Question

All problems in this homework assignment are to analyze a two-span continuously supported bridge with each span length of 116 ft. The LRFD HL-93 Design Truck, Design Lane loading, LRFD Design Tandem, and influence line charts needed to compute can be found in the lecture notes.

Answer 1

In this problem, the number of unknown reactions = 3 (Ra, Rb, Rc)

Number of equilibrium equation = 2 (Summation of vertical forces = 0 and Moment of all the forces about any point = 0).

The Degree of static indeterminacy, DSI = 3 - 2 = 1

Hence, one compatibility condition will be required to solve this problem.

We will use the method of superimposition to solve this problem and write the compatibility condition. The whole beam is assumed to be made up of two beams: 1. The loading structure (Part 1), and 2. The redundant structure (Part 2).

COMPATIBILITY Conditioned 2000000 Pe Pe P3 P4 Ps POPz Po Part lo RA Part 20 1 Pre

In Part 1, the external loads are acting on a simply supported beam and will cause a downward deflection of DL at point B in the beam.

In Part 2, the reaction force Rb will cause an upward deflection of DR at point B in the beam.

COMPATIBILITY CONDITION

Since point B has roller support in the original beam, there will be no vertical movement of this point.

Hence, Net displacement of point B = DL + DR  = 0

To solve part 1 and calculate the value of DL let us again use the principle of superposition and calculate the deflection of point B due to loads P1, P2 ........P8 separately and add then them to get total DL

Tha=29.4 L = 232 D = Pa (31² -4 Q²) = 1838894.4 EI. 4881 , P2 =26-6. A. a=29.4 +13= 42.4. Bp a Ba 48 € 1 (312-40%) = 3625088-3 €1. ( 13 =266 a = 46. De = 3930544.25 EI 1 Py = 26.6 a = 50-ye De > H2261263 EL

Deflection of central point of a beam due to an eccentric load, P acting at a distance of a from support is given as

D= (Pa / 48 EI ) (3L2 - 4a2)

DL = Dp1 + Dp2 + Dp3 + Dp3 + Dp5 + Dp6 + Dp7 + Dp8

= ( 1838894.4+3625088.3+3930544.25+4226126.3+5179150.6+533110.5+5467675.8)/EI = 35186707.05/EI

Deflection of central point of a beam due to a central load, P is given as:

D = PL3 / 48EI

DR = - 260149.3 Rb / EI [negative because upward displacement]

Compatibility equation:

DL + DR  = 0

5186707.05/EI - 260149.3 Rb / EI = 0

Rb = 135.3 kips

Equilibrium equations:

Summation of vertical forces, Ra + Rb + Rc = P1+P2+P3+P4+P5+P6+P6+P7+P8

Ra + 135.3 + Rc = 19+26.6+26.6+26.6+22.8+22.8+22.8+22.8 = 190

Ra + Rc = 54.7 ----------------(1)

Moment of all the forces about point A, Ma = 0

Rc*232 + Rb*116 - 22.8*92.4 - 22.8*88.4 - 22.8*84.4 - 22.8*80.4 - 26.6*50.4 - 26.6*46.4 - 26.6*42.4 - 19*29.4 = 0

Rc = -15.32 kips (negative means that the force is acting downwards)

Ra = 54.7 - Rc = 70 kips

Bending moment at 0.4L = 46.4': Taking moment of all the forces on the left of the section.

M = Ra * 46.4 - P1*17 - P2*4 = 70*46.4 - 19*17 - 26.6*4 = 2818.6 kips-ft

Hence, the bending moment at 0.4L distance from support A due to the given loading is 2818.6 kp-ft.