Chapter 10, Problem 10.7.8P

To determine

(a)

The design of a plate girder for the given conditions, the selection of girder cross section and the required spacing of intermediate stiffeners by using LRFD.

Answer to Problem 10.7.8P

$\frac{1}{2}\times 107$web and $1\frac{1}{2}\times 26$flanges

Four panels spaced at 58.25in.

Explanation of Solution

Given:

Span length $=100ft$

Uniformly distributed live load $=0.\text{7 kips}/\text{ft}$

Superimposed dead load $=0.3\text{ kips}/\text{ft}$

Concentrated dead load $\text{= 50 kips}$

Concentrated live load $=150\text{ kips}$

$\text{A572}$ Grade $50$ steel

Formula used:

$A_f=\frac{M_u/{\varphi }_b}{h{F}_y}-\frac{A_w}{6}$

$M_u/{\varphi }_b$ is required nominal flexural strength

$A_w$ is web area

$A_f$ is flange area

h is the depth of web

Calculation:

Assume a girder weight of $\text{450 lb/ft}$.

Determine the factored loads:

$w_u=1.2w_D+1.6w_L$

$w_D$ is uniformly distributed dead load

$w_L$ is uniformly distributed live load

$w_u=1.2w_D+1.6w_L=1.2\left(0.3\right)+1.6\left(0.7\right)=1.48kips/ft$

$P_u=1.2P_D+1.6P_L$

$P_D$ is concentrated dead load

$P_L$ is concentrated live load

$P_u=1.2P_D+1.6P_L=1.2\left(50\right)+1.6\left(150\right)=300kips$

The factored moment and shear are

$\begin{array}{l}{M}_u=524\left(50\right)-1.48\left(\frac{{50}^2}{2}\right)-300\left(25\right)=16850ft-kips\\ V_u=V_L=524\text{}kips\end{array}$

Determine the overall depth:

$\begin{array}{l}\frac{Spanlength}{10}=\frac{100\left(12\right)}{10}=120in.\\ \frac{Spanlength}{12}=\frac{100\left(12\right)}{12}=100in.\end{array}$

Use the maximum permissible depth of 110 in.

Try $t_f=1.5in.$

$h=110-2\left(1.5\right)=107in.$

To determine the web thickness, first examine the limiting values of $\frac{h}{t_w}$.

For $\frac{a}{h}\le 1.5$,

${\left(\frac{h}{t_w}\right)}_{\max }=12.0\sqrt{\frac{E}{F_y}}=12.0\sqrt{\frac{29000}{50}}=288.998$

Minimum $t_w=\frac{107}{288.998}=0.370in.$

For $\frac{a}{h}>1.5$,

${\left(\frac{h}{t_w}\right)}_{\max }=0.40\left(\frac{E}{F_y}\right)=0.40\left(\frac{29000}{50}\right)=232$

Minimum $t_w=\frac{107}{232}=0.46in.$

Try a $\frac{1}{2}in.\times 107in.$ web.

$A_w=0.5\left(107\right)=53.5in{.}^2$

Determine whether the web is slender:

$\frac{h}{t_w}=\frac{107}{0.5}=214$

$5.70\sqrt{\frac{E}{F_y}}=5.70\sqrt{\frac{29000}{50}}=137.274<214$

Therefore, the web is slender.

Estimate required flange size:

$\begin{array}{l}{A}_f=\frac{M_u}{0.9h{F}_y}-\frac{A_w}{6}=\frac{16850\left(12\right)}{0.9\left(107\right)\left(50\right)}-\frac{53.5}{6}=33.077in{.}^2\\ b_f\ge \frac{33.077}{1.5}=22.05in.\end{array}$

Try a $1\frac{1}{2}-in.\times 26-in.$ flange,

$A_f=1.5\times 26=39in{.}^2$

Girder weight = $\left[53.5+2\left(39\right)\right]\left(\frac{0.490}{144}\right)=0.447kips<ft<0.450kips/ft$ (OK)

$\begin{array}{l}{I}_x=\frac{t_w{h}^3}{12}+2A_f{\left(\frac{h+t_f}{2}\right)}^2=\frac{0.5{\left(107\right)}^3}{12}+2\left(39\right){\left(\frac{107+1.5}{2}\right)}^2\\ =280602.33in{.}^4\\ S_x=\frac{I_x}{c}=\frac{I_x}{\left(h/2+t_f\right)}=\frac{280602.33}{\left(107/2+1.5\right)}=5101.86in{.}^3\end{array}$

Compression flange:

Check flange local buckling (FLB):

$\begin{array}{l}\lambda =\frac{b_f}{2t_f}=\frac{26}{2\left(1.5\right)}=8.67\\ {\lambda }_p=0.38\sqrt{\frac{E}{F_y}}=0.38\sqrt{\frac{29000}{50}}=9.15\end{array}$

Since $\lambda <{\lambda }_p$, $F_{cr}=F_y=50ksi$

Compute the plate girder strength reduction factor:

$\begin{array}{l}{R}_{pg}=1-\frac{a_w}{1200+300a_w}\left(\frac{h_c}{t_w}-5.7\sqrt{\frac{E}{F_y}}\right)\le 1.0\\ a_w=\frac{h_c{t}_w}{b_{fc}{t}_{fc}}=\frac{107\left(0.5\right)}{26\left(1.5\right)}=1.37<10\\ R_{pg}=1-\frac{1.37}{1200+300\left(1.37\right)}\left(\frac{107}{0.5}-5.7\sqrt{\frac{29000}{50}}\right)=0.935<1.0\\ M_n=R_{pg}{F}_{cr}{S}_{xc}=0.935\left(50\right)\left(5101.86\right)=238448.716in.-kips\end{array}$

${\varphi }_b{M}_n=\frac{0.90\left(238448.716\right)}{12}=17883.654ft-kips>16850ft-kips$ (OK)

Try a $\frac{1}{2}\times 107$ web and $1\frac{1}{2}\times 26$ flanges.

Shear: At left end (end panel),

Required $\frac{{\varphi }_v{V}_n}{A_w}=\frac{524}{53.5}=9.8ksi$,

$\frac{h}{t_w}=214$

From Table 3-17a in the Manual, $\frac{a}{h}\approx 0.623$ by interpolation

$a=0.623h=0.65\left(107\right)=66.7in.$

Use $a=67in.$

This spacing will apply for the remaining distance to the centerline of the girder. This distance is

$25\times 12-67=233in.$

For a spacing a of 67 in., the number of panels is $\frac{233}{67}=3.5$

Use 4 panels at $\frac{233}{4}=58.25in.$

At $100/4=25ft$ from the left end (to the right of the concentrated load),

$V_u=524-1.48\left(25\right)-300=187kips$

Required $\frac{{\varphi }_v{V}_n}{A_w}=\frac{187}{53.5}=3.5ksi$,

For $\frac{a}{h}=\frac{25\times 12}{107}=2.8,\frac{{\varphi }_v{V}_n}{A_w}<3$

Therefore, stiffeners are needed in middle $\frac{1}{4}$.

Conclusion:

Therefore, Use a $\frac{1}{2}\times 107$ web and $1\frac{1}{2}\times 26$ flanges and four panels spaced at $58.25in$.

To determine

(b)

The size of intermediate and bearing stiffeners.

Answer to Problem 10.7.8P

2 PL $\frac{1}{2}\times 6\times 8\text{'}-8\text{'}\text{'}$for intermediate stiffeners

2 PL $\frac{5}{8}\times 8$with 1-in. cutouts for bearing stiffeners

Explanation of Solution

Given:

Span length $=100ft$

Uniformly distributed live load $=0.\text{7 kips}/\text{ft}$

Superimposed dead load $=0.3\text{ kips}/\text{ft}$

Concentrated dead load $\text{= 50 kips}$

Concentrated live load $=150\text{ kips}$

$\text{A572}$ Grade $50$ steel

Calculation:

Intermediate stiffener size:

${\left(\frac{b}{t}\right)}_{st}\le 0.56\sqrt{\frac{E}{F_y}}$

Available width:

$\frac{b_f-t_w}{2}=\frac{26-0.5}{2}=12.75in.$

Try $b=6in.$

$t\ge \frac{b}{0.56\sqrt{\frac{E}{F_y}}}=\frac{6}{0.56\sqrt{\frac{29000}{50}}}=0.4449$

To determine the required moment of inertia, use the conservative approximation from the User Note in AISC G2.3:

$\begin{array}{l}{I}_{st}=I_{st1}\\ I_{st}=\frac{h^4{\rho }_{st}^{1.3}}{40}{\left(\frac{F_{yw}}{E}\right)}^{1.5}\\ {\rho }_{st}=\max \left\{\begin{array}{l}{F}_{yw}/F_{yw}\\ 1\end{array}\right\}=1\\ I_{st}=\frac{{\left(107\right)}^4{\left(1\right)}^{1.3}}{40}{\left(\frac{50}{29000}\right)}^{1.5}=234.603in{.}^4\end{array}$

Try two $\frac{1}{2}\times 6$ plates:

$I_{st}\approx \frac{\left(3/8\right){\left(6+0.5+6\right)}^3}{12}=61in{.}^4$

Length: From Figure 10.9 in the textbook (Steel design),

$\begin{array}{l}c\ge 4t_w=4\left(0.5\right)=2in.\\ c\le 6t_w=6\left(0.5\right)=3in.\end{array}$

Assume a flange-to-web weld size of $w=\frac{3}{16}in.$ (minimum size) and $c=3in.$

Length = $h-w-c=107-\frac{3}{16}-3=103.8in.$

$c=107-104-\frac{3}{16}=2.8125in.$ (OK)

Use two PL $\frac{1}{2}\times 6\times 8\text{'}-8\text{'}\text{'}$ for intermediate stiffeners.

Design the bearing stiffeners at the supports for a load of $V_L=524kips$

Maximum stiffener width = $\frac{b_f-t_w}{2}=\frac{26-0.5}{2}=12.75in.$

Try $b=8in.$

$t\ge \frac{b}{0.56\sqrt{\frac{E}{F_y}}}=\frac{8}{0.56\sqrt{\frac{29000}{50}}}=0.593in.$

Try two plates, $\frac{5}{8}\times 8in.$, with 1-in. cutouts.

Bearing strength:

$\begin{array}{l}{A}_{pb}=\left(8-1\right)\left(\frac{5}{8}\right)\times 2=8.75in{.}^2\\ R_n=1.8F_y{A}_{pb}=1.8\left(50\right)\left(8.75\right)=787.5kips\end{array}$

$\varphi R_n=0.75\left(787.5\right)=590.6kips>V_L=524kips$ (OK)

Compressive strength: The maximum permissible length of web is

$12t_w=12\left(0.5\right)=6in.$

Compute the radius of gyration about an axis along the middle of the web:

$\begin{array}{l}I=\frac{6{\left(0.5\right)}^3}{12}+2\left[\frac{\left(5/8\right)8^3}{12}+8\left(\frac{5}{8}\right){\left(4+\frac{1}{4}\right)}^2\right]=234in{.}^4\\ A=6\left(0.5\right)+2\left(8\right)\left(\frac{5}{8}\right)=13in{.}^2\\ r=\sqrt{\frac{I}{A}}=\sqrt{\frac{234}{13}}=4.243in.\end{array}$

Compute the compressive strength:

$\frac{KL}{r}=\frac{0.75\left(107\right)}{4.243}=18.9<25$

Therefore, $F_{cr}=F_y=50ksi$

${\varphi }_c{P}_n=0.90F_{cr}{A}_g=0.90\left(50\right)\left(13\right)=585kips>524kips$ (OK)

Use 2 PL $\frac{5}{8}\times 8$ with 1-in. cutouts for bearing stiffeners at the supports.

Because there is a large difference between the reactions and the interior concentrated loads, use 2 PL $\frac{5}{8}\times 8$ with 1-in. cutouts for the interior bearing stiffeners.

Conclusion:

Use two PL $\frac{1}{2}\times 6\times 8\text{'}-8\text{'}\text{'}$ for intermediate stiffeners,two PL $\frac{5}{8}\times 8$ with 1-in. cutouts for bearing stiffeners at the supports and two PL $\frac{5}{8}\times 8$ with 1-in. cutouts for the interiorbearing stiffeners.

To determine

(c)

The design of the all welds

Answer to Problem 10.7.8P

$\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$16$}\right.in$. continuous fillet welds for the first$20ft$

$\frac{3}{16}in.\times 1\frac{1}{2}in.$E70 fillet welds spaced at 3 in. c-c for the intermediate stiffeners

$\frac{3}{16}in.\times 1\frac{1}{2}in.$E70 fillet welds spaced at 5 in. c-c for bearing stiffener at support and at interior portion.

Explanation of Solution

Given:

Span length $=100ft$

Uniformly distributed live load $=0.\text{7 kips}/\text{ft}$

Superimposed dead load $=0.3\text{ kips}/\text{ft}$

Concentrated dead load $\text{= 50 kips}$

Concentrated live load $=150\text{ kips}$

$\text{A572}$ Grade $50$ steel

Calculation:

Design the flange-to-web welds.

The shear flow is

$Q=A_f\left(\frac{h}{2}+\frac{t_f}{2}\right)=\left(1.5\times 26\right)\left(\frac{107}{2}+\frac{1.5}{2}\right)=2115.75in{.}^3$

At the support,

$\frac{V_{u}Q}{I_x}=\frac{524\left(2115.75\right)}{280602.33}=3.95kips/in.$

Minimum weld size = 3/16 in. (AISC Table J2.4)

Minimum length = $4\left(\frac{3}{16}\right)=0.75in.<1.5in.$

Use 1.5 in.

Use E70 electrodes, $\varphi R_n=1.392D$ kips/in.

where D is weld size in sixteenths.

Try an $\frac{3}{16}-in.\times 1\frac{1}{2}-in.$ intermittent fillet welds.

For two welds,

Weld strength = $2\times 1.392\left(3\right)=8.352kips/in.$

Base metal shear yield strength (web plate controls) is

$0.4F_{y}t=0.4\left(50\right)\left(\frac{1}{2}\right)=10kips/in.$

Shear rupture strength is

$0.3F_{u}t=0.3\left(65\right)\left(\frac{1}{2}\right)=9.75kips/in.$

Weld strength controls.

For a 1.5-in. length,

$\varphi R_n=1.5\left(8.352\right)=12.53kips$

Required spacing:

$\begin{array}{l}\frac{\varphi R_n}{s}=\frac{V_{u}Q}{I_x}\\ \frac{12.53}{s}=3.95\\ s=3.172in.\end{array}$

Since this is less than twice the length of the weld, use a continuous weld.

For $s=2\left(1.5\right)=3in.$

$V_u=\frac{\varphi R_n{I}_x}{sQ}=\frac{12.53\times 280602.33}{3\left(2115.75\right)}=553.932kips$

This occurs when $\begin{array}{l}524-1.48x=553.93\\ x=20.22ft\end{array}$

Maximum clear spacing: From AISC E6,

$d\le 0.75\sqrt{\frac{E}{F_y}}{t}_f=0.75\sqrt{\frac{29000}{50}}\left(1.5\right)=27.1in.$ (or 12 in.; 12 in. controls.)

Maximum $s=12+1.5=13.5in.$

For $s=13.5in.$,

$\begin{array}{l}\frac{\varphi R_n}{s}=\frac{V_{u}Q}{I_x}\\ \frac{12.53}{13.5}=\frac{V_u\left(2115.75\right)}{280602.33}\\ V_u=123.096kips\end{array}$

Shear at first interior load, left of load, = $524-1.48\left(25\right)=487kips$

So maximum spacing will not be used in the first quarter of the span.

Spacing required at left side of first interior load is

$\frac{\varphi R_n{I}_x}{V_{u}Q}=\frac{12.53\left(280602.33\right)}{487\left(2115.75\right)}=3.4123in.$

Check middle fourth of span. Shear on right side of load is

$487-300=187kips$

$s=\frac{\varphi R_n{I}_x}{V_{u}Q}=\frac{12.53\left(280602.33\right)}{187\left(2115.75\right)}=8.8866in.<13.5in.$ (OK)

Welds for intermediate stiffeners $\left(\frac{1}{2}\times 6\right)$ :

Minimum weld size = 3/16in. (AISC Table J2.4)

Minimum length = $4\left(\frac{3}{16}\right)=0.75in.<1.5in.$

Use 1.5 in.

Use E70 electrodes, $\varphi R_n=1.392Dkips/in.$

where D is weld size in sixteenths.

Try $\frac{3}{16}in.\times 1\frac{1}{2}in.$ intermittent fillet welds.

For four welds, the weld strength is

$4\times 1.392\left(3\right)=16.7kips/in.$

The base metal shear yield strength is

$0.6F_{y}t=0.6\left(50\right)\left(\frac{1}{2}\right)\times 2=30kips/in.$

Shear rupture strength is

$0.45F_{u}t=0.45\left(65\right)\left(\frac{1}{2}\right)\times 2=29.25kips/in.$

Weld strength controls.

For a 1.5-in. length,

$\varphi R_n=1.5\left(16.7\right)=25.05kips$

The shear to be transferred is

$\begin{array}{l}f=0.045h\sqrt{\frac{F_y{}^3}{E}}=0.045\left(107\right)\sqrt{\frac{{\left(50\right)}^3}{29000}}=9.9966kips/in.\\ \frac{25.05}{s}=9.9966\\ s=2.506in.\end{array}$

A center-to-center spacing of 3 in. is equal to twice the length of the weld segment, so

either a continuous weld or an intermittent weld can be used. Use intermittent welds.

Maximum clear spacing: From AISC E6,

$d\le 0.75\sqrt{\frac{E}{F_y}}{t}_f=0.75\sqrt{\frac{29000}{50}}\left(1.5\right)=27.1in.$ (or 12 in.; 12 in.; controls.)

Maximum $s=12+1.5=13.5in.$

Use $\frac{3}{16}in.\times 1\frac{1}{2}in.$ E70 fillet welds spaced at 3 in. c-c for the intermediate stiffeners.

Welds for bearing stiffeners at the supports $\left(\frac{5}{8}\times 8\right)$ :

Minimum weld size = 3/16in. (AISC Table J2.4)

Minimum length = $4\left(\frac{3}{16}\right)=0.75in.<1.5in.$

Use 1.5 in.

Use E70 electrodes, $\varphi R_n=1.392Dkips/in.$

where D is weld size in sixteenths.

Try $\frac{3}{16}in.\times 1\frac{1}{2}in.$ intermittent fillet welds.

For four welds, the weld strength is

$4\times 1.392\left(3\right)=16.7kips/in.$

The base metal shear yield strength is

$0.6F_{y}t=0.6\left(50\right)\left(\frac{1}{2}\right)\times 2=30kips/in.$

Shear rupture strength is

$0.45F_{u}t=0.45\left(65\right)\left(\frac{1}{2}\right)\times 2=29.25kips/in.$

Weld strength controls.

For a 1.5-in. length,

$\varphi R_n=1.5\left(16.7\right)=25.05kips$

The shear to be transferred is

Reaction $÷$ length available for weld = $\frac{524}{107-2\left(1.0\right)}=4.99kips/in.$

$\begin{array}{l}\frac{25.05}{s}=4.99\\ s=5.02in.\end{array}$

Use $\frac{3}{16}in.\times 1\frac{1}{2}in.$ E70 fillet welds spaced at 5 in. c-c for bearing stiffener at support and at interior portion.

Conclusion:

Use 3/16 in. continuous fillet welds for the first 20 ft, $\frac{3}{16}in.\times 1\frac{1}{2}in.$ E70 fillet welds spaced at 3 in. c-c for the intermediate stiffeners and $\frac{3}{16}in.\times 1\frac{1}{2}in.$ E70 fillet welds spaced at 5 in. c-c for bearing stiffener at support and at interior portion.