiting compactness ratio for the web $\lambda_{pw} = 137$. Because $h_c/t_w > \lambda_{pw}$, the web is slender for flexure and a bending strength reduction factor $R_{pg}$ must be calculated. The calculated $R_{pg} = 0.92$.

A symmetrical built-up plate girder requires continuous fillet welds connecting the flanges to the web. The maximum factored shear force at the support is $V_u = 350 \text{ kips}$. The moment of inertia of the entire cross-section is $I = 25000 \text{ in}^4$. The flange dimensions are $b_f = 16 \text{ in}$ and $t_f = 1.25 \text{ in}$, with the centroid of the top flange located $y = 35.625 \text{ in}$ from the neutral axis. Determine the required horizontal shear flow ($q$).

Determine the nominal shear strength ($V_n$) of a plate girder with an unstiffened web. The web dimensions are $h = 48 \text{ in}$ and $t_w = 0.5 \text{ in}$. $F_y = 50 \text{ ksi}$. Because it is unstiffened, Tension Field Action is not permitted, and $k_v = 5.34$.

A plate girder has a web depth $h = 80 \text{ in}$ and $t_w = 0.4 \text{ in}$. Stiffeners are spaced at $a = 280 \text{ in}$. Determine if Tension Field Action (TFA) can be utilized to calculate the shear strength. $F_y = 50 \text{ ksi}$.