Question

Draw the enveloped shear diagram and enveloped moment diagram for a simply supported beam with a uniform dead load of 1.00kip/ft, a uniform live load of 1.50kip/ft and a concentrated live load of 175 kip. the dead load is applied to te entire span. the live load may be positioned anywhere on the span.

Answer 1

the span of the beam, length of uniform live load is not given hence it is assumed as:

1. the span of the beam is 15 ft

2. the uniform live load is applied for full length of the beam

3. the concentrated live load is applied/ postioned at the center of the beam

solution:

[NOTE: 1 . 'X' indicates multiplication sign

2. Refer the figure attached below for the refrence of load posions, shear and bending moment diagrams]

Step I: Support Reactions:-

RA+RB = (1.0x15)+(1.5x15)+175

RA+RB = 212.50 kip

RA = RB = 106.25 kip (since the loading is symmetrical)

Step II: Shear Force Calculations:-

1. Shear Force at A:

SFA = 106.25 kip

2. Shear Force at B (just before the concentrated load):

SFB1 = 106.25-(1.0X7.5)-(1.5X7.5) = 87.5 kip

3. Shear Force at B (below the concentrated load):

SFB2 = 106.25-(1.0X7.5)-(1.5X7.5)-(175) = -87.5 kip

4. Shear Force at c:

SFc = 106.25-(1.0X15)-(1.5X15)-(175) = 0 kip

[check: since SFc = 0, which means shear force calculations are correct]

{All the values of shear forces are pollted the below attached figure}

Step III: Bending Moment calculatuions:-

1. Bending Moment at A:

BMA = (106.25X0) = 0 kip-ft   (since the distance from support reaction is zero)

2. Bending Moment at B:

BMB = (106.25X7.5)-(1.0X7.5X7.5/2)-(1.5X7.5X7.5/2)-(175X0) = 726.563 kip-ft

[Also it is Maximum Bending Moment]

3. Bending Moment at C:

  BMC = (106.25X15)-(1.0X15X15/2)-(1.5X15X15/2)-(175X7.5) = 0 kip-ft

[Check: the calculated BMC = 0, hence okay]

{All the values of Bending Moments are pollted the below attached figure}