SHEAR FORCES AND BENDING MOMENTS

Shear Forces and Bending Moments Due To Dead Loads

Dead Loads [LRFD Art. 3.3.2]

DC = dead load of structural components and nonstructural attachments

Beam self weight = 0.878 kip/ft.

Diaphragm weight

(\frac{8}{12}) (\frac{(48 - 10)}{12} \times \frac{(42 - 11)}{12} - 4 (\frac{1}{2}) (\frac{3}{12}) (\frac{3}{12}))) 0.15) = 0.81 k i p / d i a p h r a m

Self-weight of diaphragms is considered as concentrated load acting at quarter points.

Barrier weight = (2)(0.300 kip/ft.) = 0.6 kip/ft.

DW = dead load of wearing surface and utilities

Weight of 3 in. bituminous wearing surface = 0.140 kcf [LRFD Table 3.5.1-1]

D W = (\frac{3}{12} f t) (0.140 k c f) = 0.035 k s f

LRFD Art. 4.6.2.2.1 states that permanent loads (barrier and wearing surface loads) may be distributed uniformly among the beams if certain criteria are met. Since these criteria are satisfied, the barrier and wearing surface loads have been equally distributed among the eight beams. The dead load distribution factor is

(\frac{1}{8}) = 0.125

Unfactored Shear Forces and Bending Moments Due To Dead Loads

Using simple static, values of shear forces and bending moments due to the self-weight of beam, barriers, diaphragms, and wearing surface are calculated. For these calculations, the span length is the design span (100 ft).

Shear Forces and Bending Moments Due To Live Loads

Live Loads

LRFD states that vehicular live loading on the roadway of bridges, HL-93, consists of a combination of: [LRFD Art. 3.6.1.2.1]

Design truck or design tandem with dynamic allowance.

The design truck is the same as the HS-20 design truck specified by AASHTO Standard Specifications Art. 3.6.1.2.2. The design tandem consists of a pair of 25.0 kip axles spaced 4.0 ft apart, Art. 3.6.1.2.3.
Design lane load of 0.64 kip/ft without dynamic allowance.
[LRFD Art. 3.6.1.2.4
]

Live Load Distribution Factors For A Typical Interior Beam

The live load bending moments and shear forces are determined by using the simplified distribution factor

formulas [LRFD Art. 4.6.2.2]. To use the simplified live load distribution factor formulas, the following

conditions must be met [LRFD Art. 4.6.2.2.1]:

Width of slab is constant	OK
Number of beams, $N_{b} \geq 4 (N_{b} = 8)$	OK
Beams are parallel and approximately of the same stiffness	OK
The roadway part of the overhang, $d_{e} < 3.0 f t . (d_{e} = 0.0)$	OK
Curvature in plan is less than specified by Art. 4.6.1.2 (Curvature = 0.00)	OK

For precast cellular concrete box with shear keys and with or without transverse post-tensioning, bridge type is (g) [LRFD Table 4.6.2.2.1-1].

In order to use live load distribution factor formulas, it is required to determine the number of lanes. Number of design lanes equals the integer part of the ratio w/12, where w is the clear roadway width, in ft, between the curbs [LRFD Art. 3.6.1.1.1]. Referring to Figure 18, Bridge Cross-Section, on page 34:

w = 29 f t

Number of design lanes = integer part of $\frac{29}{12}$ = 2 lanes

Distribution Factor For Bending Moments

For all limit states except Fatigue Limit State.

For two or more lanes loaded, if sufficiently connected to act as a unit:

D F M = k {(\frac{b}{305})}^{0.6} {(\frac{b}{12.0 L})}^{0.2} {(\frac{I}{J})}^{0.06}

[LRFD Table 4.6.2.2.2b-1]

Provided that:

$35 \leq b \leq 60$	b=48in	OK
$20 \leq L \leq 120$	L=100ft	OK
$5 \leq N_{b} \leq 20$	N_b=8	OK

where:

DFM = Distribution factor for moment for interior girder

b = Beam width, in

L = Beam span, ft

I = Moment of inertia of the beam, in⁴

$k = 2.5 {(N_{b})}^{- 0.2} = 2.5 {(8)}^{- 0.2} = 1.649 \geq 1.5$ OK

N_b= Number of beams

J = St. Venant torsional constant, in⁴

For closed, thin-walled shapes [LRFD Eq. C4.6.2.2.1-3]

J \equiv \frac{4 A_{0}^{2}}{Σ \frac{s}{t}}

Where:

A₀ = area enclosed by centerlines of the elements of the beam=(48-5)(42-5.5)=1569.5 in²

s = length of an element of the beam, in

t = thickness of an element of the beam, in

J = \frac{4 \times {(1569.5)}^{2}}{2 (\frac{48 - 5}{5.5}) + 2 (\frac{42 - 5.5}{5})} = 3252876 {i n}^{4}

Therefore:

D F M = 1.649 {(\frac{48}{305})}^{0.6} {(\frac{48}{12.0 \times 100})}^{0.2} {(\frac{203088}{325876})}^{0.06} = 0.281 l a n e s / b e a m

For one lane loaded, if sufficiently connected to act as a unit: [LRFD Table 4.6.2.2.2b-1]

D F M = k {(\frac{b}{33.3 L})}^{0.5} {(\frac{I}{J})}^{0.25} = 1.649 {(\frac{48}{33.3 \times 100})}^{0.5} {(\frac{203088}{325876})}^{0.25} = 0.175 l a n e s / b e a m

Distribution Factor For Shear Forces

For two or more lanes loaded: [LRFD Table 4.6.2.2.3a-1]

D F M = {(\frac{b}{156})}^{0.4} {(\frac{b}{12.0 L})}^{0.1} {(\frac{I}{J})}^{0.05}

Provided that:

$35 \leq b \leq 60$	b=48in	OK
$20 \leq L \leq 120$	L=100ft	OK
$5 \leq N_{b} \leq 20$	N_b=8	OK
$25000 \leq J \leq 610000$	J=325876in⁴	OK
$20000 \leq J \leq 610000$	I=203086in⁴	OK

Where:

DFV = Distribution factor for shear for interior beam.

D F M = {(\frac{48}{156})}^{0.4} {(\frac{b}{12.0 \times 100})}^{0.1} {(\frac{203086}{325876})}^{0.05} = 0.44 l a n e s / b e a m

For one design lane loaded:

D F M = {(\frac{b}{130})}^{0.15} {(\frac{I}{J})}^{0.05} = {(\frac{48}{130 \times 100})}^{0.15} {(\frac{203088}{325876})}^{0.05} = 0.42 l a n e s / b e a m

Thus, the case of two lanes loaded controls, DFV = 0.44 lanes/beam.

Dynamic Allowance [LRFD Art. 3.6.2]

IM = 33% for all limit states except for Fatigue Limit State [LRFD Table 3.6.2.1-1]

where:

IM = dynamic load allowance, applied to design truck load or tandem only.

Unfactored Shear Forces And Bending Moments

Unfactored HL-93 shear forces and bending moments per beam due to design truck are:

VLT = (shear force per lane) (DFV) (1 + IM)= (shear force per lane) (0.44) (1 + 0.33)= (shear force per lane)(0.585) kips

MLT = (bending moment per lane) (DFM) (1 + IM)= (bending moment per lane) (0.281) (1 + 0.33)= (bending moment per lane)(0.374) ft-kips

Unfactored HL-93 shear forces and bending moments per beam due to design truck are:

VLL = (lane load shear force) (DFV)= (lane load shear force) (0.44) kips

MLL = (lane load bending moment) (DFM)= (lane load bending moment) (0.281) ft-kips

Note: The dynamic allowance is not applied to the design lane load.

Load Factors and Load Combinations [LRFD Art. 3.4]

Total factored load, Q, is taken as: [LRFD Eq. 3.4.1-1]

Q = η Σ_{i} q_{i}

where

η	=	a factor relating to ductility, redundancy, and operational importance. For this example, η is considered 1.0. [LRFD Art. 1.3.2.1]
γ	=	load factors [LRFD Table 3.4.1-1]
q_i	=	specified loads

Investigating different limit states given in LRFD Art. 3.4.1, the following limit states are applicable:

Service I: to check compressive stresses in prestressed concrete components
Q = 1.00 (DC + DW) + 1.00 (LL + IM) [LRFD Table 3.4.1-1]
Service III: to check tensile stresses in prestressed concrete components
Q = 1.00 (DC + DW) + 0.80 (LL + IM) [LRFD Table 3.4.1-1]

This load combination is a special combination for service limit state stress check that applies only to tension in prestressed concrete structures to control cracks.
Strength I: to check ultimate strength [LRFD Tables 3.4.1-1 and 2]
Q _maximum = 1.25 (DC) + 1.50 (DW) + 1.75 (LL + IM)

Q _minimum = 0.90 (DC) + 0.65 (DW) + 1.75 (LL + IM)

Note: For a simple span bridge, Q maximum controls.