Example 02: Notched beam with concentrated load

Problem
A 150 mm by 300 mm wooden beam having a simple span of 6 meters carries a concentrated load P at its midspan. It is notched at the supports as shown in the figure. For this problem, all calculations are based on shear alone using the 2010 NSCP specification given below. Allowable shear stress of wood, Fv = 1.0 MPa.
 

ex-02-notched-beam.jpg

 

  1. If P = 30 kN, calculate the maximum allowable depth (millimeters) of notches at the supports.
    1. 88
    2. 62
    3. 238
    4. 212

     

  2. If the depth of notches is 100 mm, what is the safe value of P (kiloNewton) the beam can carry.
    1. 26.67
    2. 17.78
    3. 8.89
    4. 13.33

     

  3. If P = 25 kN and the depth of notches is 150 millimeters, what is the shear stress (MegaPascal) near the supports.
    1. 0.83
    2. 6.67
    3. 1.67
    4. 3.33

 

NSCP 2010 Section 616.4: Horizontal Shear in Notched Beams
When rectangular-shaped girder, beams or joists are notched at points of support on the tension side, they shall meet the design requirements of that section in bending and in shear. The horizontal shear stress at such point shall be calculated by:
 

$f_v = \dfrac{3V}{2bd'}\left( \dfrac{d}{d'} \right)^2$

Where:

$d$ = total depth of beam.
$d'$ = actual depth of beam at notch

 

Example 01: Safe uniform load for beam notched at tension side of support

Problem
A 75 mm × 150 mm beam carries a uniform load wo over the entire span of 1.2 m. Square notches 25 mm deep are provided at the bottom of the beam at the supports. Calculate the safe value of wo based on shear alone.

Allowable shear parallel to grain = 1.40 MPa
Allowable shear normal to grain = 1.85 MPa

 

ex-01-notched-beam-uniform-load.jpg

 

Notching on Beams

NSCP 2010
When rectangular-shaped girder, beams or joists are notched at points of support on the tension side, they shall meet the design requirements of that section in bending and in shear. The horizontal shear stress at such point shall be calculated by:
 

$f_v = \dfrac{3V}{2bd'}\left( \dfrac{d}{d'} \right)^2$

Where:

$d$ = total depth of beam.
$d'$ = actual depth of beam at notch.

 

Example 04: Required depth of rectangular beam based on allowable bending, shear, and deflection

Problem
A beam 100 mm wide is to be loaded with 3 kN concentrated loads spaced uniformly at 0.40 m on centers throughout the 5 m span. The following data are given:

Allowable bending stress = 24 MPa
Allowable shear stress = 1.24 MPa
Allowable deflection = 1/240 of span
Modulus of elasticity = 18,600 MPa
Weight of wood = 8 kN/m3
  1. Find the depth "d" considering bending stress only.
  2. Determine the depth "d" considering shear stress only.
  3. Calculate the depth "d" considering deflection only.

 

ex-04-beam-with-uniform-load.jpg

 

Spiral Curve or Transition Curve

Spirals are used to overcome the abrupt change in curvature and superelevation that occurs between tangent and circular curve. The spiral curve is used to gradually change the curvature and superelevation of the road, thus called transition curve.
 

fig-04_spiral-curve-full.gif

Compound and Reversed Simple Curves

Compound Curves
A compound curve consists of two (or more) circular curves between two main tangents joined at point of compound curve (PCC). Curve at PC is designated as 1 (R1, L1, T1, etc) and curve at PT is designated as 2 (R2, L2, T2, etc).
 

fig-03_compound-curves.gif

 

Problem 03 - Symmetrical Parabolic Curve

Board Problem
A grade line AB having a slope of +5% intersect another grade line BC having a slope of –3% at B. The elevations of points A, B and C are 95 m, 100 m and 97 m respectively. Determine the elevation of the summit of the 100 m parabolic vertical curve to connect the grade lines.
 

prob-03_symmetrical-parabolic-curve.gif

 

Problem 02 - Symmetrical Parabolic Curve

Problem
A descending grade of 6% and an ascending grade of 2% intersect at Sta 12 + 200 km whose elevation is at 14.375 m. The two grades are to be connected by a parabolic curve, 160 m long. Find the elevation of the first quarter point on the curve.
 

prob-02_symmetrical-parabolic-curve.gif

 

Problem 01 - Symmetrical Parabolic Curve

Problem
A grade of -4.2% grade intersects a grade of +3.0% at Station 11 + 488.00 of elevations 20.80 meters. These two center gradelines are to be connected by a 260 meter vertical parabolic curve.

  1. At what station is the cross-drainage pipes be situated?
  2. If the overall outside dimensions of the reinforced concrete pipe to be installed is 95 cm, and the top of the culvert is 30 cm below the subgrade, what will be the invert elevation at the center?

 

prob-01_symmetrical-parabolic-curve.gif

 

Symmetrical Parabolic Curve

Vertical Parabolic Curve
Vertical curves are used to provide gradual change between two adjacent vertical grade lines. The curve used to connect the two adjacent grades is parabola. Parabola offers smooth transition because its second derivative is constant. For a downward parabola with vertex at the origin, the standard equation is
 

$x^2 = -4ay$   or   $y = -\dfrac{x^2}{4a}$.

 

fig-02_symmetrical-parabolic-curve.gif

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