## 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, *F _{v}* = 1.0 MPa.

- If
*P*= 30 kN, calculate the maximum allowable depth (millimeters) of notches at the supports.- 88
- 62
- 238
- 212

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

- If
*P*= 25 kN and the depth of notches is 150 millimeters, what is the shear stress (MegaPascal) near the supports.- 0.83
- 6.67
- 1.67
- 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:

Where:

$d'$ = 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 *w _{o}* 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

*w*based on shear alone.

_{o}Allowable shear normal to grain = 1.85 MPa

## 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:

Where:

$d'$ = 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 shear stress = 1.24 MPa

Allowable deflection = 1/240 of span

Modulus of elasticity = 18,600 MPa

Weight of wood = 8 kN/m

^{3}

- Find the depth "
*d*" considering bending stress only. - Determine the depth "
*d*" considering shear stress only. - Calculate the depth "
*d*" considering deflection only.

## 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 (*R*_{1}, *L*_{1}, *T*_{1}, etc) and curve at *PT* is designated as 2 (*R*_{2}, *L*_{2}, *T*_{2}, etc).

## 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.

- At what station is the cross-drainage pipes be situated?
- 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?

## 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