
(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
Four panels spaced at 58.25in.
Explanation of Solution
Given:
Span length
Uniformly distributed live load
Superimposed dead load
Concentrated dead load
Concentrated live load
Formula used:
h is the depth of web
Calculation:
Assume a girder weight of
Determine the factored loads:
The factored moment and shear are
Determine the overall depth:
Use the maximum permissible depth of 110 in.
Try
To determine the web thickness, first examine the limiting values of
For
Minimum
For
Minimum
Try a
Determine whether the web is slender:
Therefore, the web is slender.
Estimate required flange size:
Try a
Girder weight =
Compression flange:
Check flange local buckling (FLB):
Since
Compute the plate girder strength reduction factor:
Try a
Shear: At left end (end panel),
Required
From Table 3-17a in the Manual,
Use
This spacing will apply for the remaining distance to the centerline of the girder. This distance is
For a spacing a of 67 in., the number of panels is
Use 4 panels at
At
Required
For
Therefore, stiffeners are needed in middle
Conclusion:
Therefore, Use a
(b)
The size of intermediate and bearing stiffeners.

Answer to Problem 10.7.8P
2 PL
2 PL
Explanation of Solution
Given:
Span length
Uniformly distributed live load
Superimposed dead load
Concentrated dead load
Concentrated live load
Calculation:
Intermediate stiffener size:
Available width:
Try
To determine the required moment of inertia, use the conservative approximation from the User Note in AISC G2.3:
Try two
Length: From Figure 10.9 in the textbook (Steel design),
Assume a flange-to-web weld size of
Length =
Use two PL
Design the bearing stiffeners at the supports for a load of
Maximum stiffener width =
Try
Try two plates,
Bearing strength:
Compressive strength: The maximum permissible length of web is
Compute the radius of gyration about an axis along the middle of the web:
Compute the compressive strength:
Therefore,
Use 2 PL
Because there is a large difference between the reactions and the interior concentrated loads, use 2 PL
Conclusion:
Use two PL
(c)
The design of the all welds

Answer to Problem 10.7.8P
Explanation of Solution
Given:
Span length
Uniformly distributed live load
Superimposed dead load
Concentrated dead load
Concentrated live load
Calculation:
Design the flange-to-web welds.
The shear flow is
At the support,
Minimum weld size = 3/16 in. (AISC Table J2.4)
Minimum length =
Use 1.5 in.
Use E70 electrodes,
where D is weld size in sixteenths.
Try an
For two welds,
Weld strength =
Base metal shear yield strength (web plate controls) is
Shear rupture strength is
Weld strength controls.
For a 1.5-in. length,
Required spacing:
Since this is less than twice the length of the weld, use a continuous weld.
For
This occurs when
Maximum clear spacing: From AISC E6,
Maximum
For
Shear at first interior load, left of load, =
So maximum spacing will not be used in the first quarter of the span.
Spacing required at left side of first interior load is
Check middle fourth of span. Shear on right side of load is
Welds for intermediate stiffeners
Minimum weld size = 3/16in. (AISC Table J2.4)
Minimum length =
Use 1.5 in.
Use E70 electrodes,
where D is weld size in sixteenths.
Try
For four welds, the weld strength is
The base metal shear yield strength is
Shear rupture strength is
Weld strength controls.
For a 1.5-in. length,
The shear to be transferred is
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,
Maximum
Use
Welds for bearing stiffeners at the supports
Minimum weld size = 3/16in. (AISC Table J2.4)
Minimum length =
Use 1.5 in.
Use E70 electrodes,
where D is weld size in sixteenths.
Try
For four welds, the weld strength is
The base metal shear yield strength is
Shear rupture strength is
Weld strength controls.
For a 1.5-in. length,
The shear to be transferred is
Reaction
Use
Conclusion:
Use 3/16 in. continuous fillet welds for the first 20 ft,
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Chapter 10 Solutions
Steel Design (Activate Learning with these NEW titles from Engineering!)
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- Steel Design (Activate Learning with these NEW ti...Civil EngineeringISBN:9781337094740Author:Segui, William T.Publisher:Cengage Learning
