MCQOPTIONS
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| 1. |
The block shear strength at an end connection for shear fracture and tension yield is given by :$ |
| A. | (A<sub>vg</sub>f<sub>y</sub>/√3 γ<sub>m0</sub>)+(0.9A<sub>tn</sub>f<sub>u</sub>/γ<sub>m1</sub>) |
| B. | (A<sub>tg</sub>f<sub>y</sub>/ γ<sub>m0</sub>)+(0.9A<sub>vn</sub>f<sub>u</sub>/√3 γ<sub>m1</sub>) |
| C. | (0.9A<sub>vg</sub>f<sub>y</sub>/√3 γ<sub>m0</sub>)+(A<sub>tn</sub>f<sub>u</sub>/γ<sub>m1</sub>) |
| D. | (0.9A<sub>tg</sub>f<sub>y</sub>/√3 γ<sub>m0</sub>)+( A<sub>vn</sub>f<sub>u</sub>/γ<sub>m1</sub>) |
| Answer» C. (0.9A<sub>vg</sub>f<sub>y</sub>/‚Äö√Ñ√∂‚àö‚Ć‚àö‚àÇ3 ‚âà√≠‚Äö√¢‚Ä¢<sub>m0</sub>)+(A<sub>tn</sub>f<sub>u</sub>/‚âà√≠‚Äö√¢‚Ä¢<sub>m1</sub>) | |