# Allowable Shear Stress

## 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.

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## Example 02: Maximum Concentrated Load a Box Beam Can Carry

**Problem**

A beam is built up by nailing together 25 mm thick planks to form a 200 mm × 250 mm box section as shown. The nails are spaced 125 mm apart and each can carry a shearing force of up to 1.3 kN. The beam is simply supported for a span of 3.6 m and to carry a concentrated load *P* at the third point of the span. The allowable shearing stress of the section is 0.827 MPa.

- Determine the largest value of
*P*that will not exceed the allowable shearing stress of the beam or the allowable shearing force of the nails. - What is the maximum flexural stress of the beam for the load
*P*computed in Part (1)?

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## Example 04: Required Depth of Rectangular Timber 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.

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## Example 02: Required Diameter of Circular Log Used for Footbridge Based on Shear Alone

**Problem**

A wooden log is to be used as a footbridge to span 3-m gap. The log is required to support a concentrated load of 30 kN at midspan. If the allowable stress in shear is 0.7 MPa, what is the diameter of the log that would be needed. Assume the log is very nearly circular and the bending stresses are adequately met. Neglect the weight of the log.

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