• Cantilever Formwork with Remarkable Performances and Trustful Quality System 1
  • Cantilever Formwork with Remarkable Performances and Trustful Quality System 2
  • Cantilever Formwork with Remarkable Performances and Trustful Quality System 3
Cantilever Formwork with Remarkable Performances and Trustful Quality

Cantilever Formwork with Remarkable Performances and Trustful Quality

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Loading Port:
Tianjin
Payment Terms:
TT OR LC
Min Order Qty:
300 pc
Supply Capability:
3000000 pc/month

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1. Structure of  Cantilever Formwork Description 

 

Cantilever Formwork is  mainly used in the concrete pouring of pier, high buildings, and so on. The concrete lateral pressure is entirely supported by anchor system and the wall-through tie-rods, additional reinforcement is not needed. The construction is easy, rapid and economical.  A single pouring height is high and the finished concrete surface is perfect. Cantilever formwork has several types, the structure is similar, and has its own different characteristic. Can use them according to actual demand.

Main cantilever FormworkPJ200 & PJ240,CB-240 etc.

 

2. Main Features of  Cantilever Formwork

 

-easy to assemble 

-simple structure.

-easy, rapid and economical.

 

3.  Cantilever Formwork Images

 

 

Cantilever Formwork with Remarkable Performances and Trustful Quality

 

Cantilever Formwork with Remarkable Performances and Trustful Quality

 

 

4.   Cantilever Formwork Specifications

 

Anchor  system

Anchor system is the most important supporting part. The system is made of five parts shown below. There into, tensile bolt, anchor shoe and bowl-climbing cone can be taken out for reusing after the concrete pouring.

Tensile bolt M30/L= 110

Anchor shoe

Bowl-climbing cone M30/D15

High-strength rod D15

⑤Ancor plate D15

 

Cantilever Formwork with Remarkable Performances and Trustful Quality

 

Cantilever Formwork with Remarkable Performances and Trustful Quality

 

 

5.FAQ of  Cantilever Formwork

     

1) What about of our after-sale services?

   . Response will be carried out  in 24hours  after receiving any complain or request.

   . Any formwork cost can be refund after order is confirmed. 

   . If the products are not based on the requirements, there will be the relevant compensations made for you. 

2) What about of our after-sale service?

   . Response will be carried out in 24hours  after receiving any complain or request.

   . Steel Frame Formwork GK120 cost can be refund after order is confirmed. 

   . If the products are not based on the requirements, there will be the relevant compensations made for you.   

Q: The previous answers to this question are either wrong or sketchy. The movie, Agony and Ecstasy didn't come out of thin air - but out of some good research by Oliver Stone, who knows way more about the subject than anyone who has answered before. Further, there's actual scholarship (most of it used by Stone) to answer the question. So let's see an answer with citations.
When they moved to the ceiling, they likewise employed a system similar to Michelangelo's, which involved cantilevering a shelf outwards from the scaffolding to support a stepped and arched platform. The advantages of modern lightweight materials meant that the platform could be wheeled, facilitating easy movement along the length of the building rather than dismantling and replacement, as Michelangelo had done in 1506.
Q: How much is the adjusting range of the scaffold of the bottom
(5) bridging angle with the ground control between 45 and 60 degrees, scissors and the horizontal rod or pole cross point application of rotary fastener steel tube firmly, the length is not enough available when docking fasteners then long; the scissors must be set on the top in the end, bridging between bottom and foundation pad. In order to enhance the ability of scissors under load.
Q: Why the scaffolds need to be inspected within the preceding 7 days and after adverse weather condition?
to check for loose parts/bolts/cables/etc caused by heavy winds and storms. Also, heavy rains can sometimes cause oxidation of certain cables which weaken the scaffolding structure over time
Q: on a scale of 1-10
Are you seriously asking that question?
Q: Bowl buckle scaffold is not a steel pipe scaffolding
The steel pipe scaffold is mainly used to connect the steel pipe, the specific method is different
Q: A 105 kg scaffold is 6.2 m long. It is hanging with two wires, one from each end. A 520 kg box sits 2.2 m from the left end. What is the tension in the right hand side wire?(g = 9.8 m/s2)
(105 x 9.8) = weight of 1,029N. Half is supported by each wire, = 514.5N. (520 x 9.8) = 5,096N. box weight. (6.2 - 2.2) = 4m. 5,096/((4//2.2)) +1) = 1,808.258N. (5,096 - 1,808.258) + 514.5 = 3,802.242N. tension in left wire.
Q: The amount of bowl buckle scaffolding, know the area of 50000m2, height 6m, how to calculate the number of posts with the root, the number of bowl buckle on the number of how many?
These quantities can only be estimated in the absence of drawings.Height of 6 meters, the total root number = the total number of vertical pole = upper support number. Pull rod and diagonal rod according to the number X4.
Q: Bowl buckle scaffold is the whole force or local force
And bowl buckle scaffolding to build up the cost of steel pipe, and now the disc has gradually replaced the bowl buckle
Q: is this right any answers please
well id say not. go and talk to the owner of the company, there may have just been a mistake other wise go to you citezen advice and they will be able to better inform you
Q: A window washer is standing on a scaffold supported by a vertical rope at each end. The scaffold weighs 190 N and is 3.30 m long. What is the tension in each rope when the 710 N worker stands 2.30 m from one end?smaller tensionN?larger tensionN?
think of statics (in terms of forces and torques): because of fact the gadget is in equilibrium, the sum of the forces = 0 and the sum of the torques = 0 on the grounds that sum of forces = 0, then: ft(left) + ft(suitable) - Fg(scaffold) - Fg(employee) = 0 on the grounds that sum of torques = 0, then: -Set some element as your axis of rotation. for my area, I set between the ft as my axis. to that end, my torque equation will become ft(suitable)(2.ninety m) – Fg(scaffold)(a million.40 5 m) – Fg(employee)(a million.ninety m) = 0 you already know Fg(scaffold), Fg(employee), so resolve for ft(suitable). Plug this value into the tension equation and resolve for ft(left).

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