Calculation Procedure:
- Determine the magnitude of the resultant and its location
Since the member carries only concentrated loads, the maximum moment at any instant
occurs under one of these loads. Thus, the problem is to determine the position of the load
system that causes the absolute maximum moment.
The magnitude of the resultant #18^=10 + 4+15 = 29 kips (129.0 kN). To deter-
mine the location of R, take moments with respect to A (Fig. 47). Thus 2M 4 = 29AD =
4(5) + 15(17), or AD = 9.48 ft (2.890 m).
- Assume several trial load positions
Assume that the maximum moment occurs under the 10-kip (44.5-kN) load. Place the
system in the position shown in Fig. 4Ib 9 with the 10-kip (44.5-kN) load as far from the
adjacent support as the resultant is from the other support. Repeat this procedure for the
two remaining loads.
- Determine the support reactions for the trial load positions
For these three trial positions, calculate the reaction at the support adjacent to the load un-
der consideration. Determine whether the vertical shear is zero or changes sign at this
load. Thus, for position 1: RL = 29(15.26)740 = 11.06 kips (49.194 kN). Since the shear
does not change sign at the 10-kip (44.5-kN) load, this position lacks significance.
Position 2: R 1 = 29(17.76)740 = 12.88 kips (57.290 kN). The shear changes sign at the
4-kip (17.8 kN) load.
Position 3: RR = 29(16.24)740 = 11.77 kips (52.352 kN). The shear changes sign at the
15-kip(66.7-kN)load.
- Compute the maximum bending moment associated with
positions having a change in the shear sign
This applies to positions 2 and 3. The absolute maximum moment is the larger of these
values. Thus, for position 2: M= 12.88(17.76) - 10(5) = 178.7 ft-kips (242.32 kN-m). Po-
sition 3: M= 11.77(16.24) = 191.1 ft-kips (259.13 kN-m). Thus, Mmax = 191.1 ft-kips
(259.13 kN-m).
- Determine the absolute maximum shear
For absolute maximum shear, place the 15-kip (66.7-kN) load an infinitesimal distance
to the left of the right-hand support. Then Fmax = 29(40 - 7.52)740 = 23.5 kips (104.53
kN).
When the load spacing is large in relation to the beam span, the absolute maximum
moment may occur when only part of the load system is on the span. This possibility re-
quires careful investigation.
INFLUENCE LINE FOR SHEAR IN A
BRIDGE TRUSS
The Pratt truss in Fig. 48a supports a bridge at its bottom chord. Draw the influence line
for shear in panel cd caused by a moving load traversing the bridge floor.
Calculation Procedure:
- Compute the shear in the panel being considered with a unit
load to the right of the panel
Cut the truss at section YY. The algebraic sum of vertical forces acting on the truss at pan-
el points to the left of IT is termed the shear in panel cd.
