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Chapter 14Chapter 14
Roadway Bridge
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contentscontents
• Roadway Bridge Floor
• Side walks and Railings
• Bridge Bracings
• Design of lateral support at top chord of through pony bridge
• Cross Sections for wind Bracing
• End X-frame in deck bridges
• Transmission of the braking forces the bearing
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Truss Bridges
A. Types of bridge trusses
B- Determination of forces in truss members
c. Proportioning of truss members
D- Box section for bridge trusses Top chords
Lacing bars, batten plates
Bottom chords
Diagonals
Verticals
Design of compression member
Design of Tension Members
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Design of Bolted Joint Design of Battens and Diaphragms Design of End Portals
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Roadway Bridge FloorRoadway Bridge Floor
The floor of a Roadway Bridges consists of: 1. A wearing surface or Roadway Covering. 2. Sub floor transmitting the loads to the
stringers and X-girders. The sub floor is similar to the solid floor of a
ballasted Railway Bridges. It may be timber, steel floor or R.C. floor.
Back
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Timber floor (Type 1)•For bridges, generally two layers of flanks are provided. For calculating these flanks we assume that the maximum wheel load is distributed over two flanks.
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Reinforced concrete floors (R.C. floor)
It may be supported by the main girders only, the X.G. only or by stringers and X. girders. The span of the slab may be 2.5 to 3.5 m, and thickness of slab to be 20 cm nearly. The R.C. slab reinforcement, generally 12 bars are used at least per one meter.
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Wearing surface
The wearing surface for roadway covering consists of timber blocks, hard bricks, asphalt bricks, stone blocks, asphalt or concrete.
The choice of material depends on the traffic, the span of bridge, the cost and climate
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The side walks are placed either inside or outside of the main girder. If they are arranged outside, they must be supported on cantilever brackets situated in the plane of the X.G. so that the –ve bending moment of the bracket is transmitted to the X.G. The floor of these side walks should be a precast R.C. slab (6cm) thick resting on the side walk stringers. The wearing surface is a 2 cm layer of asphalt. In through bridges the curb should be at least 50 cm inside the main girder
Back
Side walks and Railings
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Take strip 1 m and statical system as continuous beam supported on side walk stringer (one way slab), take t = 8 cm and get As. The applied loads are considered 500
kg/m2 or one concentrated load 5 t.
Hand railings and brackets withstand the effect of a transverse horizontal force of 150 kg/ m’ in cases of Railway bridges, Roadway bridges, and foot bridges, supposed acting at top level of hand rail. This horizontal is transmitted from the hand rail to the main posts and from their connections to the cantilever brackets.
Side walks parts
1. Slab
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2. Hand rail
Simple beam span distances between two brackets (take angle or channel X. sec.).
1. Side walk stringer
Simple beam span distance between two brackets (take channel X. sec.)
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4 Post
Cantilever beam (take 2-angle or 2-channel X. sec.).
yP = M
Kg L501 = P
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5 Connections
Double shear bolts
yP = M
Kg L501 = P
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1d
dM = F
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6 Bracket
Calculate M& N& Q at center of bracket.
In case of beam loaded alone we must calculate Fl.t.b and
the check that the actual stress Fc is less than the
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allowable stress Fp.b.
For ST 37
If 100
If 100
2c 000065.040.1F
22c /
7500F cmt
= l/ i , where I for compression flange only.
for bracket l/ b 2 l/ b
assume unequal angle 80120
Bolts subjects to shear
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1d
dM = F 2
221
2 dd2 = d
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Bolts subjects to tension
A.Nt I
yM =
0.8 Ft
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Bridge BracingsBridge Bracings
The bridge is provided with horizontal and vertical bracings:-
1. The stringers are connected together by stringer bracing given before.
2. The chords of the main girders are jointed together by an upper and lower horizontal bracing called wind bracings.
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•These transmit to the bearings of the bridge;•a. The lateral forces due to wind.•b. The lateral shock 6t.•c. Centrifugal force.
3 - Special horizontal bracings for the braking forces.
4-Two vertical and transverse bracing called X-frames or portals (in case of through) transmitting reaction of the upper lateral bracing to the bearings of bridge.
5- Some intermediate vertical transverse bracings called intermediate X-frames or intermediate portals for the rigidity of the structure.
It isn’t necessary to find all these bracings in every bridge, there existence depend upon the type of the bridge, the span and the floor.
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I-The Deck Bridge
The upper wind bracing transmits the wind pressure WT on the train & WF on the floor & ½ WG
on the wind ward side of the main girder.
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The lower wind bracing transmits the wind pressure ½ WG on the wind ward side of the main girder.
W = 100 kg/ m2 in case of loaded bridge
W = 200 kg/ m2 in case of unloaded bridge
* The wind pressure WT on the train produces in addition
to horizontal loading of the upper wind bracing.
a vertical loading, to the main girder. b
eW = V T
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In case of a truss bridge, only the exposed area of the members is considered. This area is equivalent to 40 % of the hole area of the surface of the truss. In all bridges with an upper and a lower wind bracing, their shall be providedat each end a X-frame to transmit to the bearings,the horizontal reactions of the upper wind bracing. The horizontal reactions of the lower wind bracing are transmitted directly to the bearings
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The end X-frames in deck bridges shall be of rigid type. In all railways and in roadway deck bridges there shall be intermediate transverse bracing at least at every third panel point to increase the stiffness of the bridge. These intermediate X-frames will release the end X-frame from a part of the horizontal reaction of the upper wind bracing. Yet it is recommended not to consider that release unless the bridge as the space structure.
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ii-The Through Bridge
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In through bridge two horizontal wind bracings should be arranged if possible. In the plate girder through bridges we can’t arrange an upper wind bracing in the bony truss Roadway Bridge we have only a lower wind bracing which transmits all the wind loads to bearings. The force WF on the floor will be considered to act on a
solid surface as the plate girder. The through Railway Bridge shown above is provided with two horizontal wind bracings.
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The upper wind bracing transmits the wind pressure ½ WG on the wind ward side of the main girder.
The lower wind bracing transmits the wind pressure ½ (WG), WT on the train & WF on the floor & on the wind
ward side of the main girder.At the connection of each X-G to the main girder, stiffness bracket shall be arranged.
Back
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Design of lateral support at Design of lateral support at top chord of through pony top chord of through pony
bridgebridge C = force in flange = AfFt The U-frame formed by the two vertical stiffeners
and the horizontal stiff X-girder is acted upon by a horizontal transverse force = C/100 at the centroid of compression flange as well as the wind pressure between two consequence X-girders. The maximum stressed section is mn. The compression stress at point n ≯ Fltb. If the stress isn’t safe, we either increase the thickness of the bracket plate or add a stiffening angle.
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The connection between the X-girder and bracket is designed on the shearing force A that between X-girder and the bracket and horizontal gusset of wind bracing on force B. if the X-G is built up section the bracket connection is designed as a web splice.
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Back
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Cross Sections for wind Cross Sections for wind BracingBracing
The diagonal of wind bracing
The diagonal of wind bracing shall have stiff section to prevent vibration and to help in reducing the deflection of main girder due to eccentric loading (space frame treatment). The section should have a depth not less than L/40. The recommended sections are given in Fig.(5.).
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The choice of the section depends more or less upon the span of the diagonal, the two channel section is convenient for too long spans executed in Banha and Samanood bridges. The two channels are connected together by latticing or batten plates.
≯ in compression & ≯ in tension
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In case of one diagonal member only
Case of the warren system which designed on a force S;
forcen compressiofor 85.0
forcen for tensio 85.0
= S
call
tall
all
FF
FF
FSin
Q
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Case of the N-system which designed on a force S;
forcen for tensio 85.0
= S
tall
all
FF
FSin
Q
40
Case of the K-system , Rhombic And Multiple which designed on a force S;
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allFSin
Q
2 = S
If the bracing is made up of crossed diagonal and struts, the calculation is made under the assumption that the tension diagonal are only acting. The struts here receive compression force. If the multiple systems of wind bracing a further reduction of 20 % in the allowable stresses given before, shall be made to account for approximation of solution that both systems (tension and compression member) equally share the lateral loads. in case of one angle in compression the allowable compression stress shall be reduced by 40% of Fc.
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Hence, the approximations in allowable stresses are;
n compressioin only) sections ic(unsemmetraction angle onefor 0.6 stresses Reduce
ancyindeterminfor ion approximat todue 0.80 stresses Reduce
effect space todue 0.85 stresses Reduce
A & forcen compressiofor 6.08.085.0
3
3AA & forcen for tensio 8.085.0
eff.
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11 eff
gressbpall
tall
AFF
AA
AFF
Back
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End X-frame in deck bridges
The compression diagonal is assumed in acting and we design the tension diagonal, also we assume that the X-frame is resting at a movable support at one end and the hinge support at other end.
Back
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Transmission of the braking forces the bearing
In Railway bridges especial bracing should be arranged to transmit the longitudinal forces from the stringers to the panel points of the main girder, hence; they are transmitted through the main girder to the hinged bearings. Some times a bracing is arranged at every panel point. But generally two bracings at the quarter points of bridge are sufficient. The braking bracing system shown in sketch is statically indeterminate but the loads are symmetric about perpendicular axes to X-X. Therefore the diagonal B’n & c’n are zeros since they correspond to themselves. Also, the loads are antisymmetric about axes Y-Y and thus members mb’ & mc’ & nb & nc are zeros, If special bracing of the longitudinal forces is omitted, these forces are transmitted
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from stingers to main girder by bending of the X-girder.
Back
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Truss Bridges
b L/ 20, b h/ 3
Where, b = bridge width = distance between center lines of two main girders
L = span of bridge
The depth of trusses shall be chosen in such away that the elastic deflection due to L.L (without dynamic effect) shouldn’t exceed L/800 for Railway bridges and L/600 for Roadway bridges.
Back
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h≮ (simple ) & (continue) Road.
h≮ (simple ) & (continue) Rail.
10
span
10
span12
span
8
span
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Either both chords are straight and parallel or only one of them. In a through bridge the upper chord may polygonal, in a deck bridge the lower chord may be polygonal. Curved chord should not be used in bridge trusses on a account of the additional bending stresses. The loads are transmitted to the panel point of the truss by a system of stringers and cross girder. No load except the own weight of the truss members should act between the panel point.
A. Types of bridge trusses
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They suitable for span up to 60 m. the joints are simpler than in trusses with polygonal chords. The depth is h L/ 8 for Railway Bridge, or h L/ 10 for Roadway Bridge. For continuous and cantilever trusses the depth may be taken h L/ 10 for Railway bridges, h L/ 12 for Roadway bridges. Some times a greater depth is used to allow an upper wind bracing. The arrangement of web members may be N-system or warren system. The warren system trusses require generally less material than the N-shaped trusses, since the vertical members have smaller forces, the number of joints and changes of cross section in warren system are also less. Shop work for warren trusses will be cheaper.
1.Trusses with horizontal chords
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They are used for spans up to 60 m. the economical depth at middle is h = L/7. The web system is either N-shaped or warren. The economical inclination of the diagonal to horizontal = = 40 - 60. A polygonal chord trusses lighter than a truss with horizontal chords since the forces in the diagonals are smaller. On the other hand the shop work is more complicated which means a higher cost.
2-Trusses with polygonal chords
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3-Trusses with subdivided panels (e), with Rhombic diagonal (f) and K-system
These kinds are economical for span over 80 m. The panel length is reduced in all this system and thus the cost of the floor is less, but the increased number of joint increase the cost of shop work. A truss with Rhombic diagonal has a good appearance; a truss with subdivided panels has big secondary stresses. K-system trusses have the smaller secondary stress.
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4- Trusses with multiple web system
These were used in past where the tension diagonals consist of flat bars. Now they are again used for main girders, but type h with crossed diagonals is frequently used for wind bracing. For approximate calculation, the common assumption is that the truss may be divided up into two or more component trusses with the same chords but with different web system. The loads also are divided and placed upon this component trusses. Then the stress in a web member is determined as its stress calculated in the truss of which it is a part. The chords are a part of all component trusses, hence the stresses in a chord member is obtained by adding it partial stress from each component truss.
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5- Trusses with 3 chords
The arch truss with a tie (k) and the truss reinforced by a hinged arch (Pow string truss) are supported on a hinged bearing at one end and a movable bearing at the other. They are therefore externally static determinant but internally they are static indeterminate. These trusses have good appearance but they more expansive than trusses with two chords.
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Back
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b- Determination of forces in truss members
We determine the forces in the truss members on the assumption that the member are connected by hinges, so that loads applied at the panel point produce only axial forces in the truss member. The secondary stress which are the bending stress induced by the rigidity in connection, are generally neglected. In our specification it is required to calculate the secondary stress in the following cases:
1- For trusses with subdivided panels.
2-For member whose width in the plane of the truss is more than 1/10 of its length.
3-For loads acting between the panel point.Back
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c. Proportioning of truss members
For the chord member we can use either sections with one web plate (T section) or sections with two web plates (Box section). T-sections are used only for small bridge. Box sections have grates moment of inertia about axis y-y and are better used for the connection of gusset plate. The sections of all chords and web member should be symmetrical about axis y-y in the plane of the truss. Diagonals and verticals are usually symmetrical about axis x-x also. The required area of the chords change at every panel point of the truss and in choosing the different cross section we must try to get simpler connection and splices at the panel point. Back
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D- Box section for bridge trusses
Top chords
The minimum section consists of a horizontal plate and 2 channels or a horizontal plate, 2 vertical plates and four angles.
Depth of top chord h = (1/12 – 1/15) of the panels length ≯
Width a = (0.75 – 1.25) h 10
L
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To avoid local buckling, the minimum thickness of web and cover plates should be as follows;
The unsupported width of a plate measured between adjacent lines of rivets or welds connection the plate to other parts of the section should not exceed:
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t = thickness of a single plate or of 2 or more plate provided that this plates are adequately tacked together.
YFt
b 64
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yFt
d 301
Only excess over this width should not be included in the effective sectional area in computing direct compressive stresses. The center of gravity axis x-x for the different section should not change too much. In drawing the truss we use an average value y = (y1 + y2 + y3 + ....)/ n.
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It is good practice to use a cover plate over the whole length of the top chord even if the end members have excessive cross section.
Back
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Lacing bars, batten plates
The two plates of the compression members shall be connected together by diaphragms and the open side of the box section shall be provided with batten plate close to the gusset plate and with intermediate batten plates or lacing bars to avoid lateral buckling of their component parts. The slenderness ratio of each component part between consequent connections of lacing bars or batten plates shall be not more than 50 (and 2/3 l/i of the whole section).
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Bottom chords
The depth of the bottom chord is equal to that of the top chords h = (1/12 – 1/15) of the panels length, or slightly more (2 – 4 cm). No horizontal plate is provided at the bottom of the section to avoid water packets. In continuous and cantilever bridges where some bottom chord members are in compression, horizontal plate may be used and it must be provide with drainage holes (4 – 5 cm) . The two component parts of tension member shall be connected together by diaphragms and batten plates similar to these of the compression members, but their thickness may be taken 25 % lighter (t2 lmember/ 15).
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bottom with cover plate (holes)
Diagonals
For appearance the width of the diagonal should not be more than that of the top chord and should decrease from the end to the middle of the bridge. The compression diagonal at end of the warren truss has a section similar to that of the top chord. Back
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Verticals
In trusses with a N-shape web system, the vertical have similar sections as diagonals. In warren trusses the vertical may consist of a web plate + 4 flange plates or an I-beam (B.F.I.B).
For diagonal or vertical tension member;
t2 lmember/ 15 ( D & V )
t2 (lmember)/ 30 Railway bridges ( D & V & C )
t2 (lmember)/ 35 Roadway bridges ( D & V& C )
t2 (lmember)/ 10 (C )
D → Diagonal & V→ Vertical & C → Chord
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Back
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Design of compression member
The slenderness ratio l/ i of compression member of main girder shall not exceed 90 for Railway bridges and 110 for Roadway bridges.
E. Effective buckling lengths
Table 4.5
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MemberIn planeOut of plane Compressio
n chord
Effectively braced
Compression chord
Unbraced
Chords 0.85 l 0.85 l0.75 span (Clause 4.3.2.2 or equation 4.2 if using U-frames)
Diagonals- Single triangulated web system
0.70 l0.85 l1.20 l
Table (4.5)
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- Multiple Intersected web rectangular system adequately connected
- K-system
0.85 l/20.75 ll
0.90 l1.20 l1.50 l
Vertical members- Single triangulated web system
- K-intersected web system
0.70 l0.85 l1.20 l
l0.35(0.75+0.25Ns/Nl)L
(0.9+0.3Ns/Nl)l
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Effective buckling lengths
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Unbraced compression chords
a- For simply supported truss, with laterally unsupported compression chords and with no cross-frames but with each end of the truss adequately restrained (Figure 4.1), the effective bucking length (kL), shall be taken equal to 0.75 of the truss span, (clause 4.3.2.2).
b- For a bridge truss where the compression chord is laterally restrained by U-frames composed of the cross girders and verticals of the trusses, the effective buckling length of the compression chord (Lb) is:
aa..I.E2.50 = L 4 yb
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E = the Young’s modulus = 2100 t/cm2
Iy = the moment of inertia of the chord member about the Y-
Y axis.
a = the distance between U-frames (cm) = S
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2
22
1
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EI*2
*d
EI*3
d =
B
d1 = the distance from the centroid of the compression
chord nearest face of the cross girder of the U-frame = dw –
Hx.G.
d2 = the distance from the centroid of the compression
chord to the centroidal axis of the cross girder of the U-frame = dw – Hx.G./2
I1 = the second moment of area of the vertical member
forming the arm of the U-frame about the axis of bending.
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I2 = the second moment of area of X-G about the axis of
bending = IX
B = the distance between centers of consecutive Main Girders connected by the U-frame
Back
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Tension members shall always be of rigid construction and their slenderness ratio l/ i shall not exceed 160 for Railway bridges and 180 for Roadway bridges. The effective net section area shall be taken for all tension members. This area shall be the least that can be determined from any plane or planes cutting each component plate or sections to its axis,
Design of Tension Members
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diagonally or following zig-zag line through adjacent rivet holes. In each case all holes of line to meet with shall be deducted from the gross sectional area where any portion of the sectional area is measured for a diagonal plane adding (S2/ 4g) for each gauge space. The minimum sectional area should not be less than that obtained by assuming all the holes to be in one perpendicular plane.
Back
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Design of Bolted Joints
Connection of web member to gusset plate “and splices of chord member” shall have a strength equal to the maximum strength of the connected members. The bolts shall be arranged symmetrical about the center line of the member. The connection to either direct or part of it is indirectly connected by:-
1- Splice plates or lug angles.
2- If breaking is along section S-S bolts (1 single + 2 double + 3 double) shear, must carry the load.
3- The strength of the splice plate should be enough to carry a force corresponding to bolts (4 + 5) single shear.
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Connections for members
The connection shall be designed for a capacity based on the maximum of:-
1- The average between the actual force and the maximum strength of the member of not less than 0.75 the maximum strength of the member.
2 The bolts between the chord and the gusset plate must correspond to the algebric sum of the horizontal components of the strength of the diagonals S1, S2
L = S1Cos + S2Cos
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3. The number of bolts should correspond to the effective strength of the two diagonal, i.e., the number of = (number of bolts in member 1 + number of bolts in member 2)Cos .
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Example
L=750
In a lower chord panel, if the maximum force in chords and diagonals are as given, design suitable cross section, connections and splices.
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Member S1 = + 94 t
2 [ No 24
Bolts M22, = 24
Anet = 2(42.3 – 2.40.95 - 22.41.3) = 67.56 cm2
Fact = 94/ 67.56 = 1.39 t/ cm2 < 1.6 t/ cm2
Maximum force = 67.561.60 = 108.10 t
Rleast = Rs. sh
qb = 0.25 Fub For bolts of grade ( 8.8),
Rs. sh = qb As n = 0.258.03.03 1 = 6.06 t
(n = No. of shear planes)
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Bolts connecting diagonal to gusset = (Maximum force/ Rleast) 1.15
= (108.1/ 6.06)1.15 = 20.52 bolts
Use 24 bolt M22
Member S2 = - 70 t
2 [ No 22 & Bolts M22, = 24
Agross = 2(37.40) = 74.80 cm2
y = ly/ iy = (0.85750/ 11) 1.20 = 69
(1.20 due to lacing bars)
x = lx/ ix = (0.70750/ 8.5) 1.20 = 61
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max = 69 Fp.b = 1.60 – 0.000085 (l/ i)2 = 1.119 t/ cm2
Fact = 70/ 74.80 = 0.94 t/ cm2 < 1.119 t/ cm2
Maximum force = 74.801.119 = 83.70 t
Rleast = Rs. sh
qb = 0.25 Fub For bolts of grade ( 8.8),
Rs. sh = qb As n = 0.258.03.03 1 = 6.06 t (n =
No. of shear planes)
Bolts connecting diagonal to gusset = (Maximum force/ Rleast) 1.15
= (83.7/ 6.06)1.15 = 15.88 bolts
Use 16 bolt M22
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Member S3 = + 126 t
2 [ No 30 & Bolts M22, = 24
Anet=2(58.80 – 22.41.00 - 22.41.6) = 92.64 cm2
Fact = 126/ 92.64 = 1.36 t/ cm2 < 1.6 t/ cm2
90
Member S4 = + 196 t
2 [ No 30 + 2 PL 24012 & Bolts M22, = 24
Anet for 2[ = 2(58.80 – 22.41.00 - 22.41.6) = 92.64 cm2
Anet for 2PL = 2(24 – 22.4) 1.2 = 46.08 cm2
Anet for 2[+2PL = 92.64 + 46.08 =
138.72 cm2
Fact = 196/ 138.72 = 1.413 t/ cm2
< 1.6 t/ cm2
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Force to be transmitted from gusset plate to bottom chord = Fmax
Fmax = (S1 + S2)Cos = (67.561.6 + 74.801.119) 1/2
= 135.62 t
Bolts connecting diagonal to gusset = (135.62/ 6.06)1.15 = 25.74 bolts
Use 28 bolt M22
The bolts in the framing angle are not counted as they used to transmit the reaction of the cross girder.
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Splice of chord
Splices of tension or compression chord shall be designed on the maximum strength of the member. For straight chord the splice shall be outside the gusset plate. For broken chord the splice will be within the gusset plate.
Member S3
2 [ No 30 & Bolts M22, = 24
Net area of flange = (10 – 23) 1.60 = 12.30 cm2
Net area of splice plate = (10 – 23) 1.60 = 12.30 cm2
Number of single shear field bolts =
(12.301.6/ 6.06) 1.15 = 3.25 bolts
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Net area of web = (30 – 21.60) 1.00 = 22.20 cm2
Use 2 splice plates = 300.80 + 240.80
Net area of 2 splice plates = (30 – 22.3) 0.80 + (24 - 22.30) 0.8 = 35.80 cm2
Rleast = RD. sh or Rb
Rb = Fb d min t
Fb = Fub
= 0.60 for end distance (S1.5d), (Table 6.2)
Fb = 0.60 8.00 = 4.80 t/ cm2
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Rb = 4.80 2.20 1.00 = 10.56
t
qb = 0.25 Fub
For bolts of grade ( 8.8),
Rs. sh = qb As n =
0.258.03.03 2 = 12.12 t
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Rb = 10.56 t
Number of double shear field bolts = (22.201.6/ 10.56) 1.15 = 3.87 bolts
Use 4 bolt M22
The splices in the chords are placed in the side of the:-
1. Smaller cross section except in cases where the erection is done by the cantilever method.
2. Gusset plate; in a polygonal chord in order to avoid the bending of plates, angles and channels the splice plate is placed at the brake of the chord on the gusset plates. 3.Gusset shall be proportion to withstand the effective forces in the web member
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The thickness of the gusset is determined from the critical section abcd. This section should be at least 15 – 20 % stronger than the diagonal it self generally all gusset are made of the same thickness. The thickness of gusset plate shall be at least 12 mm in Railway bridges and 10 mm in Roadway bridges.
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4- At the chord panel point the cover plate should be connected to a gusset plate by special connection angle to make the center of gravity of the rivet group between gusset plate and top chord nearer to the center of gravity of the top chord.
101Back
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Design of Battens and Diaphragms
The two parts of the box section must be connected together in such away that they act ass one unit. For compression member stronger details are necessary than for ten member.
Diaphragms
Diaphragms are transverse plates or channel connected to the two webs of the box section by angles. They are necessary to assume the rectangular shop of the box section. For the chords wee have at least one diaphragm between the two panel points. In the diagonals, we arrange at least one diaphragm near each end.
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Batten plate
One the open sides of the box section we have batten plates as close to the gusset plate as possible one intermediate batten plate or lacing bars to avoid lateral buckling of the unsupported flange for the calculation of the lattice bars of a compression member we assume a transverse force = 2 % of the longitudinal force in the member. If there is a continuous plate at the upper side of the box section, latticing will be in the lower side only and transverse force will be according to cross section of the lower side only. In tension member a lattice system and batten plates 25 % lighter is used.
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Battening of compression member
The number of batten is such that we get at least 3 bays batten shall be of plates, channels, I-section bolted or welded to resist the following forces:-
The member as a whole can be considered as a vierandeen girder or we can assume hinges at mid distances and change it to statically determine system.
Shear in batten plate = Qd/ a
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Bending moment in batten = Qd/ 2
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Design of End Portals
End transverse bracing are called portals. The portals are placed either in the vertical plan of the end post (1), in the plan of the first vertical (2), or in the inclined plan of the end diagonal (3). In case (2) the first panel of the lower wind bracing is affected by the reaction transmitted by the end portals. The arrangement of end portal in case (2) is stiffer than in case (3). The portal must not interfere the clearance line.
Back
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The shop of the portal depends on the depth and on the clearance line. The portals are generally static indeterminate closed frames in which the post over subject to bending stresses.
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Approximate calculation of portals
The point of inflection of the post are situated according to the relative stiffness of the cross girders, post, and the upper strut at height between 1/3 – 1/2 of the force height h’. We can replace the point of inflection by two hinges at C and C’ each of them transmits ½ W1. Then the portal
can be calculated as static determine frame. If the portal is in the plan of inclined end diagonal, the points of inflection C, C’ should net be more than h’/ 3 from A and A’.
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