A large piece of equipment is being delivered via special transport to its final destination. Our firm has been hired to evaluate the adequacy of the bridges the equipment must cross. We have been assigned to evaluate a three-span overpass. One of the concerns is overloading the splice that holds

Question

A large piece of equipment is being delivered via special transport to its final destination. Our firm has been hired to evaluate the adequacy of the bridges the equipment must cross. We have been assigned to evaluate a three-span overpass. One of the concerns is overloading the splice that holds pieces of the girder together. Using the concept of influence lines, we have determined where the transport must be in order to cause the largest shear on the splice: just to the right of the splice. Now we need to find the actual shear generated by the transport and equipment.

The bridge deck will distribute the point forces out to the various bridge girders across the width. A common method for determining the effect on each girder is the Distribution Factor Method. Another team member will perform those calculations, but first we need to determine the net shear at the splice due to the equipment and transport.

The senior engineer on the team wants to be sure we have reasonable results, so one engineer is assigned to use the Force Method to find the shear at C. Our job is to use the Moment Distribution Method. We are also tasked with finding the four vertical reactions.

Accepted Answer

Detailed Analysis:
Determine the distribution factors:
Stiffness of each member:

[latex]K_{A B} = K_{B A} = \frac{4 E I}{30 \mathrm{ft}}[/latex]

[latex]K_{B E} = K_{E B} = \frac{4 E I}{60 \mathrm{ft}}[/latex]

[latex]K_{E F} = K_{F E} = \frac{4 E I}{30 \mathrm{ft}}[/latex]

Distribution factors:

[latex]D F_{A B} = 1 \quad ext { (single member at end with roller support) }[/latex]

[latex]D F_{B A} = \frac{4 E I / 30 \mathrm{ft}}{4 E I / 30 \mathrm{ft} + 4 E I / 60 \mathrm{ft}} = 0.667[/latex]

[latex]D F_{B E} = \frac{4 E I / 60 \mathrm{ft}}{4 E I / 30 \mathrm{ft} + 4 E I / 60 \mathrm{ft}} = 0.333[/latex]

[latex]D F_{E B} = \frac{4 E I / 60 \mathrm{ft}}{4 E I / 30 \mathrm{ft} + 4 E I / 60 \mathrm{ft}} = 0.333[/latex]

[latex]D F_{E F} = \frac{4 E I / 30 \mathrm{ft}}{4 E I / 30 \mathrm{ft} + 4 E I / 60 \mathrm{ft}} = 0.667[/latex]

[latex]D F_{F E} = 1 \quad( ext { single member at end with roller support })[/latex]

Calculate the fixed-end moments:

For the members with no load between the nodes:

[latex]F E M_{A B} = F E M_{B A} = F E M_{E F} = F E M_{F E} = 0[/latex]

For member BE, use formulas from inside the back cover and superposition:

[latex]F E M_{B E} = -\frac{20 \mathrm{k}(12 \mathrm{ft})(48 \mathrm{ft})^2}{(60 \mathrm{ft})^2} - \frac{20 \mathrm{k}(38 \mathrm{ft})(22 \mathrm{ft})^2}{(60 \mathrm{ft})^2} = -255.8 \mathrm{k} \cdot \mathrm{ft}[/latex]

[latex]F E M_{E B} = +\frac{20 \mathrm{k}(12 \mathrm{ft})^2(48 \mathrm{ft})}{(60 \mathrm{ft})^2} + \frac{20 \mathrm{k}(38 \mathrm{ft})^2(22 \mathrm{ft})}{(60 \mathrm{ft})^2} = +214.9 \mathrm{k} \cdot \mathrm{ft}[/latex]

Distribute the moments:
Stop when the peak UM is no more than 2.1 k•ft.
Moment distribution is shown in Table 13.10

Resulting unbalanced moments:

[latex]U M_A = 2.1 \mathrm{k} \cdot \mathrm{ft}[/latex]

[latex]U M_B = -0.5 \mathrm{k} \cdot \mathrm{ft}[/latex]

[latex]U M_E = 0.0 \mathrm{k} \cdot \mathrm{ft}[/latex]

[latex]U M_F = -1.1 \mathrm{k} \cdot \mathrm{ft}[/latex]

All are at or below our stopping threshold of 2.1 k•ft.

∴ Okay to stop

Evaluation of Results:
Satisfied Fundamental Principles?
Check equilibrium of the net moments:

[latex]\Sigma M_A = 2.1 \mathrm{k} \cdot \mathrm{ft} \approx 0[/latex] ✓

[latex]\Sigma M_B = 186.6 \mathrm{k} \cdot \mathrm{ft} - 187.1 \mathrm{k} \cdot \mathrm{ft} = -0.5 \mathrm{k} \cdot \mathrm{ft} \approx 0[/latex] ✓

[latex]\Sigma M_E = 166.3 \mathrm{k} \cdot \mathrm{ft} - 166.4 \mathrm{k} \cdot \mathrm{ft} = -0.1 \mathrm{k} \cdot \mathrm{ft} \approx 0[/latex] ✓

[latex]\Sigma M_F = -1.1 \mathrm{k} \cdot \mathrm{ft} \approx 0[/latex] ✓

Table 13.10

NODE	A	B		E		F
MEMBER	AB	BA	BE	EB	EF	FE
K	4EI/30	4EI/30	4EI/30	4EI/30	4EI/30	4EI/30
DF	1	0.667	0.333	0.333	0.667	1	-UM
FEM (k•ft)			-255.8	214.9
Release B		170.5	85.3				255.8
Carryover	85.3			42.6
Release E				-85.8	-171.7		-257.5
Carryover			-42.9			-85.8
Release F						85.8	85.8
Carryover					42.9
Release A	-85.3						-85.3
Carryover		-42.6
Release B		57.0	28.5				85.6
Carryover	28.5			14.3
Release E				-19.1	-38.1		-57.2
Carryover			-9.5			-19.1
Release F						19.1	19.1
Carryover					9.5
Release A	-28.5						-28.5
Carryover		-14.3
Release B		15.9	7.9				23.8
Carryover	7.9			4.0
Release E				-4.5	-9.0		-13.5
Carryover			-2.2			-4.5
Release F						4.5	4.5
Carryover					2.2
Release A	-7.9						-7.9
Carryover		-4.0
Release B		4.1	2.1				6.2
Carryover	2.1			1.0
Release E				-1.1	-2.2		-3.3
Carryover			-0.5			-1.1

Total M	2.1	186.6	-187.1	166.3	-166.4	-1.1

Find the shear at C:
FBD of BE:

[latex]\begin{aligned}+\circlearrowright \Sigma M_E =& 0 = -187.1 \mathrm{k} \cdot \mathrm{ft} \&+ V_{B E}(60 \mathrm{ft}) - 20 \mathrm{k}(48 \mathrm{ft}) \&- 20 \mathrm{k}(22 \mathrm{ft}) + 166.3 \mathrm{k} \cdot \mathrm{ft}\end{aligned}[/latex]

[latex]V_{B E} = 23.7 \mathrm{k}[/latex]

FBD of BC:

[latex]+\uparrow \Sigma F_y = 0 = 23.7 \mathrm{k} - V_{C B}[/latex]

[latex]V_{C B} = 23.7 \mathrm{k}[/latex]

Find the vertical reactions:
FBD of AB:

[latex]+\circlearrowright \Sigma M_B = 0 = A_y(30 \mathrm{ft}) + 186.6 \mathrm{k} \cdot \mathrm{ft}[/latex]

[latex]A_y = -6.2 \mathrm{k} \quad(+\uparrow)[/latex]

FBD of AE:

[latex]\begin{gathered}+\circlearrowright \Sigma M_E = 0 = (-6.2 \mathrm{k})(90 \mathrm{ft}) + B_y(60 \mathrm{ft}) - 20 \mathrm{k}(48 \mathrm{ft}) \- 20 \mathrm{k}(22 \mathrm{ft}) + 166.3 \mathrm{k} \cdot \mathrm{ft}\end{gathered}[/latex]

[latex]B_y = 29.9 \mathrm{k}(+\uparrow)[/latex]

FBD of EF:

[latex]+\circlearrowright \Sigma M_E = 0 = -166.4 \mathrm{k} \cdot \mathrm{ft} - F_y(30 \mathrm{ft})[/latex]

[latex]F_y = -5.5 \mathrm{k} \quad(+\uparrow)[/latex]

FBD of BF:

[latex]\begin{aligned}+\circlearrowright \Sigma M_B = 0 =&- 187.1 \mathrm{k} \cdot \mathrm{ft} + 20 \mathrm{k}(12 \mathrm{ft}) + 20 \mathrm{k}(38 \mathrm{ft}) \&- E_y(60 \mathrm{ft}) - (-5.5 \mathrm{k})(90 \mathrm{ft})\end{aligned}[/latex]

[latex]E_y = 21.8 \mathrm{k} \quad(+\uparrow)[/latex]

Summary of reactions:

Evaluation of Results:
Satisfied Fundamental Principles?
Check equilibrium:

[latex]+\uparrow \Sigma F_y = -6.2 \mathrm{k} + 29.9 \mathrm{k} - 20 \mathrm{k} - 20 \mathrm{k} + 21.8 \mathrm{k} - 5.5 \mathrm{k} = 0[/latex] ✓

[latex]\begin{aligned}+\circlearrowright \Sigma M_A =&-29.9 \mathrm{k}(30 \mathrm{ft}) + 20 \mathrm{k}(42 \mathrm{ft}) + 20 \mathrm{k}(68 \mathrm{ft}) \&- 21.8 \mathrm{k}(90 \mathrm{ft}) - (-5.5 \mathrm{k})(120 \mathrm{ft}) \ =& 1 \mathrm{k} \cdot \mathrm{ft} \approx 0\end{aligned}[/latex] ✓

Satisfied Fundamental Principles?
Our predictions of the shear at C and the vertical reactions match the
Force Method results from Example 12.4 exactly. ✓

Conclusion:
These observations are strong evidence that our results from the Moment Distrubution Method are reasonable.

Comparison with computer aided analysis results:
The reactions match the results of the computer aided analysis to within ½%, which is within roundoff error. ✓

Reaction	Moment Distribution	Computer Aided	Difference
[latex]A_y[/latex]	- 6.2 k	- 6.2 k	0%
[latex]B_x[/latex]	0 k	0 k	0%
[latex]B_y[/latex]	29.9 k	29.9 k	0%
[latex]E_y[/latex]	21.8k	21.9 k	0.5%
[latex]F_y[/latex]	-5.5 k	-5.5 k	0%

Question 13.1: A large piece of equipment is being delivered via special tr...

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