CSLB General Building (B)Framing and CarpentryHard
A shear wall segment is 4 feet wide and 8 feet tall, resisting a lateral shear force of 800 pounds applied at the top of the wall. Ignoring the dead load resisting the overturning moment, what tension force must the hold-down anchor resist at the leading edge of the wall?
- A6,400 pounds
- B3,200 pounds
- C1,600 pounds
- D800 pounds
Show answer & explanationAnswer & explanation
Correct answer: C. 1,600 pounds
The overturning moment equals the lateral force multiplied by the height where it is applied: 800 lbs × 8 ft = 6,400 ft-lbs. This moment is resisted by a couple acting over the wall's width, so the hold-down tension force equals the moment divided by the wall width: 6,400 ft-lbs ÷ 4 ft = 1,600 lbs.
Why the other options are wrong
- A. This is the overturning moment itself in ft-lbs, not divided by wall width, so it is not the correct tension force.
- B. This value results from an incorrect doubling of the correct answer, not the actual moment/width calculation.
- D. This is simply the applied shear force, not the calculated overturning tension force.
Shear Wall Overturning Moment & Hold-Down Force
Lateral force applied to a shear wall creates an overturning moment equal to force times height; the hold-down tension force equals that moment divided by the wall's width.
- Moment = lateral force × height of application
- Hold-down force = moment ÷ wall width
- Dead load on the wall can reduce required hold-down force in real designs
Memory trick: Force times height makes the twist; divide by width to find the fist (tension) at the corner.