[amp_mcq option1=”Is safe in shear” option2=”Is safe with stirrups” option3=”beam of 6 m span is 30 cm wide and has a lever arm of 55 cm. If it carries a U.” option4=”Needs revision of section” correct=”option2″]<\/p>\n
The correct answer is: B. Is safe with stirrups<\/strong><\/p>\n The shear force at the critical section is $V = 12 \\times 0.6 = 7.2$ kN. The shear stress is therefore $\\tau = \\frac{V}{bd} = \\frac{7.2}{0.3 \\times 0.55} = 3.4$ kg\/cm$^2$. This is greater than the allowable shear stress of 5 kg\/cm$^2$, so the beam is not safe in shear without stirrups.<\/p>\n Stirrups are steel bars that are placed perpendicular to the main reinforcement in a beam. They help to resist shear forces by providing a mechanism for the concrete to transfer the shear stress to the main reinforcement.<\/p>\n If stirrups are added, the shear stress in the beam will be reduced. The new shear stress can be calculated using the following equation:<\/p>\n $$\\tau = \\frac{V}{bd} + \\frac{As_v f_y}{s}$$<\/p>\n where:<\/p>\n In this case, the new shear stress is:<\/p>\n $$\\tau = \\frac{7.2}{0.3 \\times 0.55} + \\frac{0.005 \\times 400}{0.1} = 2.75$$<\/p>\n This is less than the allowable shear stress of 5 kg\/cm$^2$, so the beam is safe in shear with stirrups.<\/p>\n","protected":false},"excerpt":{"rendered":" [amp_mcq option1=”Is safe in shear” option2=”Is safe with stirrups” option3=”beam of 6 m span is 30 cm wide and has a lever arm of 55 cm. If it carries a U.” option4=”Needs revision of section” correct=”option2″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[640],"tags":[],"class_list":["post-5707","post","type-post","status-publish","format-standard","hentry","category-rcc-structures-design","no-featured-image-padding"],"yoast_head":"\n\n